The Monty Hall problem

[Rolfe]

Yup. And both staying and switching win exactly half of the cases when the game is not over.

Nope. Count it and see.

You're not calculating your probabilities correctly.

Just write a simulation if you don't believe it. Or just examine and re-use the MATLAB one I wrote here, to confirm the probabilities I calculated here: in the case of a clueless Monty revealing a random door that's not the player's, what you do when it happens to be a goat is irrelevant to your chances. (Of course, just what happens when he shows a car makes a difference, but that's not the disputed case.)

"[The host] forgot which door hides the car! So he says a little prayer and opens No. 3. Much to his relief, a goat is revealed."

Look, if you're saying that the problem is incompletely defined because we still don't know how Monty picked his door, I agree with you: in that case we can't say much of anything at all. But there are two interesting subcases: (1) Monty picks a random door, and (2) Monty picks a random door that's not the player's. I've considered both of these in the second post linked above. I guess one could make even more possible assumptions compatible with the statement in Marylin's column, but (2) is actually the most natural.

That's what I thought. I was thinking Humber was saying something subtly different, but I still don't see how your explanation can be wrong for the situation where the choice of second door is random.

One third of the iterations, you were right first time, so switching was a bad move.
One third of the iterations, Monty reveals the car and says, goodbye and thanks for playing.
One third of the iterations, you were wrong first time, Monty reveals the other goat, and switching gets you the prize.

The second of these three possibilities is off the table because we know in the iteration we're considering, Monty revealed a goat. So it has to be the first or the third. These are of equal probability, so 50/50. I'm sure you could write a simple computer simulation to show this, just as you can write one to show that you double your chances of winning by switching if every iteration of the problem has a goat revealed behind the door Monty opens.

So I'm confused by the paper Humber linked to, which says this.

Proposition 3. If initially all doors are equally likely to hide the car and if the host is equally likely to open either door when he has a choice, then, given the door initially chosen and the door opened, switching gives the player the car with conditional probability 2/3, whatever door was initially chosen and which door was opened (Morgan et al., 1991a,b).

I can only assume I'm misunderstanding this, as the author seems to be some sort of expert.

Rolfe.

 

[Rolfe]

Not "Smart Alecs". The criticisms of many puzzles of this type, including Vos Savant's, are valid. Those puzzles work around the alternative cases that you may make, and mislead. Many, like the wave example, are factually wrong.

The wave example suggests she is brain-dead. I assume this is not the case, though. In this case, the criticism seems to be merely that the problem was sloppily presented in the first instance, whether by vos Savant or by Chris Whittaker seems unclear. The original players appear merely to have been posing a simple probability puzzle, and assuming that all iterations of the scenario would play out identically up to the point where the contestant if given the choice of switching or not.

Others have noted that the assumption is not stated, and thus other possibilities may be considered. Fair enough, but that doesn't make the original players wrong in fact, merely careless in formulating the presentation.

There are more objections than simply sniping. I have a book where the author does a real hatchet job on her, and that may perhaps be spleen.

Well to be honest, I don't think Marilyn's IQ is especially interesting. It's the problem that's interesting, not the person who is associated with its first public performance.

The problem was posed long ago: [snip exposition of the problem, as we all know it]

It appears not to be a general question, but a specific procedure. The advantage in swapping can be found from strategy alone, and needs no probability at all. I think it is about information and reaction to unfamiliar situations. Much as Mark6 wrote.
http://www.internationalskeptics.com/forums/showpost.php?p=7063207&postcount=51

I did not say that other possibilities were not worthy of exploration.

But, I think that the puzzle, as presented for public consumption, has only one solution that is remotely interesting. As Roger wrote;

"You don't get much traction on the "where do they bury the survivors" plane crash puzzle by arguing that perhaps in this county they do bury live people."

Nor, the prize being revealed, nor all doors opened etc...

I don't really follow you. Are you saying, it wouldn't be good TV if Monty opened his chosen door and revealed the car, so that wouldn't happen and the possibility can be discounted? I dispute that strongly. The possibility that Monty might do exactly that is what would give suspense and drama to that door-opening moment. And Monty reveals the car! Too bad, thanks for playing!

It's the same moment as someone getting kicked out of the Big Brother house. Of course it would be good TV. I cannot see any a priori reason for discounting the possibility that a random choice between the two remaining doors is exactly the game Monty might be playing.

Rolfe.

