Talk:Monty Hall problem/Arguments

Falsifiable

(This paragraph has been moved from the article talk page to the Arguments page)

The "explanation" for this theory bases itself on the claim that the middle option out of three is statistically likely to be a much, much higher chance of being the car. If the options are completely random, this is completely false.

The only way for the options not to revert to 50/50 is if, somehow, a completely random door generation leaves the middle door as having the car much higher than is statistically possible. 124.168.241.91 (talk) 11:32, 6 September 2013 (UTC) Sutter Cane

Please try to understand the intended scenario of this world famous unintuitive paradox that is based on the premise that the host knows what's behind the doors, and in opening of a door will always show a goat but never the car. I repeat: the host will never open a door that hides the prize. So it is NOT on the 2/3-subset only, where the host by randomly opening of an unselected door happened to show a goat simply by chance. The remaining 1/3-subset of ALL events, representing 1/2 of all "winning events", may not be discarded.

• We have to consider that only in the "lucky guess scenario", where the contestant by chance initially picked the door with the car (be it door #1, #2 or #3) the host is really free to open any of this two remaining doors at random, as both non-selected doors hide goats. Only in this 1-out-of-3 case where the host may open an unselected door at random to show a goat, sticking wins the car and swapping to the door offered will hurt, as it hides the second goat.
• But in the remaining two "wrong guess scenarios", where the contestant initially picked a wrong door hiding either goat A or a wrong door hiding goat B, the host never can nor will open one of the two unselected doors at random, as one of them definitely will hide the car. In that 2/3 of all events where swapping will win the car for sure, the host is forced to show nothing but the other goat and must offer a swap to his door that hides the car. In these 2-out-of-3 cases swapping wins the car for sure.

The correct answer to this famous paradox is that players who stick have a chance of 1/3 only to win the car, whereas players who swap to the door offered as an alternative have double chance of 2/3 to win the car. Please read section "The paradox" and have a look to University of California San Diego, Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game.

Regards, Gerhardvalentin (talk) 22:21, 6 September 2013 (UTC)

Sigh. It is irrelevant whether or not the "host" knows what's behind which door. The problem claims that the host always opens the third door after the contestant picks the first door. The host then asks if the contestant wants to switch doors. The claim is that, somehow (magically apparently), the first door will have a 66% chance of being a goat, so the player should switch.

The host isn't switching the "doors" around. So given that the 2 goats and a car are RANDOMLY placed in their perspective areas, if the third door is ALWAYS a goat then the first and second doors have a 50/50 chance of being a goat or a car.

In fact I mention this as I have recently finished doing this problem after reading about the ridiculous, illogical reasons people come up with as to how a randomly selected three door pick, in which the first door is always picked and the third door is always a goat, would magically significantly statistically come up with the first pick as a goat every time. I've just finished doing this 500 times over the last few days. I couldn't be bothered making it to a thousand, even after the ridiculous amount of people claimed (after hilariously changing it from 100) it "totally proves it".

I'm sorry but it's effectively 50/50. In fact, as i'll point out below, the claim you should switch is statistically incorrect.

Taking 3 cards, a black (car) and two reds (goats), I shuffled them repeatedly (5 times quickly per deck to create a truly random system). I would then lay them out into a row of three.

- Card(1) Card(2) Card(3)

I then lifted the third card, which if black I put back and repeatedly shuffle again until I lift a red, and when red I flip over facing up.

- (Card(1) Card(2) Red(3)

Now, following the scenario, I choose the first card. I ask myself if I want to choose the second card, obviously picking no, and I flip both cards 1 and 2.

Example: Black(1) Red(2) Red(3)

Now a slight discrepancy is just a statistical likelihood, it explains what I found, but this ridiculous claim of 66% likelihood of hitting goat if you don't switch is moronic.

Out of 500, FIVE HUNDRED, attempts I ended up with 54 percent of attempts of not switching coming up with black (the car).

54% FIFTY FOUR PERCENT. In which NOT switching comes up with the car.

The claim that the choice between the first and second card after the third is shown to be a goat is 66% (1/3) in favour of the first choice being a goat if you don't switch is flat out false. It's 50/50. 124.168.241.91 (talk) 01:16, 7 September 2013 (UTC) Sutter Cane

If Monty opens a remaining door at RANDOM to show a goat, then things are different. This time you will be left with two doors, each with 50% chance of the car. Here is why. In 2/3 of all cases Monty will show you a goat (by symmetry) and in 1/3 of all cases you have picked the car correctly initially (also by symmetry). P(door 1 has car given that a goat was shown) =P(door 1 has a car and a goat is shown) / P(a goat is shown) = (1/3) / (2/3) = 1/2. Note that P(door 1 has a car and a goat is shown) = 1/3 because the chances of door 1 having a car is 1, and in that case the chances of showing a goat has to be 1 (since there are then only goats to show).

So if he shows you a goat, you are on a 50/50, if he shows you a car, you are on a 1 (100% chance of winning) - you only get the car revealed 1/3 of times, so it works out at a 2/3 chance of winning. This 2/3 chance of winning makes sense because two doors are opened, but in this 'monty has no knowledge' version, the final door is effectively chosen at random (with a 1/3 chance of the game terminating with the car behind it). Gomez2002 (talk) 15:27, 17 October 2013 (UTC)

Sorry, you are not talking about the intended scenario of the famous *paradox*, where the host strictly avoids to show the prize in the 2-out-of-3 cases where the contestant picked one of the two goats. Although it is a clear premise of the famous paradox that the host can only show a goat but never the prize, because he knows what is behind the doors and he intentionally avoids to show the prize.

But in contrast, by your three cards you only show the remaining subset of 2/3 of cases where the host just only "by luck" opened a door with a goat behind, after you have discarded all events (1/3 !) where the host by accident happened to open a door with the car behind, instead of showing the second goat as per the the clear premise of the famous paradox, where switching necessarily will win. You have just silently discarded "those 1/3 winning events" of ALL events and are referring only to the remaining subset of 2/3 of ALL events, meaning that in effect you discarded exactly "one half of all winning events". See the link that I gave above to UCSD. They show that it is quite another problem but no more the famous paradox, if the host does NOT know where the car is located.

If Monty (the host) in a plain winning event should happen to open the door with the car behind by accident, instead of a goat, and you interpret this to mean that – if the car is revealed – then the game is over (winning event discarded) and the next contestant plays the game, then this is no more the scenario of the famous paradox. Then it is quite "another problem", where the chances "staying:switching" are only "1:1". Regards, Gerhardvalentin (talk) 08:18, 7 September 2013 (UTC)

Sutter, the Monty Hall problem is famous because so many people, like yourself, believe that the odds of winning by switching are 50:50. The correct answer is that the odds are 2:1 in your favour if you switch. This fact has been proved by many people using various forms of mathematical analysis and it has been conclusively verified by numerous simulations, some using computers some not.

The question to be asked is why you are not convinced by any of the arguments given in the article. If you have a distrust of mathematicians and statisticians you can indeed to a simple simulation of the problem with a pack of cards or three cups and a pea but, as Gerhard points out, you must do it correctly, that is to say, in a way that corresponds to what happens in the problem.

Here is how to do the simulation correctly with cards.

1. Take three cards, two red, representing goats, and one black representing the car.
2. Have a friend shuffle the cards and place then face down on the table so that you both have absolutely no idea which card is which.
3. Chose one card but do not turn it over.
4. Have your friend look at the two remaining cards, without letting you see either of them, and always turn a red card face up.
5. After your friend has done this, decide whether to stick with the card you originally chose or to switch to the other face-down card on the table,
6. Repeat the experiment many times and keep a record of how many times you win by sticking and how many by switching.
7. Let us know your results. Martin Hogbin (talk) 11:33, 7 September 2013 (UTC)

Ha! Martin Gardner's three shells problem, which Marylin vos Savant clearly knew about! Marilyn seems to know instinctively that the three shells problem and the three doors problem are the same. Actually they are different, but the latter can be reduced to the former by arguing that by symmetry, the actual door numbers in any particular case don't matter.

I would suggest that one explains the easier problem (the three shells problem) first. After all, the first problem on wikipedia is to get people to realize that it's not 50-50. If you can explain people through a simulation experiment, start with the most simple possible simulation experiment. Don't keep track of any door numbers. After they have got that, you can start to look at the problem in different ways.

Three cards get placed face down next to three numbers. The host takes a peek. The player may choose any door, the host secretly tosses a coin to determine which different card to turn over in case he has a choice. (He should make a big show of tossing his coin every time, also on those occasions when he doesn't need it). You can study the success rate of switching for each of the six possible situations the player can be in, at the last moment before they would have to decide between switch or stay.

It's a more complicated experiment but it might be helpful if you want to explain to people the difference between unconditional and conditional probabilities. A difference which in the standard MHP turns out not to be very important. Something which one could have realized in advance. Richard Gill (talk) 17:20, 10 September 2013 (UTC)

As you know we disagree about the MHP and its 'correct' interpretation and solution but I do very much agree with your approach. Start by explaining the solution to the shell (unconditional or whatever else you want to call this) in order to convince the listner that the answer is 2/3 not 1/2 then, for those interested, the complications of conditional probability can be discussed ad infinitum. Martin Hogbin (talk) 20:01, 10 September 2013 (UTC)

Martin - as you well know, this doesn't quite match the description of the problem. The cards need to be put next to numbers 1,2,3 on the table, and you should record only those cases where the initial pick is #1 and the card the host turns up is #3. This will go faster if in step 3 the initial pick is always #1. Gerhard's point is that in step 4 the host MUST turn over a red card (which means if the initial pick is #1 the host CANNOT always turn over #3). And now, if you do the experiment you still might not see the 2:1 odds in favor of switching - particularly if the host looks at #3 first and turns it over if it's a red card (in the same spirit of speeding things up as the player always picking #1). The issue here is you need another constraint in addition to the host always turning over a red card - specifically, that the host MUST pick which card of the remaining two to turn over randomly if they are both red cards (i.e. the host can't just look at #3 and turn it over if it's a red card but must look at both and if both are red cards then must flip a coin or something to decide which one to turn over).

So, here's the revised experiment:

1. Take three cards, two red, representing goats, and one black representing the car.
2. Shuffle the cards and place them face down on the table next to numbers 1,2,3 so that you both have absolutely no idea which card is which.
3. Choose one card but do not turn it over (always pick #1).
4. Have your friend look at both of the two remaining cards, without letting you see either of them, flip a coin, and always turn a red card face up. Either there is only one red card, in which case that is the one your friend must turn up, or there are two red cards. If there are two, which one is turned up is determined by the coin flip (for example, turn up the lower numbered one if the result of the coin flip is heads).
5. After your friend has done this, decide whether to stick with the card you originally chose or to switch to the other face-down card on the table,
6. Repeat the experiment many times and keep a record of how many times you win by sticking and how many by switching in the cases where you picked #1 and your friend turned over #3.
7. Let us know your results.

