quick 'monty hall problem' question...

Factual Questions
41

But prisoner C still has the same 1/3 chance as he did previously. His odds of execution didn’t increase to 2/3 just because Prisoner A performed his inquiry. That’s the part I can’t understand.

Suppose prisoner C had made the same request to the guard and was also informed that Prisoner B had been given the letter. Then he rightfully concludes that he still has a 1/3 chance of being executed, but that A now has a 2/3 chance.

Someone isn’t right here because the “other” prisoner still has the same 1/3 chance as the original prisoner no matter the disclosure.

42

You have to be careful to distinguish between the probability of the event that C will be executed and the probability that C subjectively assigns to the event that C will be executed. These would have to match up if everyone had the same information, but given that they don’t, the two probabilities can differ.

43

He would be right based on what he knows.

It might help to make a chart of possibilities. I’ll make six cases, equally likely. The column labeled A’s letter lists which prisoner receives A’s letter, and similar for the column labeled C’s letter. If A is guilty, it’s assumed to be equally likely for the guard to give the note to either B or C.



Case  guilty   A's letter C's letter
1       A         B         B
2       A         C         B
3       B         C         A
4       B         C         A
5       C         B         A
6       C         B         B

Prior to finding out who his letter was given to, for both A and C, all six cases were equally likely. After A finds out B received his letter, three cases were eliminated, and in two of the remaining cases, C is guilty.

Similarly, when C finds out B received his letter, three cases are eliminated from C’s point of view, and in two of those, A is guilty.

If C and A could compare what they knew, only cases 1 and 6 would remain, and they would then conclude they each had a 50/50 chance at that point.

44

I think that this simply highlights the paradox. A and C have received information that they already knew from the outset: at least one of the other two prisoners was innocent. By taking this already known information and comparing it, their chances of execution increased from 33% to 50%.

45

I don’t see a paradox. When they compare notes, they know that B is innocent. Look at the chart. There were originally the six, equally-likely possibilities. After comparing notes, there are only two of those six remaining. How is that not new information?

46

Probability is straightforward when approached properly. As ultrafilter points out, probabilities are not absolute but depend on the particulars of the subject’s information (or, equivalently, on his ignorance). Two people with different information will, correctly, have different probability estimates for the same event. Much confusion would go away if this fact were kept in mind.

The prisoner with a 1/3 chance will still have the same 1/3 chance since the guard’s response tells him nothing he didn’t know about his situation. If two prisoners communicate their guard conversations to each other, each has acquired new information.

Similarly, if box-has-car was a 1/3 chance before Monty exposes a goat, the chance is still 1/3 if you know Monty will always expose a goat. (But in fact, Monty Hall has allegedly claimed in interview that he took player’s personality and car location into consideration when making offers.)

47

True. In the Classic Monty Hall problem, the host is explicitly so constrained. The real Monty Hall never was.

The outcome is the same - the contestant saw a goat revealed - but the problem is not the same because new options were added* after the contestant chose*.

In this example, the contestant only has two choices to pick from. The third revealed goat was something else that did not affect either the door chosen nor the door not chosen. The host might as well have said “By the way, I have a banana in my pants.”

The difference is that in the classic problem, the contestant has three options to begin with, and Monty must reveal a goat from the 2 doors that the contestant does not initially pick.

What makes the classic probabilities work the way they do is the constraint that Monty must offer the opportunity to swap doors, and that Monty must know where the car and goats are, and that Monty must reveal a goat from the options the contestant did not originally select. With those three constraints, the host has effectively not revealed anything new, but rather offered the contestant to swap from the first door he picked to the two remaining doors, one of which is worthless*.

If you are saying that 2/3 is not a large enough advantage that it makes the results obvious, then I agree. Changing to 51/52 for a deck of cards or 99/100 for a stack of numbers would make the results more obvious.

But the big issue is displayed that people are not, in fact, seeing the 2/3 increase in probability. They see it as a 50/50 choice. Thus, there is very little incentive to swap from their initial decision. Psychologically, they see they had a 1/3 chance first, now they have a 50/50 shot. Sticking in that situation is psychologically more comforting. If they knew the odds set at 2/3 in their favor if they swapped, that would be significant enough to affect the decision, though in that case you would be correct, it isn’t such a definitive shift to convince everyone.

Especially if you only get one shot. If you’re playing the game over and over, then playing the odds makes sense. If you only get one shot, do you play the odds or play your intuition? Many people will take intuition over odds.

  • The goat was intended as a novelty booby prize that the contestant didn’t actually keep. It was more fun than just an empty door. The contest could be restated that the winning prize is a goat and the losing prize is nothing.
48

This has the right idea, but the explanation is wrong. You need to distinguish between the probability that an event E occurs, which is absolute, and the degree of belief that any person assigns to the statement “E will occur”, which can depend on the information that a person has. In the Bayesian literature the latter is often known as credence.

This is subtle, so let me give an example. First, I need to point out that the analysis of the Three Prisoners problem that was linked relies on the unstated assumption that, in the event that prisoner A is slated for execution, the guard is equally likely to give the letter to prisoner B or prisoner C. If that’s not the case, then A can learn something about his own chances for execution by learning that the guard gave the letter to prisoner B. I was hoping that someone else would find it, as learning to read for unstated assumptions is a vital part of probabilistic modeling.

Suppose that A is indeed slated for execution, and that the guard finds prisoner C repulsive and would never approach him unless he had no choice. Prisoner A gives the letter to the guard, and is told that the guard gave the letter to prisoner B. In this case, based on the information he has, prisoner A should believe that the probability that he will be executed is 1/2. Prisoner B should also believe that the probability that A will be executed is 1/2, as he knows that he will survive. Prisoner C has no information beyond the fact that one of the prisoners will be executed, so he should assign the execution of prisoner A a probability of 1/3. However, the guard knows that A has been chosen, and so he assigns the event that A is executed a probability of one.

The important thing to understand is that each player in this little drama is making the correct inference based on the information that he has. The differences are entirely due to the fact that they all have access to different information.

You have no idea how badly I wish this were true.