 

[Rolfe]

Tim Mann mann@pa.dec.com wrote to point out that Marilyn's analysis does not require that the game show host offer every contestant the opportunity to switch. All that is required is that the host's decision about whether to offer the contestant an opportunity to switch is independent of whether the contestant's initial choice is correct. For example, the host could make the decision whether to offer the switch before the contestant chooses a door, or the host could make the decision based on the flip of a coin, the day of the week, or some other variable independent of the contestant's initial choice.

I was thinking further about this, and specifically about the mechanism for denying the contestant the opportunity to switch.

The host could simply open the door the contestant had already chosen, and go with that choice - good or bad. Or he could open the door concealing the car, assuming that wasn't the one already chosen, and say too bad, thanks for playing. Both of these moves deny the opportunity to switch, and so may be regarded as identical. If Monty's mental coin-flip says "this time, no choice", he can choose either means of terminating the game on the initial choice.

If Monty's mental coin-flip says "this time, he gets the chance to switch", then he has to open the (or an) unchosen door, to reveal a goat. His decision to do this is random, unrelated to whether the contestant has already chosen the right door.

How do the probabilities work out in that scenario? That's the part I haven't got my brain round yet.

Rolfe.

 

[Vorpal]

So I'm confused by the paper Humber linked to, which says this.

I can only assume I'm misunderstanding this, as the author seems to be some sort of expert.

Rolfe.

It says "when he has a choice," implying he knows (in that case he would have only one poss if the player picked a goat). So that's not the case of a clueless Monty that forgot or otherwise does not know. If he was clueless, he'd always have a choice.

 

[humber]

The wave example suggests she is brain-dead. I assume this is not the case, though.

Fair enough, but that doesn't make the original players wrong in fact, merely careless in formulating the presentation.

It does in the case of Savant's version, and her responses to it.

This is the way Marilyn first explained it: "Suppose there were 100 doors and Monty opened 98 of them. You'd switch pretty fast then, wouldn't you?"

Really?

You are on a game show with three doors. A car is behind one; goats are behind the others. You pick door No. 1. Suddenly, a worried look flashes across the host’s usually smiling face. He forgot which door hides the car! So he says a little prayer and opens No. 3. Much to his relief, a goat is revealed. He asks, “Do you want door No. 2?” Is it to your advantage to switch?
—W.R. Neuman, Ann Arbor, Mich.

Nope. If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch.


Obviously wrong.

I don't really follow you. Are you saying, it wouldn't be good TV if Monty opened his chosen door and revealed the car, so that wouldn't happen and the possibility can be discounted? I dispute that strongly. The possibility that Monty might do exactly that is what would give suspense and drama to that door-opening moment. And Monty reveals the car! Too bad, thanks for playing!

Perhaps takes an E-Prime view of the words "host" and "contestant"
You cited an example of a machine that opened the door without all the drama. What did it do? I expect if would open the remaining door that has the goat behind it.

It's the same moment as someone getting kicked out of the Big Brother house. Of course it would be good TV. I cannot see any a priori reason for discounting the possibility that a random choice between the two remaining doors is exactly the game Monty might be playing.

In Russia, door picks you?

 

[Rolfe]

It says "when he has a choice," implying he knows (in that case he would have only one poss if the player picked a goat). So that's not the case of a clueless Monty that forgot or otherwise does not know. If he was clueless, he'd always have a choice.

I still don't see it in context, because I don't understand what the author means by "when he has a choice". Does he mean that he mustn't reveal the car, so has no choice when the contestant has chosen a goat first time? If so, that's so trivial it's hardly worth saying.

Here's the proof he offers for that proposition.

Prop. 3: If all doors are equally likely to hide the car then by independence of the initial choice of the player and the location of the car, the probability that the initial choice is correct is 1/3. Hence the unconditional probability that switching gives the car is 2/3. If the player's initial choice is uniform and the two probability distributions involved in the host's choices are uniform, the problem is symmetric with respect to the numbering of the three doors. Hence the conditional probabilities we are after in Proposition 3 are all the same, hence by the law of total probability equal to the unconditional probability that switching gives the car, 2/3.

I think maybe you're right. This isn't very impressive stuff.

Rolfe.

 

[marplots]

Interesting question. The answer is that it doesn't matter. The probability is split 50:50 between the door you picked and the other unopened door.

(This was not my first guess, but then I worked it out with pencil and paper.)

ETA: Simulation confirms the 50:50 answer.

I agree.

Then I said, "If your odds are reduced with this procedure instead of the regular one, isn't it counter-intuitive that him showing you your choice was wrong sometimes actually hurts your chances? "

I don't understand this question.

Well, when Monty isn't overtly helping (the regular game) you have a 2/3 chance if you switch. When Monty peeks as I described, he is going to re-mix because your choice was not the car. It seems, on first blush, that you are getting a bye -- and you are. But you are also giving up the ability to switch, which you would have done. So, Monty's help actually hurts you.