Rick Block (talk) 17:04, 7 September 2013 (UTC)

Rick, you're describing not the Monty Hall problem but something crucially different. Martin above has a more accurate description of the Monty Hall problem. Where in any description of the MH problem does it say that the host/ the friend is constrained to turn over #3? MartinPoulter (talk) 17:36, 7 September 2013 (UTC)

The host is not constrained to turn over #3 (as the original poster is apparently thinking - which I agree is not the Monty Hall problem), but we're told the case to think about is where the player has picked #1 and the host has revealed #3. The point is you are deciding whether to stay or switch in one of the 6 possible cases of two closed doors (one you originally picked and the other one) and one open door - and we're using the case where you pick #1 and the host opens #3 as the example (and presumably the intent is that the odds will be the same in all 6 of these cases). In Martin's problem, you're effectively deciding to switch before seeing which door the host opens (i.e. if you initially pick #1, you win by switching if the car is behind either #2 or #3 - in which case, of course switching has a 2:1 advantage!). It's not until after the host opens a door that the car must be behind only 2 doors you're deciding between (your original pick, say #1, and the other unopened door, say #2). The difference between these two problems is fairly subtle, and with the constraints that the host MUST always open a door and MUST pick randomly between two goats (if this comes up) they have the same answer - but they're still slightly different problems (distinguished by numerous sources, see the "Criticism of the simple solutions" section of the article). -- Rick Block (talk) 18:37, 7 September 2013 (UTC)

Once again the whole point of the problem (that people think the answer is 1/2 when really it is 2/3) is lost because of a conjuring trick by a bunch of probability professors. This has been discussed her endlessly before and the consensus on every occasion has been to ignore the Morgan scenario until after the basic problem has been solved. Martin Hogbin (talk) 21:11, 7 September 2013 (UTC)

The original poster here is clearly thinking about the conditional probability the car is behind door 2 given the player originally picked door 1 and the host opened door 3, as opposed to the overall probability of winning by a strategy of staying vs. a strategy of switching (look at the experiment he claims to have done - which indeed shows a 50% chance of winning by switching). Why slyly switch problems on him (without even mentioning you're doing this)? Why not, instead, show him how to correctly simulate the problem he's actually thinking about - which necessitates bringing up the additional constraint on the host? -- Rick Block (talk) 22:29, 7 September 2013 (UTC)

Perhaps the 124.168.241.91| would be kind enough to let us know how they interpreted the question.

You will, no doubt recollect that the original question does not specify that the host opened door 3. It says, 'opens another door, say No. 3'. This might be ambiguous but for the fact that we know for sure that the original composer of the question (vos Savant) did not intend to specify door numbers. Martin Hogbin (talk) 08:59, 8 September 2013 (UTC)

We know for sure vos Savant's published wording includes door numbers and that this is the most well known statement of the problem. Are you perhaps referring to Whitaker's original question which is all but unknown? -- Rick Block (talk) 17:19, 8 September 2013 (UTC)

No, vos Savant said that she added the door numbers and that this was a big mistake since she did not want to specify door numbers. Martin Hogbin (talk) 18:19, 8 September 2013 (UTC)

OK - how about if we wait for the original poster here to say how he interpreted the question. -- Rick Block (talk) 19:04, 8 September 2013 (UTC)

To be more specific, everyone here is saying the general rules of the show are:

1. 2 goats and a car are hidden behind three doors, with the location of the car selected randomly (e.g. the host rolls a die and if it comes up 1 or 4 hides the car behind door 1, if it comes up 2 or 5 hides the car behind door 2, and if it comes 3 or 6 hides the car behind door 3)
2. the player initially selects a door, which is noted but not opened
3. the host must now (after the player selects a door) open another door showing a goat, and must make the offer to switch

At this point there are at least two possible interpretations for what question is being asked. Is it

a) What is best, a strategy of switching or a strategy of staying, i.e. if you are intending to go on this show should your strategy be to pick a door and stay with your original choice, or pick a door and then switch?

Or is it

b) Consider the case where you pick door #1 and then see the host open door #3. Should you stay with your original choice of door #1 or switch to door #2?

The question to the original poster of this thread is which of these questions do you think is being asked? -- Rick Block (talk) 15:58, 10 September 2013 (UTC)

I am perfectly happy for you to ask this question and I would be very interested to hear the answer. The problem is though that most people on seeing the question will not see any difference between the two cases you give (and if course we know the numerical answer is in fact the same for both cases). I doubt that we will hear from the OP again since his original point has now been lost in what many see as an irrelevant detail. As I have said before, I have nothing against studying this detail but this can only be done after a person has come to terms with the fact that the answer is 2/3 not 1/2. Martin Hogbin (talk) 16:22, 10 September 2013 (UTC)

Given the experiment he describes doing (way above at this point), it seems fairly obvious to me that he's attempting to answer (b). I agree he may be unlikely to respond, but let's give it a while. -- Rick Block (talk) 00:08, 11 September 2013 (UTC)

As I said to Richard above, I have nothing against discussing the details of conditional probability with anyone, indeed I would be happy to help explain the Morgan argument to newcomers. I think though that to pile straight into this complication before individuals have come to terms with the fact that the answer is 2/3 not 1/2 in unhelpful. We know that the belief that the probabilities for the original and unchosen doors are 50:50 is extraordinarily pervasive and until that belief can be dispelled no progress can be made in issues of conditional probability. I think the additional confusion that it creates probably drives many away. I therefore suggest that you refrain from discussing conditional probability with newcomers until they have accepted the 2/3 answer. After they have understood why this answer is correct I am perfectly happy to leave you to explain the Morgan solution to them without interruption. Martin Hogbin (talk) 09:01, 11 September 2013 (UTC)

Martin, Martin, after all these years, and all the discussion, you still don't get it. Nijdam (talk) 09:51, 11 September 2013 (UTC)

On the contrary...Martin Hogbin (talk) 14:16, 11 September 2013 (UTC)

Per this section above, it seems to me discussing the conditional probability with newcomers leads directly to accepting the 2/3 answer. While (per this section) attempting to get someone to accept the 2/3 answer by ignoring the conditional situation (which was your approach) fails miserably. I therefore could just as legitimately (more so, since we have two data points supporting my approach neither of which supports yours) suggest that you refrain from discussing "simple" solutions with newcomers (on this, the Arguments, page) until they have understood the conditional answer is 2/3. And, are you continuing to respond here in a deliberate attempt to prevent the original poster from responding? I've suggested we wait for a response from the OP twice now. Are you willing to wait yet, or are you going to continue to argue that your approach (reflecting your POV about how the problem should be understood) is the one and only approach that newcomers will understand? -- Rick Block (talk) 16:10, 11 September 2013 (UTC)

Rick, I was waiting for the OP to respond. Martin Hogbin (talk) 21:53, 11 September 2013 (UTC)

What problem are we talking about?

"Chose one card but do not turn it over" what? Just...what? That would indeed be a 1-in-3 chance since none of the variables have changed. It's still 3 cards unflipped. That's not the question I posed.

The question i'm specifying is literally what I stated earlier.

---

Taking 3 cards, a black (car) and two reds (goats), I shuffled them repeatedly (5 times quickly per deck to create a truly random system). I would then lay them out into a row of three.

- Card(1) Card(2) Card(3)

I then lifted the third card, which if black I put back and repeatedly shuffle again until I lift a red, and when red I flip over facing up.

- (Card(1) Card(2) Red(3)

Now, following the scenario, I choose the first card. I ask myself if I want to choose the second card, obviously picking no, and I flip both cards 1 and 2.

Example: Black(1) Red(2) Red(3)

---

In that question the outcome between whether or not you should change cards between the two card options left IS 50/50. In that question, you may change the cards around and STILL get the 50/50 outcome. Be the 2 card picked first, followed by the 1 and 3 or the 1 card picked first followed by the 3 and 2 etc. etc.

This is the question posed both on here and other sources (including the "pop culture" references section that uses this) and the exact problem specified on a variety of sources stating the Monty Hall problem. Which does not, demonstratively, result in a "66% chance" but a 50/50 chance on whether the "door" you originally chose will be a goat or a car regardless of it flipping (which i'm just going to blatantly assume here that people are not trying to claim that the choices will magically change places based on your decision to stay or change).

Which question of the previous stated 1/2/a/b options does the above problem i've stated pertain to, Rick?

If one and it is, somehow, not 50/50 in the proposed explanation then the proposed explanation is demonstratively false. If it applies to none of them, then we have a few issues (which I sought to address with my original post) that I shall specify:

1) The problem is poorly worded. It views exactly how I have stated above (in fact a variety of people defending the 66% chance equation stating that the method I used would demonstrate how it works, which as i've shown is false) and the issue here is addressing it's wording to REMOVE the suggested equation (which I used above) and replace it with something more specific.

2) The problem is unsound and is relying on poor research to come to it's erroneous conclusions (which I doubt is what occurred here).

3) The problem is poorly worded but doesn't rely on the equation shown but mitigating, vague factors that can't be processed specifically in text.

4) The problem is poorly worded but is ignoring real word applications in a situation in which psychological factors and other such non-objective equations are at play, which defies the claim of 66% (it could be a 1/4 chance, 1/37 chance, given the mitigating factors).

My monies on number one, in that the equation is poorly worded and is relying on different scenarios while using a SPECIFIC example to test with that ISN'T (somehow) meant to be repeated. 124.168.241.91 (talk) 09:28, 19 September 2013 (UTC) Sutter Cane

Sutter Cane, you are right in saying that the article still is a mess, despite years of efforts to show the clean *paradox*:

• One open door showing a goat, and two doors still closed, one of them with certainty hiding the second goat, and one of them with certainty hiding the car. But the odds in favour of switching to the door offered are not 1:1, but they are 2:1 in favour of switching doors. That's the *paradox*, but the misty article still gets bogged down in details that are not addressing this famous paradox. Thank you for your comments, it will help to make the article more intelligible. Despite some authors here who are of the opinion that this Wikipedia article is NOT to make the paradox better intelligible. Gerhardvalentin (talk) 10:08, 19 September 2013 (UTC)

@Sutter Cane: Where did you get the notion that the host reveals a goat before the player's initial pick? Here's the wording as vos Savant published it in Parade:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Isn't the sequence of events pretty clearly, in order:

1. Car is hidden behind one of 3 closed doors.

2. Player makes an initial pick (example given is player picks door #1)

3. Host opens a door showing a goat (example given is host opens door #3)

4. Player is asked whether he/she wants to switch to the other door (example given is switching from #1 to #2).

In your problem, you've flipped steps 2 and 3. I agree the problem is not worded very well (for example it implies, but does not explicitly state, that in step 3 the host must open a door showing a goat, which means if the player has picked door #1 the host cannot always open door #3), but I think it at least makes the sequence of events clear and this sequence is different from your interpretation. Is this interpretation from some source, or is it your own analysis (here on the "Arguments" page we can talk about original analysis - I'm just curious where this came from)? -- Rick Block (talk) 16:08, 19 September 2013 (UTC)

@Rick Block: Please read his comment again, he did not say "before", he just said that in his "test" the contestant always first selects door #1, and that – in his test – he makes sure that the host is "avoiding to show the car" in just discarding 1/2 of all winning events where the unselected door #2 contains the second goat and the unselected door #3 contains the prize. By his method the the host discards all those winning events where door #3 hides the car. And I say that's because the still messy article is not clear enough in distinguishing the two scenarios "host knows" and "host doesn't know". In the latter case the chances indeed are 1:1. But MvS said that there’s no way the host can always open a losing door by chance! Gerhardvalentin (talk) 08:01, 20 September 2013 (UTC)

Actually, he did say "before". His "host" flips card 3 first (and, if it's the winner this causes a re-deal), and then his "player" selects card 1. I'm expecting him to say it doesn't matter whether the player picks first or not, because his host is never flipping card 1 (if card 3 is the winner then this is a case where the host could have flipped card 2, so wouldn't fit the given example where the player picks #1 and the host reveals #3). To me it's obvious he's attempting to simulate the conditional case posed as an example by vos Savant where the player initially picks door #1 and the host opens door #3. However, he's doing it in a way where the host always opens door 3 if possible, i.e. does not pick randomly between #2 and #3 if #1 is the winner (and the way he's doing it indeed makes the odds 50/50). I think you and I agree he's not following the "normal" rules where the player picks first and then the host must reveal a loser from the other unselected two, and must pick randomly between the remaining two if the player initially selects the winner. I also think he and I agree that the problem is poorly worded - for example the requirement that the host pick randomly between two losers (if this comes up) is not mentioned.

There are actually two "fixes" to his experiment. One is to do as I suggest above. After the player picks #1, have the host look at both #2 and #3, reveal #3 if #2 is the winner (and vice versa), and flip a coin to decide which of #2 or #3 to reveal if #1 is the winner. If #3 is revealed, count how many times staying with #1 wins vs. switching to #2. The other "fix" (suggested by Martin) is to ignore the example case and count every "deal" regardless of which card the host ends up flipping (and, with this approach, it doesn't matter how the host decides which of #2 of #3 to reveal if #1 is the winner). I think the OP has already acknowledged that in this case (player picks #1, wins by staying if this is the winner and wins by switching if either #2 or #3 is the winner) staying wins 1/3 of the time and switching wins 2/3 - so it's clear to me he's thinking about the example case described in the problem (player picks #1 and host reveals #3). The 2:1 odds in favor of switching necessarily occur in this example case only if the host picks randomly which of #2 or #3 to reveal if #1 is the winner (as you know, the odds of winning by switching in this case can vary between 1/2 [as the OP has shown] to 1 depending on how the host picks which loser to reveal if both #2 and #3 are losers). Picking which loser to reveal randomly (if there's a choice) is as crucial a condition as that the host knows where the winner is (and avoids revealing the winner). -- Rick Block (talk) 15:27, 20 September 2013 (UTC)

No Rick, you got it wrong. On Sept. 7 above he clearly said "The problem claims that the host always opens the third door after the contestant picks the first door." Meaning the host opens door #3 after the contestant in any case picked door #1 before. And he clearly said "The host then asks if the contestant wants to switch doors." Meaning to swap from door #1 to door #2.

In his "test" the contestant ALWAYS (!) picks door #1 (card #1), but the host does NOT ALWAYS open #3, but only if #3 is a looser. So the host "afterwards" is checking the contents of #3, but will show the contents of #3 and offer a swap to #2 ONLY if door #3 (card #3) is NOT THE PRIZE. If #3 hides the prize: game over in such clear winning event. In fact that's the pure "host does not know" scenario. Please read it again. The article still isn't clear enough in this respect. Gerhardvalentin (talk) 16:23, 20 September 2013 (UTC)

Per the description at the beginning of this section ("I lay out the cards ... I then lifted the 3rd card ... Now, following the scenario, I choose the first card"). If the host always reveals #3 (if possible) it doesn't matter if the player picks before or after the host reveals #3, the chances end up 50/50. Yes, it eliminates a winning "switch" if #3 is the winner, but if #3 is the winner then if the player picks #1 the host cannot reveal #3 but must reveal #2. If you're interested in the odds in the case where the player picks #1 AND the host reveals #3 (as he clearly is), then this case (where #3 is the winner) is irrelevant (which is why he simply re-deals in this case). His problem is that he's NOT randomizing which loser to show in the case #1 is the winner. He needs to re-deal if #3 is the winner (in which case #2 must be revealed) and also if the random choice when #1 is the winner selects #2 (i.e. he's counting all cases where #1 is the winner rather than only half of them). -- Rick Block (talk) 18:32, 20 September 2013 (UTC)

@Sutter Cane: In fact we should know what we are talking about. Above there is my example with "joker 1" and "joker 2". "Joker 2" is the well formed problem with a 2/3 solution, "joker 1" has a 1/2 solution. I think you'll agree.--Albtal (talk) 17:36, 21 September 2013 (UTC)

It seems obvious to me he's interested in the specific case where the player has picked #1 and the host has revealed #3 (at which point there are only two possibilities for the location of the car, not three). @Sutter Cane: is this correct? -- Rick Block (talk) 19:28, 21 September 2013 (UTC)

We all are interested in the (not specific) situation where the host has opened a door with a goat and where two doors remain. And the question in the joker example is: Will "joker 1" yield to another chance in the two doors situation than "joker 2"?--Albtal (talk) 10:40, 22 September 2013 (UTC)

You may be interested in the non-specific situation, but the question is what is Sutter Cane interested in? Once again, it seems obvious to me he's interested in the specific case where the player has picked #1 and the host has opened #3. The odds in this specific case with your joker 1 and joker 2 can be equivalent (if joker 2 always opens #3 if possible - indeed, this is what Sutter Cane's procedure above is showing). With joker 2 if you pick #1 you win by staying with probability 1/3 (if the car is behind #1) and you win by switching with probability 2/3 (if the car is behind either #2 or #3). But these are probabilities measured at the beginning of the game, before the host has opened a door. It's something rather different than saying if you do this 100 times or so, you'll see #1 win about 1/3 of the time and #2 win about 2/3 of the time after the host has revealed #3. Saying this requires the host to pick which loser to show randomly if the player initially picks the winner. -- Rick Block (talk) 17:57, 22 September 2013 (UTC)

---

The difference between Sutter Cane's experiment and the usual interpretation is in which outcomes are discarded. Sutter's version discards those times when the car is behind door #3 and then has the host always open door #3 in the remaining cases. The standard interpretation is that, once the contestant picks a door, the host picks either door #2 or door #3 to open, choosing uniformly at random if both have goats; otherwise choosing the one with the goat when the other has the car. We then discard the times when the host picked door #2 - which are all the times when the car was behind door #3 and (by assumption) half the times the car was behind door #1.

There are two reasons why Sutter's interpretation isn't the standard one: the answer of 1:1 is less interesting than 1:2; and, probably more importantly, by guaranteeing up front that the car could never have been behind door #3, it contradicts the usual assumption that the car is equally likely to be behind any of the three doors.

The standard interpretation assumes that the door numbers are not special during the game, and it's only when we come along and choose which games to count for our statistics that it's narrowed down to games where the contestant happened to choose door #1, and the host happened to open door #3 - in particular, we want the answers to be the same (to within experimental error) if we, from the same simulation run, choose the games where the contestant chose door #3 and the host opened door #2 - something that never happens in Sutter's experiments.

Rmsgrey (talk) 00:03, 2 November 2013 (UTC)

I completely agree. However, there are folks arguing here that the "right" way to fix the experiment is to force the host to always open a "goat door" (selected between #2 and #3, without specifying how the host chooses which of #2 or #3 to open if the car is behind #1) and then count all outcomes, not just the times the host opens #3. This "fix" changes the probabilities the experiment is simulating from

A: P(car behind door #1|player picks #1 and host opens #3) vs. P(car behind door #2|player picks #1 and host opens #3)

to

B: P(car behind door #1|player picks #1 and host opens either #2 or #3) vs. P(car behind #2 or #3|player picks #1 and host opens either #2 or #3)

Since the host must always open either #2 or #3 if the player picks #1, B can be mathematically simplified (regardless of how the host decides to pick between #2 and #3 if the car is behind #1) to

C: P(car behind door #1|player picks #1) vs. P(car behind door #2 or #3|player picks #1)

Note that C (and since B can mathematically simplified to C, B as well) is simply the probability the car is behind #1 vs. the probability it is behind #2 or #3, before the host opens a door - which is obviously 1/3 vs. 2/3. My contention is that it seems fairly clear Sutter Cane (and, IMO, nearly anyone reading the standard problem description) is interested in A (the probability after the host opens a specific, known, door that the car is behind door #1 vs the probability it is behind one other door, not both other doors), not B (or C). -- Rick Block (talk) 19:18, 2 November 2013 (UTC)

I think a lot of people assume that it's natural to say that A and A' (like A but where the host opens door #2) should have equivalent answers so it doesn't matter whether you look at A, A', or combine them to give B. And, of course, B is easier to solve since it doesn't require you to make any assumptions about how the host chooses between two goats. 81.156.216.158 (talk) 23:02, 2 November 2013 (UTC)

Did someone just say 'chooses between two goats'? Martin Hogbin (talk) 00:13, 3 November 2013 (UTC)

Clearly meaning choosing which of #2 or #3 to open if the car is behind #1. -- Rick Block (talk) 00:28, 3 November 2013 (UTC)

It is entirely natural to say A and A' should have equivalent answers. The insight that this means they both are numerically the same as B is (I think) beyond most people - indeed, I believe most people (including some who regularly comment here) do not have a clear idea of what B is actually saying. On the other hand, I think nearly everyone understands there is a difference between C and A - just not the difference they usually assume. -- Rick Block (talk) 00:28, 3 November 2013 (UTC)

I really like comparing the approaches by comparing

A: P(car behind door #1|player picks #1 and host opens #3)

A': P(car behind door #1|player picks #1 and host opens #2)

C: P(car behind door #1|player picks #1)

If one simulates MHP then the probabilities A, A' and C will be estimated by different relative frequencies, because they talk about conditional probabilities given different events. It is only if we assume that in the simulations, the host chooses a goat door completely at random when he does have a choice, that the three relative frequencies will be more or less equal. Simulation will confirm that C is 1/3 however the host makes his choice, but A and A' will only be equal to one another and equal to 1/3 when the host's choice if completely at random.