 

[Rolfe]

It does in the case of Savant's version, and her responses to it.

This is the way Marilyn first explained it: "Suppose there were 100 doors and Monty opened 98 of them. You'd switch pretty fast then, wouldn't you?"

Really?

Yes, really. Actually, that was the way of thinking it out I came to myself, independently. That example demonstrates whether or not the host is deliberately avoiding the prize.

In that example, there is a 1 in 100 chance you picked right to start with, a 1 in 100 chance he is opening doors at random and simply missed the prize every time by chance, and a 98 in 100 chance he is deliberately avoiding the prize.

Common sense says you switch, obviously.

You are on a game show with three doors. A car is behind one; goats are behind the others. You pick door No. 1. Suddenly, a worried look flashes across the host’s usually smiling face. He forgot which door hides the car! So he says a little prayer and opens No. 3. Much to his relief, a goat is revealed. He asks, “Do you want door No. 2?” Is it to your advantage to switch?
—W.R. Neuman, Ann Arbor, Mich.

Nope. If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch.

Obviously wrong.

It's not obvious to me that this is wrong. Quite the contrary.

Imagine that happened on multiple iterations. It's the same as I outlined above.

• One third of the iterations, you were right first time, so switching was a bad move.
• One third of the iterations, Monty reveals the car and says, goodbye and thanks for playing.
• One third of the iterations, you were wrong first time, Monty reveals the other goat, and switching gets you the prize.

If he's forgotten and is guessing, then it's the same as if he's selecting either unchosen door at random and not deliberately avoiding the prize. One third of the iterations reveals the car. Of the remaining iterations, where the contestant does get the chance to switch, it's 50/50.

You cited an example of a machine that opened the door without all the drama. What did it do? I expect if would open the remaining door that has the goat behind it.

It does what you programme it to do. You can programme it to open a remaining door with a goat behind it if you like. If you do that, you will find that switching doubles your chances of winning. As we all know.

Or you can programme it to open either of the two remaining doors at random. In that version, one third of the time, the car is revealed. Sorry, you lost. Good game! If that's the version you're playing, then switching (if you haven't already lost, which we know you haven't because that's the scenario we're examining) confers no advantage.

The point about Monty's intentions is in effect to demonstrate which version of the game you're playing. If it's the former, switching is an advantage. If it's the latter, switching is neutral. So if you don't know which it is, but you do know it's one or the other, switch anyway, because it can't harm your chance, and might improve it.

It's counter-intuitive to believe that Monty's motivation might alter the probability. But in fact Monty's motivation defines the pool of possible outcomes you are selecting from.

Rolfe.

 

[69dodge]

I was thinking further about this, and specifically about the mechanism for denying the contestant the opportunity to switch.

The host could simply open the door the contestant had already chosen, and go with that choice - good or bad. Or he could open the door concealing the car, assuming that wasn't the one already chosen, and say too bad, thanks for playing. Both of these moves deny the opportunity to switch, and so may be regarded as identical. If Monty's mental coin-flip says "this time, no choice", he can choose either means of terminating the game on the initial choice.

If Monty's mental coin-flip says "this time, he gets the chance to switch", then he has to open the (or an) unchosen door, to reveal a goat. His decision to do this is random, unrelated to whether the contestant has already chosen the right door.

How do the probabilities work out in that scenario? That's the part I haven't got my brain round yet.

The probabilities are the same as when Monty always shows a goat and lets you switch: switching has probability 2/3 of getting the car.

 

[69dodge]

Then I said, "If your odds are reduced with this procedure instead of the regular one, isn't it counter-intuitive that him showing you your choice was wrong sometimes actually hurts your chances? "

Well, when Monty isn't overtly helping (the regular game) you have a 2/3 chance if you switch. When Monty peeks as I described, he is going to re-mix because your choice was not the car. It seems, on first blush, that you are getting a bye -- and you are. But you are also giving up the ability to switch, which you would have done. So, Monty's help actually hurts you.

I see.

But Monty does help in the regular game, by eliminating a goat and allowing you to switch. This help increases your chances of winning from 1/3 to 2/3.

Monty also helps in your new game, by (sometimes) re-mixing when you picked a goat. This help increases your chances of winning from 1/3 to 1/2.

So he helps in both cases, but more in the first than in the second.

 

[JoelKatz]

Monty also helps in your new game, by (sometimes) re-mixing when you picked a goat. This help increases your chances of winning from 1/3 to 1/2.