Regarding the question, which probability are we supposed to calculate, it is clear that opinions in the literature are divided: the maths professors tend to go for A and A', the amateurs tend to go for C. Since the numerical answers are the same (assuming an unbiased host) I'm afraid that the lay people simply aren't going to care much. I don't see why this should be a problem for a wikipedia article. In a wikipedia article one reports neutrally what is "out there" and one of the things that is out there is a lot of disagreement about what constitutes a solution to MHP. Richard Gill (talk) 14:16, 6 November 2013 (UTC)

We're not talking about the article here - we're talking about understanding. And the question is what probability is Sutter Cane (presumably an amateur, also presumably unconvinced by the usual lay explanations showing C is 1/3) thinking about? -- Rick Block (talk) 17:57, 6 November 2013 (UTC)

Sample space

Lets talk here (rather than on the main talk page) about the sample space. There's one car and two goats hidden behind three doors. The player initially picks a door. The host subsequently opens a door revealing a goat. So, the set of outcomes consist of triples indicating which door the car is hidden behind, which door the player picks and which door the host opens. As has been mentioned before, there are 12 possible outcomes (with the rule that the host cannot open the same door the player picks and must open a door showing a goat). These are as follows where (x,y,z) indicates the car is behind door x, the player initially picks door y, and the host opens door z.

(1,1,2)   (1,2,3)   (1,3,2)
(1,1,3) 
----------------------------  
(2,1,3)   (2,2,1)   (2,3,1)
          (2,2,3)   
----------------------------                     
(3,1,2)   (3,2,1)   (3,3,1)
                    (3,3,2)

I've arranged these in three rows corresponding to where the car is located and three columns corresponding to the player's initial pick of door - so (for example) the first row shows all possible outcomes for when the car is behind door 1 (the player picks door 1 and the host opens door 2, or the player picks door 2 and the host opens door 3, and so on), while the first column shows all possible outcomes if the player initially picks door 1 (the car is behind door 1 and the host opens door 2, the car is behind door 1 and the host opens door 3, and so on). Note that there are 4 possible outcomes in each row and in each column.

Unlike many probability problems, in this one the outcomes are not all equally likely. This is because the host sometimes but not always has a choice for which door to open. In particular, if the player's initial pick happens to be the same as the door hiding the car the host has a choice (usually assumed to be 50/50) but if the player's initial pick is not the door hiding the car the host has no choice. When the host has a choice, there are two outcomes that together have the same probability as one of the outcomes where the host has no choice.

At the start of the game, before the player initially picks a door, the probabilities of the six "no choice" outcomes are all 1/9 while the probabilities of the six "choice" outcomes are all 1/18. The total probability must be 1, and indeed 6/9 + 6/18 = 1.

After the player initially picks a door, there are always 4 possible outcomes (one of the columns from above). For example, if the player picks door 1 these are (1,1,2), (1,1,3), (2,1,3), and (3,1,2). These outcomes still are not equally probable with the "choice" outcomes (the first two) half as likely (probability 1/6) as the latter two (probability 1/3). Again, these sum to 1 (1/6 + 1/6 + 1/3 + 1/3 = 1).

After the host finally opens a door, there are only 2 possible outcomes. For example, if the player picks door 1 and the host opens door 3 the only possibilities are (1,1,3) and (2,1,3). As before, these outcomes are not equally probable with the "choice" outcome (the outcome where the car is behind door 1) half as likely (probability 1/3) as the "no choice" outcome - the outcome where the car is behind door 2 (probability 2/3). Again, these sum to 1 (1/3 + 2/3) = 1. -- Rick Block (talk) 01:54, 5 January 2014 (UTC)

Thank you for advancing the dialog. However, the diagram which I want to start with, as one of two potential diagrams, is this: What does the sample space look like BEFORE any choice is made? for this first diagram, I do not want to include any information about the possible permutations available as a consequence of that choice. Rather, I only want to illustrate a visual representation, in the form of blocks, as to what the sample space looks like before the first choice. If you agree that's what we're discussing there, then ok. If not, then why are we here?Tweedledee2011 (talk) 04:00, 5 January 2014 (UTC)

This illustrates sample space size reduction from door removal

This is the block diagram I want to discuss. If it's not accurate, state the reasons why. If it is accurate, let's discuss how to use it in the article. Tweedledee2011 (talk) 04:02, 5 January 2014 (UTC)

The "1st choice" diagram does not include any information about what door the host opens, which is in turn influenced by what door the player initially picks - so it is not the sample space for the MHP (it's a sample space for "I've hidden a car and two goats behind three doors, pick one"). The "2nd choice" diagram might be a continuation of the first ("now I've opened a door showing a goat, pick one of the two that are left") - also having little, if anything, to do with the MHP.

The probabilities involved in the MHP arise because of the rules pertaining to the host which are related to the door the player initially picks - specifically

1. the host can't open the door the player picks
2. the host must show a goat (i.e. the host knows what is behind the doors and deliberately reveals a goat)
3. the host must pick between 2 goat doors evenly if the player's initial pick is the door hiding the car

If any of these conditions are not met, the probabilities of winning by staying and switching may not be 1/3 and 2/3. If the sample space ignores the host's actions as influenced by the player's initial pick, it isn't relevant to the MHP. In particular, the fact that the door the player picks initially has a 1/3 chance of hiding the car does not mean that the probability this door hides the car remains 1/3 regardless of what happens (of course assuming the car is not moved to a different door). As Ruma Falk [1] puts it (emphasis in the original): "Truly, Monty can always open one of the two other doors to show a goat, and the probability of door No. 1 remains unchanged subsequent to observing that goat, still, it not because of the former that the latter is true." With this comment she's criticizing the slightly more sophisticated argument (put forward by vos Savant and many others) that the probability the car is behind door 1 does not change when the host opens door 3 because with two doors to choose from the host can always open one of them to reveal a goat. Even this is not accurate. The probability of where the car is depends on how the host decides what door to open, which in the usual interpretation includes all three of the conditions mentioned above. Omit any one, and you have a different problem with different probabilities. -- Rick Block (talk) 05:34, 5 January 2014 (UTC)

Rick -For a minute please think about and try to answer only this question: Yes or no, is the 9 blocks an accurate visual representation of the game's sample space at the moment just prior to the first choice? I need a clear answer to that point before I can effectively dialog with you. Tweedledee2011 (talk) 06:31, 5 January 2014 (UTC)

No. As I've explained above the sample space for the MHP just prior to the player's initial choice consists of 12 potential outcomes, not 3 as your figure shows. -- Rick Block (talk) 07:30, 5 January 2014 (UTC)

My block diagram shows nine possible results to the first choice, not three. Now presuming you are correct, can you make a block diagram the same as I have, showing the 12? Tweedledee2011 (talk) 07:58, 5 January 2014 (UTC)

Note: any diagram Rick presents is a Block diagram. Nijdam (talk) 14:03, 5 January 2014 (UTC)

No, your diagram shows only 3 outcomes, i.e. the rows labeled "Layout 1", "Layout 2", "Layout 3". Each one of these rows is a possible outcome of the initial step of hiding the car and two goats behind 3 doors - which is the situation the player finds herself in at the beginning of the game (not after having made her initial selection or after the host opens a door). I think I may have tumbled on to what one of the issues here might be. Can you let me know what problem you're attempting to address (both involving 3 doors, 1 car and 2 goats, a player who makes an initial selection, and a host who must show a goat and make the offer to switch)?

A. You are going on this show and want to know whether you are more likely to win the car by picking a door and staying with it, or by picking a door and switching to whichever door the host does not open. For example, should you pick door 1 and stay with it, or pick door 1 and switch to whichever of door 2 or door 3 the host does not open.

B. You are on this show, made your initial selection, have seen which door the host has opened, and are now deciding whether you are more likely to win the car by staying with your initial selection or by switching to the other unopened door. For example, you picked door 1 and the host opened door 3 so you are deciding whether to stay with door 1 or switch to door 2 while looking at a goat behind door 3.

These sound quite similar, and they are indeed related but they are not the same problem. Are you attempting to address (A) or (B)? -- Rick Block (talk) 17:36, 5 January 2014 (UTC)

Rick - You keep over-thinking and getting too far ahead. What I am trying to do is resolve (1) simple question. Then, after doing that, I want to think some more and come back for more discussion. As I see it, there are only three places for the car to be located. Based on that fact, there are only three possible ways to configure the layouts, if the goats are fungible. Of they are not, then we must distinguish between them. And if so, that adds some additional possible initial layouts. And mind you, I am talking about the initial layout of the doors. Again, I think the vernacular is what's tripping up this dialog. Allow me to be as precise as a I can, using links to emphasize terms which have particular meaning in probability. That said, here goes: I see nothing in the description of the game which says anything about how soon, after the player makes the first choice, that the host will open a door. This being the case, it's possible to freeze time and examine where are we in the logic of this process at an exact point in time. From the player's perspective, because the instructions do not say that the player knows ahead of time that the host will open a door, or that the player knows (ahead of time) he will be offered a second choice, then as far as the player knows, he is facing (1) car (2) goats and (3) doors. The instructions do not say if the goats are distinguishable, therefore, because this is an unknown, we can examine it either way. This particular walk-through assumes they are fungible. And if we get through this, we can always go back and look as if the goats are not fungible. But for now, with two indistinguishable goats, what the player knows at the start is on this: There are three Doors. There is one Car. There are two Goats. Based on that knowledge alone - which is all the player has access to at the start, there are only three possible ways to configure the doors (as per my diagram). Now the player is told to choose a door, but at the point in time of that choosing, that player would reasonably expect that the chosen door is his choice - because as of yet, he's not been told he'll get a second choice. Therefore, at the point just before the first choice, the first choice can only be seen by the player as an experiment, which consists of a Singleton event. And that should, from the players perspective, lead to the outcome of the door he chooses becoming an opened door. However, from our perspective (outside the game) we can see that the player doesn't get to the point of immediately opening the door. Rather, that information which would result from that event outcome is stopped by the host saying something like "before you open that door, would you like to keep it or switch...". So then, the outcome of the first choice is not immediately known. In other words, the player does not at that point find out the result of his door selection. The experiment was to find out what's behind the door. The door is chosen. But the player does not at that point find out. That information is not yet revealed. So then, up to this point, before the host opens a door and offers a new choice, the only possible outcomes of the first choice, were nine - just like my block diagram shows. This is what I want you to discuss with me. DO NOT YET introduce the host's opening of a door or offer to switch - and any effects those have on the composition of the sample space. Instead, look at it from the perspective of the player. Up until the door is opened and the offer to switch is made, I see the player's sample space as containing nine possible outcomes. It's not three, because we don't know which door the car is behind, so we have to think it could be any. And if it's any, then there are nine possible outcomes to player's the first choice. Three possible outcomes at any of three doors. Now please, think about this and tell me if what I am saying is true. At the point in time I am specifying, is the total number of the player's first choice possibilities nine? Tweedledee2011 (talk) 22:41, 5 January 2014 (UTC)

So, you're at this point wanting a sample space describing only the possible outcomes after the car is hidden and after the player's initial choice? As you say, there are 3 possible locations for the car and 3 possible doors the player may pick which results in a set of 9 possible outcomes. Generally, outcomes are specified by showing the values of the variables involved. In this case there are two relevant variables, the location of the car and the door the player picks (this assumes we don't care at all about the goats - they're simply zonks after all). If you want to show this as a 2-dimensional array it looks like this (the cross product of the possible locations for the car and the possible player picks):

Car=1 & Pick=1	Car=1 & Pick=2	Car=1 & Pick=3
Car=2 & Pick=1	Car=2 & Pick=2	Car=2 & Pick=3
Car=3 & Pick=1	Car=3 & Pick=2	Car=3 & Pick=3

Note that this differs from your table in that each cell shows the combination of car location and player pick rather than the car/goat "outcome". It's important to keep track of these if we're next going to talk about the door the host opens (which I assume we are). If you want, you could annotate these with the car/goat result, perhaps like this:

Car=1 & Pick=1 Result=Car	Car=1 & Pick=2 Result=Goat	Car=1 & Pick=3 Result=Goat
Car=2 & Pick=1 Result=Goat	Car=2 & Pick=2 Result=Car	Car=2 & Pick=3 Result=Goat
Car=3 & Pick=1 Result=Goat	Car=3 & Pick=2 Result=Goat	Car=3 & Pick=3 Result=Car

Is this more or less what you're looking for? -- Rick Block (talk) 06:25, 6 January 2014 (UTC)

Is your 9 block diagram correct for the point in time we are talking about? Tweedledee2011 (talk) 06:29, 6 January 2014 (UTC)

What about this look, is this correct? Tweedledee2011 (talk) 06:38, 6 January 2014 (UTC)

This is door #1 Car is located here Player picks this door Result: Win	This is door #2 Goat is located here Player picks this door Result: Lose	This is door #3 Goat is located here Player picks this door Result: Lose
This is door #1 Goat is located here Player picks this door Result: Lose	This is door #2 Car is located here Player picks this door Result: Win	This is door #3 Goat is located here Player picks this door Result: Lose
This is door #1 Goat is located here Player picks this door Result: Lose	This is door #2 Goat is located here Player picks this door Result: Lose	This is door #3 Car is located here Player picks this door Result: Win

Correct? Well, not really. The text "This is door #n" in each cell is distinctly odd. The cells are not doors. They are combinations of the two variables "where is the car hidden" and "what is the player's initial pick of door". It's like you're rolling two dice with the cells corresponding to a particular combination of the value of one die and the value on the other die. And you're omitting information in the cells that will be important later. For example, the cell in the first column and second row says it's door #1 and it has a goat and the player picked it - but not that the car is (in this case) behind door #2 - in fact it looks identical to the cell below in which case the car is behind door #3. These are not the same outcome even though they look identical in your table. The location of the car is critically important and must not be lost. For example, for all the cells in the second row the car is behind door 2 - each cell in this row should say this. Why are you resistant to the table I suggested? -- Rick Block (talk) 07:17, 6 January 2014 (UTC)

Rick - I am not "resistant" to your diagram. Rather, I am trying to develop a diagram which explicitly explains the significance of each possible choice to a total novice in this subject. As for what the cells "are", they can indeed been seen as doors. The first row represents the first possible configuration, the second row the next and the third, the final. There are only three possible ways to set the initial configuration of the doors, and this diagram shows them all. Look above at my first version on this page - that version called the rows "layouts". This current version (modified from your diagram) is both a logical map and fully fleshed-out physical diagram of the three possible layouts. And by my count, there are only nine (9) possibilities at the start of the game. Do you see that? Do you agree that at the start of the game, there are only nine (9) possibilities available to the player? Tweedledee2011 (talk) 07:44, 6 January 2014 (UTC)

I'm honestly trying to help you, but if you're not going to listen to me I'm not sure why I should bother. If you're talking about the "start" of the game (before the player's initial pick) there are 3 possible layouts of 1 car and 2 goats - not 9. One layout is (C,G,G). If you'd like you can call this a sample set consisting of 3 (not 9) outcomes {(C,G,G), (G,C,G), (G,G,C)}. The expansion of the sample set to 9 outcomes happens because of the player's pick. Each of these 9 outcomes is the combination of one of the 3 possible layouts and one of the 3 possible player picks. For example the outcome "player picks door 1 in the layout where the car is behind door 1" might be represented as ((C,G,G),1) (the layout is the (C,G,G) layout and the player pick is door 1). You can arrange these outcomes visually in a 3x3 array with the rows being all the possibilities with the same layout (same car location) and with the columns being all the possibilities with the same player pick - but the individual cells are not "doors". Do you understand this? -- Rick Block (talk) 16:30, 6 January 2014 (UTC)

Rick - you are not even trying to hear me. My diagram represents ALL THREE of the possible initial layouts of the doors. Yes or no, do you concede that at the start of the game, the car can be behind any of the doors? If yes, then the possibilities I am talking about are the possibilities of WHERE THE CAR IS AT THE START. I am illustrating this in my nine block diagram. Do you understand what I am saying, yes or no? Tweedledee2011 (talk) 18:55, 6 January 2014 (UTC)

Rick - you are leaping ahead to "outcomes" of the first choice again. STOP before you get there. My diagram is a diagram of all the possibilities of the playing field BEFORE the first choice is made. Tweedledee2011 (talk) 18:58, 6 January 2014 (UTC)

Initial state of game, from player's perspective

This block represents door #1 There is a 1/3 chance the car is located here If the player picks this door (and this door is opened), the result will be: Win	This block represents door #2 Goat is located here If the player picks this door (and this door is opened), the result will be: Lose	This block represents door #3 Goat is located here If the player picks this door (and this door is opened), the result will be: Lose
This block represents door #1 Goat is located here If the player picks this door (and this door is opened), the result will be: Lose	This block represents door #2 There is a 1/3 chance the car is located here If the player picks this door (and this door is opened), the result will be: Win	This block represents door #3 Goat is located here If the player picks this door (and this door is opened), the result will be: Lose
This block represents door #1 Goat is located here If the player picks this door (and this door is opened), the result will be: Lose	This block represents door #2 Goat is located here If the player picks this door (and this door is opened), the result will be: Lose	This block represents door #3 There is a 1/3 chance the car is located here If the player picks this door (and this door is opened), the result will be: Win

This diagram shows all three possible layouts of the doors, and what would happen, if the player picks one of the three doors (each row is one of the three possible starting layouts) and then WE FREEZE TIME at that point to think. At the start of the game, the player is only told to pick a door and that's what this diagram illustrates. It shows what the outcome of that initial pick would be, if that initially picked door was opened. This is a physical diagram map which shows each and every possible result of the player's 1st choice, from the perspective of the player, BUT BEFORE the host opens a door and offers a 2nd pick. This is a moment frozen in time and it's this point in time which I want to discuss. Once the others on this board understand this moment in time, then the discussion can continue. Tweedledee2011 (talk) 19:16, 6 January 2014 (UTC)

I hope somebody else can help you. I'm done. -- Rick Block (talk) 19:58, 6 January 2014 (UTC)

@Tweedledee2011: I understand what you mean. Please continue.--Albtal (talk) 20:17, 7 January 2014 (UTC)

@Tweedledee2011: I read in your comment above:

From the player's perspective, because the instructions do not say that the player knows ahead of time that the host will open a door, or that the player knows (ahead of time) he will be offered a second choice, then as far as the player knows, he is facing (1) car (2) goats and (3) doors. ... Now the player is told to choose a door, but at the point in time of that choosing, that player would reasonably expect that the chosen door is his choice - because as of yet, he's not been told he'll get a second choice.

Indeed for the 2/3 solution being correct it is a necessary condition that the player knows ahead of time that the host will open an unchosen door with a goat and will offer a switch. And indeed this condition is not part of Marilyn vos Savant's "Monty Hall Problem" going around the world. But exactly this problem has been published plenty of times together with the asserted 2/3 solution; meaning that the great "Monty Hall Paradox" really is a joke. The problem with this critical rule of the game has a very simple solution: Suppose you first "choose" door 1. Then you will win the car if it is behind door 2 or door 3. For you know that the host now has to open door 2 or door 3 with a goat. If he opens door 2 you finally choose door 3, and if he opens door 3 you finally choose door 2. See my former comments. I have stopped discussing here. But if I understood you correctly this might be a helpful comment for you.--Albtal (talk) 21:35, 7 January 2014 (UTC)

@Albtal - I reformatted your post for clarity. No words have been changed. Here's what I am saying: The problem with this problem isn't that the math experts are solving it with probability calculations. Rather, it's that they are solving it by using legitimate math in an illegitimate manner. For instance, the very premise of the problem says "Suppose you're on a game show...". This statement requires a full stop and an adoption of the premise, before we continue. From Merriam-Webster |"Suppose: To think of (something) as happening or being true in order to imagine what might happen". The math experts are explaining this poorly, because they have not yet satisfied the first condition of the problem, which is to suppose you are playing a game show. It does not state that it's Let's Make A Deal, and it's illegitimate to impute that into the premise or that as a player, before the host announces anything, that you know he's going to open a door. Therefore, the decision-tree models which claim the starting sample space to be 12 are illegitimate, because the player, at the start of the game, knows nothing about a door to be opened by the host or a second chance to be given. It's only after the host opens a door and offers a choice, that the player knows anything other than, three doors, one prize. Based on three doors, one prize, there are three possible configurations of doors, and this means that the initial sample space is 9, not 12. Also, as soon as a door is opened, it's not choose-able anymore, so it's no longer part of the sample space. As a result, the sample space is reduced, immediately upon when the door is opened. This leaves a 2x2 (down from 3x3), which is now 4. And when the player sees that, it looks like 1:2 or 50-50. But the 2:3 can still be arrived at legitimately - though not by the means of a 12 position decision-tree calculation. Those calculations are made on a sample space which never existed in the game and are therefore illegitimate by the very definition of sample space: "the set of all possible outcomes or results of that experiment". I can still arrive at the 2:3 result and that result is correct, but it's not correct for the reasons the math experts are saying it is. The math experts are arriving at that conclusion by violating their own premise. Tweedledee2011 (talk) 23:07, 7 January 2014 (UTC)

@ Tweedledee2011: Allow me to offer some thoughts about the confusing issues here, and about why your line of inquiry is meeting so much resistance. I am trying to help. (I actually fault the way introductory courses on Probability & Statistics are typically organized, which leaves many students unclear about fundamental issues even when they have mastered the applied techniques presented.)

Note that the opening sentence of the Sample space article you cite specifically refers to "an experiment or random trial". This concept only applies to the Monty Hall problem in an abstract sense (described below) because it is not a problem of statistical sampling. The contestant's initial choice is not a sample or experiment: the door remains closed and no outcome is observed. The revelation of a goat is not a sample, experiment, or random trial: it is selective information. There is no statistical sampling going on.