Yes, you do win 1/2 the time Monty turns over a goat. But that has no effect on your overall chances of winning, which are still 1/3.

If Monty turns over the prize, you lose (whether you switch or not). So you can only win if Monty turns over a goat. Monty turns over a goat 2/3 of the time. As you noted, you win half of those times. So you win 1/2 of 2/3 of the time, or 1/3 of the time.

Rolfe: You and Richard Gill are saying the same thing. Your odds of winning overall are 1/3. Your odds of winning after Monty shows you a goat are 1/2.

 

[69dodge]

I still don't see it in context, because I don't understand what the author means by "when he has a choice". Does he mean that he mustn't reveal the car, so has no choice when the contestant has chosen a goat first time? If so, that's so trivial it's hardly worth saying.

The author means that, when Monty does have a choice because you picked the car, he opens one of the two remaining doors at random, with probability 1/2 each, rather than, say, always opening the lowest numbered available door.

This matters. See my post #44.

 

[Rolfe]

The author means that, when Monty does have a choice because you picked the car, he opens one of the two remaining doors at random, with probability 1/2 each, rather than, say, always opening the lowest numbered available door.

This matters. See my post #44.

Yes, I realise that now. Of course I know that's important. I was paying the author of the paper the compliment of thinking he was saying important things, when actually I don't think he is, he's just dressing banalities up in fancy language.

Rolfe.

 

[Rolfe]

The probabilities are the same as when Monty always shows a goat and lets you switch: switching has probability 2/3 of getting the car.

I think I need to go to bed and work that one out in the morning. If that's correct, it's a much more elegant way to formulate the original conundrum, and I think even less intuitive to solve for good measure.

Care to show your working for me?

Rolfe.

 

[Rolfe]

The probabilities are the same as when Monty always shows a goat and lets you switch: switching has probability 2/3 of getting the car.

You know, I think you're right. (Sorry, very late here.)

I like that formulation of the problem a lot. It eliminates the ambiguity, without in so doing drawing you a route map of what the answer is going to be.

Rolfe.

 

[JoelKatz]

The author means that, when Monty does have a choice because you picked the car, he opens one of the two remaining doors at random, with probability 1/2 each, rather than, say, always opening the lowest numbered available door.

This matters. See my post #44.

No, it doesn't matter. You're confusing conditional probability with overall probability. It does affect an intermediate conditional probability, but it has no effect on the overall probability.

Since you picked the car, if you switch you lose, if you don't, you win. It makes no difference how Monty shows you a goat.

Monty is like Fred in this case. Fred knows whether you're going to win or lose, and right before we open your door, he flips a coin. If it comes up heads, he tells you truth about whether you're going to win or lose. If it comes up tails, he lies to you. You win half the time Fred tells you that you're going to win.

We can calculate all kinds of complex probability surrounding Fred. What is the probability he'll tell you that you'll win? If Fred says you'll win, what are the odds Monty showed a goat? And so on. But Fred has no effect on your odds of winning.

 

[Rolfe]

I think you're confused about what we're talking about. We're talking about the paper Humber linked to, which seems not to be nearly as clever as its author thinks it is. I thought at first that proposition 3 was referring to either of the remaining doors being opened at random, irrespective of whether one concealed the prize, when in fact the author was merely confirming that if the contestant had already chosen the prize, the host could open whichever of the remaining doors he liked. Which hardly seems to merit a "proposition" all of its own I have to say.

Rolfe.

 

[69dodge]

Yes, you do win 1/2 the time Monty turns over a goat. But that has no effect on your overall chances of winning, which are still 1/3.

In marplots's version of the game, Monty always (eventually) turns over a goat.

If Monty turns over the prize, you lose (whether you switch or not).

No. If Monty sees the prize, he remixes everything and starts over, so you haven't won or lost yet.

 

[69dodge]

No, it doesn't matter. You're confusing conditional probability with overall probability.

I don't think I am.

I'm saying that if a problem asks

You pick door 1. Monty opens door 2 and reveals a goat. What the probability that switching will get you the car?​

the answer should be the conditional probability, given that Monty opened door 2. Because Monty did open door 2 in the situation presented.

I'm also saying what that conditional probability is.

 

[69dodge]

Fred knows whether you're going to win or lose, and right before we open your door, he flips a coin. If it comes up heads, he tells you truth about whether you're going to win or lose. If it comes up tails, he lies to you. You win half the time Fred tells you that you're going to win.

I'm not sure what conditions you're thinking of here.

The probability of winning given that Fred says you'll win is the same as the unconditional probability of winning. If one is 1/2, the other one is too; if not, not.