There is an abstract sense in which one may consider a statistical sample from among the set of possible universes, or instances of the game; but calling this a "sample space" rather than simply the set of possible scenarios can be confusing. Firstly, the elementary statistical events in this space are not things like "finding a car behind a particular door", as in ordinary statistical sampling, but are "ways the game might play out" – a probabilistic notion that may not be apparent to the average reader.

Secondly, and more importantly, there is no "one right way" to define outcomes for analyzing the problem. Your choice of two possible outcomes for the final question (car behind door 1 and car behind door 2) is not wrong, (any win/lose game may be modeled with exactly two outcomes, "you win or you lose"), but this is not a very informative way to define outcomes. Rick's model for ways the game might play out is much more informative because it is decomposed into more elementary events that can be summed to show the composite win/lose outcomes are not equally probable.

In terms of making the solution understandable for our readers, your emphasis on the size of the sample space is not useful when the outcomes you define are not equally probable. Simply counting such outcomes leads people directly to the naïve conclusion that if the car is behind one of two doors then they must be equally likely.

As far an analyzing the situation before/during/after the initial choice of door #1, there is no need to belabor the point using sample spaces or any other type of analysis. Everybody understands that this is a blind choice among three alternatives with equal probabilities of success.

Finally, and very importantly in terms of Wikipedia policy, if you make the extraordinary claim that all of the math experts are wrong about this then you should cite some very credible expert publications from some field that specifically say they are wrong about the particular point you are criticizing. Otherwise, claiming that you know better than the experts will only give people the definite impression that you don't know what you are talking about.

I hope some parts of this have helped you understand why your approach is not being received well, and has given you some food for thought about probabilistic reasoning. ~ Ningauble (talk) 20:35, 12 January 2014 (UTC)

@Ningauble - First, let me stipulate there is no argument from me that switching results in a win 2:3 times. That's not the point of my dialog. Rather, my point is that the article is only explaining this math from the orthodox math-proof direction and is not actually thinking it through as a novice would. It's my view that the article could benefit from an examination of the perspective I am trying to offer. And to achieve that, I am trying a three step approach: 1) Explain the perspective, 2) Obtain some consent that the way I want to explain it can fit in the article and 3) Enlist team support to help me find enough reliable sources that the naysayers here will accept the edits I wish to make. Now, since I am only trying to enhance the article, and because I've NOT jumped the gun and rushed my edits into the article, it's obvious that people should try more to dialog and listen, rather than fight me here. So, once again, here goes:

1) Sample space does not mean "sample" the way you have used the term

2) Your misuse of that term in this context proves my point - the vernacular is fraught with traps

3) At the point where the player makes his first choice, from the player's perspective, that choice is indeed an experiment

4) That the payer does not immediately see the result of that experiment, is not relevant to the point I am making.

5) At the point in time when the player makes his first choice, the player has not yet witnessed the host open one of the "lose" doors

6) At the point in time when the player makes his first choice, the player has not yet been told that he will get a 2nd choice

7) Therefore, based on 5 & 6, at the point when the player makes his first choice, the total size of the sample space is 3x3 or 9 - that's it

8) Then, at the very moment that the host opens a door, that door is no longer choose-able by the player, and it is therefore, not part of the sample space of the player's 2nd choice.

10) The player, for his 2nd choice, is allowed to pick only from 2 doors.

11) The reason why switching works is not because of the 12 point decision tree model shown in the MIT .pdf which I linked to on the talk page

12) No, the reason why switching works is because by opening a door, the host reduces the amount of doors behind which the car can be.

13) In the first choice, the player has a best chance of 1:3 - and that never changes because the car's location is fixed

14) When the host opens a door, the odds of switching must be 2:3 because the total must equal 3:3 (1) and the original choice is fixed at 1:3

15) This can indeed be reasoned out via the decision tree model, but only if one cheats and stretches the literal meaning of sample space, which is the set of all possible results of an experiment.

When the host opens a door, the sample space shrinks - this is indisputable. The only real argument here is whether or not it starts at 9, 12 or 15. I say it's 9, unless we distinguish between goats, then it's 15. However, the explanations which say it's 12 are simply wrong - because they rely on the logical consequences of what would happen to the array of possible outcomes when the player chooses again. But adopting this logic violates the verbal premise of the problem which requires the problem solver to "Suppose you're on a game show". A player in the game, when facing his very first choice, is not facing a sample space of 12 outcomes. The reasoning which proves 12 relies on the player being given a 2nd choice, and therefore, until the player is actually given that choice, it's logically impossible to build the 12 count sample space decision tree. But, when the player is given the choice, it's only after a door has been opened. And because a door has been opened, that door is not choose-able any more. And because it's not choose-able, it can not hold any results for the player's second choice. And if it can't hold any results, it's not part of the sample space. The sample space starts at 9 and shrinks to 4. But you can still use a decision tree to arrive at 2:3, but you can't call that decision tree a diagram of the sample space of the game from the player's perspective, because it's not. That, is my objection to how this article is written - it omits the fact that the experts are explaining this wrong. They are using correct math, but are sloppily applying the vernacular. But as a result of how we have written this article, anyone who's read an example of the 12 result decision tree model and comes here for more information, will not find anything explaining this point.

Tweedledee2011 (talk) 06:32, 14 January 2014 (UTC)

I still do not see how the perspective you are trying to explain will help our readers to understand why it is better to switch, particularly since the inferences you are drawing from that perspective are not correct.

1:  Your argument does not support the assertion [your point #11] that the decision tree in the course notes of Meyer and Rubinfeld at MIT (and by implication, the related tree in the "Conditional probability by direct calculation" section of the present Wikipedia article) is invalid.

1.a:  If you are suggesting that the tree on page 5 fails to account for the state of affairs at the point of the contestant's initial choice [your points #5 through 7], this is simply not the case: read the diagram from left to right. This is made explicitly clear where Meyer and Rubinfeld construct the tree progressively and show, on page 4, the specific situation of the player's initial choice without reference to subsequent events.

1.b:  If you are suggesting that at the point of final decision, i.e. the actual question posed by the Monty Hall problem, the tree improperly includes situations where the contestant chooses the goat that has already been revealed [your points #8 through 10], this is simply not the case: the tree enumerates the possible situations immediately before the final choice. One could add another level to the tree showing the contestant's final "stay or switch" decision, labeled A, B, or C with the constraint that it not have the same label as its parent node (the door that was revealed). Each of these nodes can be tagged with Car or Goat to indicate the resulting prize awarded. In no case is this the goat that was revealed before the question was posed.

1.c:  If you simply do not like Meyer and Rubinfeld's use the term "sample space" for the labeling generated by their decision tree, it is really not an issue here because the present Wikipedia article doesn't even use the term. I do not particularly like their use of the term "outcome" in this context to refer to a state prior to observing the result of the contestant's decision, but this is completely immaterial and does not invalidate their logic. I assure you that the decision tree methodology is not a cheat.

2:  Your conclusions about the reason switching works are not correct: the fact that switching is better does not follow from your observations about the sample space. To see why this is so, consider this variation of the problem, in which the same observations are true but the resulting probabilities are different:

• The Mary Hale problem is identical to the Monty Hall problem in all material respects except one: the host does not know or care where the car is. Mary picks one of the unchosen doors to open, and there happens to be a goat behind it. (Monty, on the other hand, picks an unchosen goat, and opens the door it happens to be behind.)

2.a:  In both problems "the amount of doors behind which the car can be" [your point #12] is exactly the same. In the Mary Hale problem the probability of winning by switching is only 1/2, not 2/3 as in the Monty Hall problem. Therefore, the the amount of doors is not, as you state, the reason why switching works.

2.b:  In both problems the car does not move [your points #13 through 14]. In the Mary Hale problem the probability that the car is behind the initially chosen door after another door has been opened is 1/2, not 1/3 as it was beforehand. Therefore, the reason it is 1/3 at both points of time in the Monty Hall problem is not, as you state, because the car's location is fixed.

Unless you can provide a clear demonstration of how your formulations of the sample space logically determine the correct probabilities, i.e. how the correct result can be deduced from that starting point, I think you should not be focusing on how to include this perspective in the article, but should instead focus on gaining an understanding for yourself of the reason it is better to switch. ~ Ningauble (talk) 20:31, 14 January 2014 (UTC)

This illustrates sample space size reduction from door removal

@Ningauble - There appears to no longer be any point in discussing this with you. The few replies I get, such as yours, are so tangential to what I am saying, that it's an entirely different debate. If you won't concede that the sample space, prior to a door being opened, and prior to the 2nd choice being offered; contains only 3x3 (a total of 9) possibilities, then there is no point in discussing this. Tweedledee2011 (talk) 07:27, 15 January 2014 (UTC)

@Tweedledee2011: The mere fact that you're talking about the sample space, prior to a door being opened, and prior to the 2nd choice being offered, shows you're not very experienced on the matter. Reading a book on probability theory doesn't help. A sample space contains the possible outcomes of a probability experiment. Experiment is the general term for situations where chance plays a role. You have to ask yourself what the possible outcomes are in the MHP. That's not very difficult, although you seem to have some trouble with it. We have the possible positions C of the car, the possible first choices X of the player, and the door H opened by the host. As each of C, X and H may have the values 1, 2 and 3, this leads to 3x3x3=27 possible outcomes. Some of these outcomes will have probability 0. It may simplify the sample space if we leave them out. The outcomes left are (CXH): 112, 113, 123, 132, 213, 221, 223, 231, 312, 321, 331, 332. If you count them well, they are 12. These 12 outcomes are necessary to describe the MHP. The sampole space is not symmetric, i.e. not all outcomes have the same probability. I leave the probability distribution to you. Nijdam (talk) 11:51, 15 January 2014 (UTC)

@Nijdam - Have you even read any of my comments, are just going to come in, guns blazing, repeating what we already know? What I am trying to help the others see here, is why so many people get the explanation of this game wrong. Part of that reason is people like you, who do nothing but repeat the obvious - yet ignore the plain language of the problem. The problem says "Suppose you're on a game show, and you're given the choice of three doors". At that exact moment in time, a player on the show would be facing a sample space of 3x3 - this is an irrefutable fact. The 12 space decision tree model is not in existence until after a door is opened and after the player is given another choice. In fact, the very descriptions of the 12 space tree make it very clear that the tree is intended to show the possibilities of switch/not switch - after a door is opened. That you can't see this (or refuse to) makes your comments pointless. What I am trying to help the math experts see, is that the novice player's intuition isn't all that wrong. Rather, it's that they don't know enough about probabilities to know which sets of information they can use. A novice, when told to Suppose you're on a game show does just that - and they go into the mindset of playing a game, a game which is typically based on guessing (if played by non-math people, which is mostly everyone). The opening of the door does indeed reduce the size of the sample space - and you have not disproved this assertion. The simple fact is that after a door is opened, the player can only make his 2nd choice from a 2x2 set of possibilities - one goat, one car, two doors. And yet, switching still works - but not for the reasons the experts are saying. Removing a door (and it's associated possibilities) reduces the sample space and therefore; the solutions which are based on a 12 space tree misrepresent the true size of the sample space. A door which is not choose-able does not hold any possibilities and therefore; even though the math calculation is correct, calling the 12 space decision tree a model of the sample space for the problem is a false statement. It simply is not one. Rather, it's what the sample space would be, if and only if, the open door (which is not player choose-able) still held possibilities for the player - which it does not. That you can't seem to understand this is truly astounding. Tweedledee2011 (talk) 07:39, 16 January 2014 (UTC)

Question for Nijdam

Good to see you back.

I have a question for you, nothing to do with Tweedledee. I am sure that you will see its connection with the MHP.

There are two doors each with an urn behind it. One urn contains two black balls and one white and the other two white balls and one black. You select a door and from the urn behind it remove a ball (as always with urns, at random) which proves to be white.

What is the probability that from that same urn the next door ball you draw will be white and how do you calculate this? Martin Hogbin (talk) 15:46, 15 January 2014 (UTC)

Martin - I think you meant to say "What is the probability that from that same urn the next ball you draw will be white"Tweedledee2011 (talk) 07:44, 16 January 2014 (UTC)

Yes, thanks, I have corrected it.

And to answer your question, the probability is .25 and here's why:

URN #1 has BBW and if you first choose W from this, the next choice from this urn can only be B, which is 0

URN #2 has WWB and if you first choose W from this, the next choice from this urn is between W or B, which is .50

Since you don't know which urn had what to start, if you repeat this 2 choice series enough times, you will find W on the second choice .25 of the time.

Half the times, you have W at 0 from URN #1 for a possible 2nd choice result and half the times, you have W at .50 from URN #2 for a possible second choice result.

That's why you only get W .25 of the time on your 2nd choice

Tweedledee2011 (talk) 08:21, 16 January 2014 (UTC)

Tweedledee, I did not set this problem as a trap for you but to demonstrate a completely different point to Nijdam. As it happens though you have made a mistake which helps me demonstrate the point that I want to make.

The problem lies with your step, 'Half the times, you have W at 0 from URN #1 ... and half the times, you have W at .50 from URN #2...'.

Let us suppose that you initially toss a coin to decide which door to open. Let us, purely for our own convenience, call the urns A and B. I think we all agree that before you open a door the probability that you have chosen urn A for example is 1/2.

Next you open the door to see the urn. This tells you nothing which might let you revise that probability.

Next you draw a ball from urn A, which proves to be white. This does give you information about which urn you originally chose. One urn contains two white balls but the other contains only one white ball. Given that you have in fact chosen a white ball you must now consider it more likely that you chose from the urn which had two white balls. This is a conditional probability; the probability of an event given that another event has actually occurred. This has been a much discussed subject here in the past.

To get the correct answer we need to start by considering all the events that might have happened.

[Point for Nijdam] To do this, purely for our own convenience, we may choose to label the balls W1 and W2 in one urn and W3 in the other. The other balls may be labelled K1, K2, and K3. Although all the white balls look the same to us and all the black balls look the same and we only actually see two of the balls, we know what balls there are so to get the probability right we should, in principle, label them all.

So our sample space is AW1, AW2, AK1, BW3, BK2, BK3 (if you really want to we could add BW1, BW2, BK1, AW3, AK2, AK3) but I think we can agree that this is not necessary. Each event in this space has a probability of 1/6.

Now we must condition our sample space on the event that we actually drew a white ball. This means considering only the events that conform to our condition, namely AW1, AW2, BW3. As we know for sure that one of these events actually occurred, we must adjust the probabilities so that they total 1. This makes them 1/3 each.

As you can now see, after a white ball has in fact been drawn, it is twice as likely that your original ball came from the WWB urn than from the WBB urn, that is to say twice as likely that you have chosen the urn with two white balls. Martin Hogbin (talk) 09:50, 16 January 2014 (UTC)

Hi Martin, as you see, I'm not gone, but I'm keeping watch. Concerning your experiment, you're right in your analysis. In terms of conditional probabilities: P(W2|W1)=P(W2W1)/P(W1)= (1/3) / (1/2) = 2/3. I do not see the direct connection with the MHP. Nijdam (talk) 20:17, 16 January 2014 (UTC)

My point is this. We have two black balls in one of the urns. They are, to most intents and purposes, identical. We may never even see one of the black balls if we draw only two balls from that urn. The balls are not numbered, nevertheless it helps us formulate the problem correctly if we do number the two black balls in one urn (and the two white ones in the other).

Now consider the two urns. They may be identical, we do not know because we only see one of them, but again it is useful to label the urns. One is the urn that we chose and the other is the one we did not choose In my example it matters which urn you choose. If each urn had two white and two black balls it would make no difference but it still might be good practice to label them in some way. Do you agree so far? Martin Hogbin (talk) 17:54, 17 January 2014 (UTC)

Martin - see below Tweedledee2011 (talk) 16:42, 16 January 2014 (UTC)

True answer to Martin's question

Martin - if you first pick a white ball from an Urn #1 mix of BBW, the next pick from Urn #1 can only be B, which is a probability of 0 for W. Do you concede this yes or no? Your mistake is that you think the information developed by the 1st choice is relevant - it's not. If you remove a W from a mix of BBW, there is no longer any possibility that W can come from that Urn. There are two urns and the condition you set was that the first pick "proves to be white". The odds of the first pick being white are irrelevant, because you have predefined the results of the first pick. For that reason, we are only concerned about the possible results of a 2nd pick. The possible results of a 2nd pick from a Urn #1, which starts at BBW, is only B.

Urn #1 starts as BBW

The first pick "proves to be white"

This leaves only BB in Urn #1

Pick #2 from Urn #1, can never be anything but B, if pick #1 from Urn #1 is W

Tweedledee2011 (talk) 16:32, 16 January 2014 (UTC)

Last try

You say 'Pick #2 from Urn #1, can never be anything but B, if pick #1 from Urn #1 is W'. Yes, of course.

Your problem is with this statement "Half the times, you have W at 0 from URN #1 ... and half the times', you have W at .50 from URN #2...". I have marked the problem in italic. Martin Hogbin (talk) 17:06, 16 January 2014 (UTC)

Because first pick results are predefined ...

Martin - The possible pools of choices results after the first choice are: BB and WB. There are no other combinations of possibilities available for the 2nd choice. Under no circumstances will you find a B in Urn #1 after first drawing out a W. And under no circumstances, will there be anything other than WB in Urn #2, after first drawing out a W. Do you concede this? Tweedledee2011 (talk) 16:52, 16 January 2014 (UTC)

Yes, of course. See above. Martin Hogbin (talk) 17:06, 16 January 2014 (UTC)

In a large series of first picks, the first pick will be W .50 of the time

Martin - because there is a total of WWW and BBB mixed between the jars, do you agree that 1:2 times, your first pick will be W, if you do the fist pick a large number of times? Tweedledee2011 (talk) 17:19, 16 January 2014 (UTC)

Yes. Martin Hogbin (talk) 17:43, 16 January 2014 (UTC)

In a large series of first picks, the first pick will be from Urn #1 .50 of the time

Martin - do you agree that in a large series of random first picks, your first pick with be from Urn #1 - the BBW Urn .50 of the time? Tweedledee2011 (talk) 17:30, 16 January 2014 (UTC)

Yes. That is what I say above. Please read what I have written and try to understand it. Martin Hogbin (talk) 17:44, 16 January 2014 (UTC)

Martin - if on average, over a large series of tests, for .50 of the times you run this test you must pick from Urn #1 on the first pick; then it logically follows that since your 2nd pick must be from the same urn (your rule sets this condition) .50 of the times you run this test, your 2nd pick must be from the remaining BB in Urn #1. What you don't understand is that your result of W from Urn #1 is an anomaly from Urn #1 - not the suggested proof you are in Urn #2, which you are interpreting it as. Look at your answer to this section again "Yes, that is what I wrote". Martin, if your first pick is from Urn #1 .50 of the time and if the results of the first pick are W, then for .50 of all first picks, the next result (the 2nd pick) from Urn #1 can only be B - and you have already conceded both of these points. Your predicate conditions create an anomalous subset which you are not correctly thinking about. You are the one who created this - by the way you phrased the series of predicates. Go back and re-read your own question. Tweedledee2011 (talk) 17:52, 16 January 2014 (UTC)

For any a large series of first picks, the 2nd pick will be between BB and WB, with each being .50 of the time

Martin, you can't simply extrapolate out that you are mostly going to be dealing with Urn #2 which is what you think your calculation proves. That's post hoc reasoning. You must first start with the correct mix of urns - as per the above results which are what would occur in a large series of 1st choices. Tweedledee2011 (talk) 17:35, 16 January 2014 (UTC)

No, the 2nd pick will not be between BB and WB, with each being .50 of the time. You can call this post hoc reasoning if you like. Everyone else calls it Conditional probability.

Let me try to make this easier for you. Suppose urn A contains 99 white balls and 1 black ball and urn B contains 99 black balls and 1 white ball. You chose evenly between the urns so you initially have a 1/2 chance of picking urn A and a 1/2 chance of picking urn B.

Now you draw a ball randomly from your chosen urn and this ball proves to be white. Do you claim that the probability that your chosen urn is A is still 1/2?

If you do claim that I have an idea. Let us play the game for money, real money. You bet a sum of money. You pick an urn at random and pick a ball. If it is a white ball I will give you twice your money if the next ball you pick from the same urn turns out to be black. If the first ball is white and the next ball is white I win the stake. Would you like to try to set up this game somehow? Think hard about it! Do some experiments. Martin Hogbin (talk) 18:12, 16 January 2014 (UTC)

Martin, you area again missing my point: Your conditional probability explanation relies upon a false premise - because you fail to accept that you are only dealing with a small subset - one which you cherry-picked in such a way that it could not occur as a consequence of the game you are describing. I've shown you that, but you refuse to accept it. If you concede that .50 of the time your first pick will be W and if you concede that .50 of the time your fist pick will be from Urn #1, then it's inescapable that .50 of the time (counting a full series of the 2 picks as 1 "time"), your second pick with have 0 chance of W and .50 of the time (counting a full series of the 2 picks as 1 "time") your second pick will have .50 chance of W. Tweedledee2011 (talk) 19:47, 16 January 2014 (UTC)

But I said that the first pick proves to be white. The means that the cases where the fits pick proves to be black do not count. That is what Conditional probability means. Only counting the cases where the condition applies is called conditioning the sample space. Now re-read what I write at the start. Martin Hogbin (talk) 20:03, 16 January 2014 (UTC)

If of course, you are saying that of the original WWW because WW are in Urn #2, then a larger portion of the original 1:2 chances of W are in Urn #2, then I do see what you are saying. You are saying that because it's more likely to find a W in a pool of WWB than in a pool of BBW, then when you find a W, you are more likely to be in the WWB pool, which means that your next potential results ought to take into account. This is true, but then the odds of getting W in Urn #2 on a 2nd choice are .50 And in any case, then my error on this also shows how people using normal verbal reasoning do not apply the math which you are talking about - which makes my point why how we are explaining MHP is deficient. What should be done is that we should show some ordinary normal vrbal reasoning, then show how that diverges from the conditional probability method have ably illustrated here - and contrast the different results, showing the gaps. With MHP, it's hard to show the gaps - due to the configuration of the problem itself. Also, what is the sample space of pick #2 from Urn #2 - I say it's WB. Tweedledee2011 (talk) 20:07, 16 January 2014 (UTC)

The conditional probability proof for Martin's problem does not answer the question asked

Martin laid out a puzzle (see above) and then asked this question: "What is the probability that from that same urn the next door ball you draw will be white?" And in regards to this, another editor posted a formal probability calculation which he says affirms Martin's assertion. But neither of them make any sense for the simple reason that the first pick from the urn does not affect the second pick, other than to finalize the mix of ball colors. To suggest otherwise is akin to the Gambler's fallacy. At any given moment, if there are one black and one white ball in a urn and you randomly pick one of them, the odds of getting either one is 1:2 (or 0, if you are picking from the BB urn). To suggest otherwise is like a person who tries to outguess a flipping coin. The hand reaching into the urn is a distinct random event. Nothing which comes before that pick affects the actual pick itself in any way. The only thing a previous pick does is reduce the total choices down to two. But when it's at two, the answer to the question "What is the probability that from that same urn the next door ball you draw will be white?" is 1:2 (or 0). I think that the reason why Martin's puzzle is false, is that the contents of the two urns form two distinct sample spaces such that the richness of the W population per space can help you calculate which space you are in, but once you've removed that 1st ball, there are only two left per urn, and the positions of the balls inside the urn are fungible - they are in flux. For that reason, there's nothing which carries over any information from the 1st pick. Conditional probability can't be used for this puzzle. After removing a W ball from the urns, you are left with BB and WB - just like I said above. The next pick into either of these sets can not be 2:3 - there are not enough remaining positions in either set to yield that. Once a ball is removed from the urn, it does not effect the odds of the next pick, except, as I said, to set the mix as fixed as either BB or WB. Tweedledee2011 (talk) 05:08, 17 January 2014 (UTC)

Time to put your money where you mouth is

According to you, if we were to play the urn game that I suggested. You would win half the time. I will offer you twice your bet if you win, which should be a good deal for you. If you want to play let me know, otherwise I have nothing more to say.

Several users have tried to explain probability to you but you have resisted all attempts. The fact that no one has agreed with you must surely open in your mind the possibility that it is you who is wrong. The only way you will be convinced is by a demonstration in which you lose money. If you want to do this let me know, otherwise I will not respond. Martin Hogbin (talk) 10:06, 17 January 2014 (UTC)

Martin, your phrasing says "remove a ball". In your problem, is the ball returned to the urn after seeing what color it is? If so, the word "removed" is misleading. Tweedledee2011 (talk) 16:12, 17 January 2014 (UTC)

No, it is not returned. Do you want to play? Martin Hogbin (talk) 17:38, 17 January 2014 (UTC)

Before I answer that, please answer these:

True or false - If we flip a perfect coin three times, there are 2x2x2 (8) possible outcomes

True or false - There is only a 1/8 chance that of three flips, three in a row will be heads

True or false - The odds of a 4th flip being heads is 1/2

True or false - The odds of the 3rd flip being heads was 1/2

True or false - The cumulative odds required to get to the 1/8 on the 3rd flip is not the same as the 1/2 chance on each flip

True or false - The 2/3 you are talking about, is akin to the cumulative odds of the 1/8 chance of three heads in a row

If all true, then please explain how following your decision tree and getting 2/3 odds differs from a stranger walking in, picking from your urn and getting 1/2 odds?

I ask in all seriousness because by your example, you could sequentially place all of the available W and B balls in the world into a series of urn #2's (while using up some to populate the urn #1's) and even though the distribution of balls in those #2-modeled urns would be WB (one each), because you set them all in place via the scenario of this problem, your next choice from any of those urn #2's would have a 2/3 chance of being white. But if anyone else in the world were to draw a ball from any of your #2 urns, they would face 1/2 odds of a white. And remember, all the white and black balls in the world are used up by populating urns (in decreasing quantities, by using the removed balls) and your next draw must be from a #2 modeled urn which arrived at a WB configuration because you personally removed a W from it. Please reply to this, before asking me to answer. Tweedledee2011 (talk) 19:49, 17 January 2014 (UTC)

Sorry, I am not interested in trying to explain things to you. I am happy to clarify exactly what the game is so that there are no misunderstandings and we can play the game for real money. Do you want to put your money where your mouth is? Martin Hogbin (talk) 21:02, 17 January 2014 (UTC)

Provided that the computer code we use to test is valid, I am happy to take your money. And, since you are such a good sport giving me odds and such, let's start by you paying off on the first test I ran manually (see below) Tweedledee2011 (talk) 00:24, 18 January 2014 (UTC)

This illustrates sample space size reduction from ball removal

Which urn was it?

Tweedledee, consider that if you knew which urn it was then the answer would be different. For example, if you peek in the urn before picking the second ball, and see that it contains one of each color then, knowing this, you would say the probability of picking a second white ball is 1/2.

Without cheating like this, we don't know which urn it was, not definitely. However, the color of the first ball gives us some evidence about which urn it was. So consider this question about what the evidence tells us:

• Given the evidence of having picked a white ball, what is the probability that this is the urn that started out with more white balls in the first place?

— The answer to this elementary probability question is simple: it is 2/3. If you are not sure how to calculate this then please ask.

The point of calculating this probability is this:  Yes, the mix is either BB or WB but, given the evidence, it is twice as likely to be WB. It is not correct to say "Nothing which comes before that pick affects the actual pick itself in any way": it affects which urn is actually being picked from. This is the whole point of the question asked in Martin's problem. ~ Ningauble (talk) 14:07, 17 January 2014 (UTC)

@Ningauble I do not deny that a pick from a pool of WWB is more apt to yield a W than a pick from a pool of WBB. But you are misquoting me by omitting my proviso. Here is what I am saying: At the point with any urn, where there is one W an on B in it only, the next pick is a discrete random event and as such, is 1:2. There is no difference between two urns identically configured with one W and one B. But for your explanation to be true, if just before the 2nd pick, you took the original urn #2, which now has one W and one B in it and you substituted an urn #3, which has one W and one B in it -even though the urns are identically configured, the results of pick #2 would have different probabilities between those urns. But that's nonsense. Also, the question says that on the fist pick we are to "remove a ball", so you do agree that the first ball is taken out and not returned to the urn, yes?. Also, if Martin is referring to problem, the way he stated it is not the same Tweedledee2011 (talk) 15:32, 17 January 2014 (UTC)

There are several problems with this:

1. I was not saying that a pick from WWB is more apt to yield a W than a pick from WBB, which is true. What I said is that a result of W is more likely to have come from the WWB urn than from the WBB urn. (Please re-read the bulleted question in my post.)
2. My explanation does not imply anything about any third urn. Martin's problem refers to picking a second ball "from that same urn", and that is the urn to which I refer. (Please re-read Martin's problem.)
3. It is an urn with a history, the urn that was first selected by opening a door. The Gambler's fallacy, to which you refer in opening the top level thread, applies to independent events. It is not relevant because selecting balls from the urn depends on which urn was selected. (Please read the "Non-examples of the fallacy" section of the article to which you linked.)
4. Also, I did not misquote you. The post from which I quoted contains no "proviso" indicating your words pertain to a different situation than the one posed in Martin's problem. (Please re-read your post.)

Martin's problem is much simpler than the Monty Hall problem, because it does not involve any "sneaky" selective information. It is a problem of pure random sampling where it is easy to make a statistical inference about the population (urn) from which the sample is drawn. ~ Ningauble (talk) 20:29, 17 January 2014 (UTC)

But the doors are irrelevant. What this problem is saying is that from any pool of 2-1 (2 same, 1 different) of items, the chances of picking two same items in a row are 2/3. That's nonsense because once you remove the first ball, the remaining pool is 1-1. And the next is a fully random and distinct pick. Tweedledee2011 (talk) 20:55, 17 January 2014 (UTC)

For Martin's answer to be correct...

For Martin's answer to be correct, the question must be phrased this way: If there are two urns, one which holds two white, plus one black ball and the second which holds two black, plus one white ball; when you randomly pick one ball from one of the urns and that ball proves to be white, if you replace that ball back into the urn from which you picked it, what is the probability that an immediately subsequent pick from by you from the same urn that this same urn will yield a white ball? The answer to this question is 2/3 - but not the way Martin phrased it, it's not. Tweedledee2011 (talk) 16:01, 17 January 2014 (UTC)

Results of my manual test of Martin's urn problem

I ran a manual simulation as follows:

1) I made 24, 1 of 3 picks from a BBW urn and 24, 1 of 3 picks from a WWB urn

2) This is accurate because in big test, you'd randomly select first from either urn an equal number of times

3) Results from the BBW first pick urn are 9W, 15B

4) I removed the W from the BBW mix, leaving only a BB mix

5) All possible 2nd picks from BB are B - no need to test this

6) I then made 24, 1 of 3 picks from a WWB urn

7) Results from WWB urn first picks are 14W, 10B

8) Each time a W came up first, I removed one W, leaving only a WB mix

9) I then picked a 2nd pick from this mix of WB

10) I made 2nd picks 14 times

11) Of the 14 2nd picks, exactly 7 were W, which is 50% of the 2nd picks

Martin - do you want to pay by check or paypal?

I had to double-check my counts more than once, but the numbers now posted are accurate. I can scan my notes and post them, if you want

Tweedledee2011 (talk) 00:02, 18 January 2014 (UTC)

Martin's problem is wrong because...

Any initially picked W balls are removed from the sample space after the 1st pick. This means that 1/6 of the time a first pick is made, the entire #1 urn is no longer part of the puzzle as it can't possibly yield 2 whites in a row. And, 1/3 of the time a first pick is made, the possible choices of the 2nd urn are reduced from WWB to WB. This change in the original sample space must be taken into account on the 2nd pick. The proof provided above only works if the W balls picked are looked at to see the color, but not removed from the mix (reinserted into the urn). 50% of the time a 1st pick is made, the test fails because B comes of first and 50% of the time (1/6 + 1/3) the sample space changes (because the mix of available balls is changed by the removed W's) and that must be taken into account Tweedledee2011 (talk) 00:11, 18 January 2014 (UTC)