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I was reading the Wikipedia page for the solution to Monty Hall problem using Bayes' theorem. I am very confused about this part

P(H3|X1) = 1/2 because this expression only depends on X1, not on any Ci. So, in this particular expression, the choosing of the host does not depend on where the car is, and there's only two remaining doors once X1 is chosen (for instance, P(H1|X1) = 0); and P(Ci,Xi) = P(Ci)P(Xi) because Ci and Xi are independent events (the player does not know where the car is in order to make a choice).

I still don't understand why P(H3|X1) = 1/2. I understand P(H3|X1) as the probability that the host will open door 3 given that the player chooses door 1 (and the car is inside door 1). Does the probability of 1/2 indicate that the host will either open door 2 or door 3? But I thought H3 is the event the host will open door 3? What am I missing?

John Davies
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    that part of the wiki article is extremely poorly written, you might do better to look for a different source. It seems like "$Hi$" is the event the host opens door $i$, $Ci$ is the event the car is behind door $i$, and $Xi$ is the event the player chooses door $i$ (though it's not clear to me what "C" at the end of the first sentence of the Direct Calculation paragraph means). – Matthew Towers Jan 19 '23 at 06:55
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    "does the probability of 1/2 indicate that the host will either open door 2 or door 3?" is not really answerable. It's true that given X1 occurs, the host will open 2 or 3, but the probability $P(H3|X1)$ is the probability the host opens 3 given that the player chooses 1. You are correct in the second-to-last sentence that H3 is the event the host opens 3. – Matthew Towers Jan 19 '23 at 06:58
  • @MatthewTowers I am not sure about other sources that use this similar calculation method. So how do interpret 1/2 then? The host will either open door 3 or leave all doors closed? – John Davies Jan 19 '23 at 07:08
  • In the original problem, a random door is opened , if there is a choice at all which wrong door is opened. This probability is assumed to be 1/2 for both possible choices. But a door is opened always ! – Peter Jan 19 '23 at 07:20
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    The "direct calculation" depends on an additional assumption that is not listed in the "standard assumptions" earlier on the page. Namely, it is assumed that if the other assumptions allow the host to open either of two doors, the host will not have any tendency to open one door more likely than the other. – David K Jan 21 '23 at 14:21
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    Various good explanations here: https://math.stackexchange.com/q/96826/42969 – Martin R Jan 23 '23 at 05:38

8 Answers8

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Suppose the player chooses Door 1, and the prize is indeed behind Door 1. Which of the other two doors will Monty Hall open? Each has a probability of 1/2.

David G. Stork
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Using

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we can construct three tables by collapsing/conditioning our model:

$$P(C_1) = 1$$

enter image description here

$$P(C_2) = 1$$

enter image description here

$$P(C_3) = 1$$

enter image description here

If we restore the sample space, we have the probabilities for all the entries $P(H|X)$ by scaling them with $P(C_1) = P(C_2) = P(C_3) = \frac{1}{3}$.

We have

$P(H_3|X_1) = P(H_3|X_1,C_1) + P(H_3|X_1,C_2) + P(H_3|X_1,C_3) = $
$\quad \quad \quad \quad \quad \quad 0.5 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$

We are now in business and can crank out the solution to the following problem:

What is the probability the contestant wins the car if he initially selects door number $1$ but then switches when the host opens door $3$ (with a donkey)?

${\displaystyle {\begin{aligned}P(C2|H3,X1)&={\frac {P(C2,H3,X1)}{P(H3,X1)}}={\frac {P(H3|C2,X1)P(C2,X1)}{P(H3,X1)}}={\frac {P(C2)P(X1)}{P(H3|X1)P(X1)}}={\frac {1/3}{1/2}}={\frac {2}{3}}\end{aligned}}}$

CopyPasteIt
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  • But why are the values shown as 0.5, 0.5? That seems to be the crux of OP's question. – Alex K Jan 23 '23 at 17:37
  • @AlexK If , say, the car is behind Door 1, i.e. $P(C_1) = 1$ ( see Table 1), and the contestant starts by selecting Door 1, then the host will randomly pick one of the wrong doors, i.e. $P(H_2 | X_1, C_1) = P(H_3 | X_1, C_1) = 0.5$. And the best tactic is for the contestant to 'move away' from what was actually the correct starting choice. The audience would gasp and say what a fool, but the contestant knows Bayes' rule and has no regrets. – CopyPasteIt Jan 23 '23 at 18:09
  • "then the host will randomly pick one of the wrong doors" I think for a question about a Bayesian approach, it's worth explaining why "randomly" matters – Alex K Jan 23 '23 at 21:48
  • @AlexK If the host has any bias in picking doors and it is modelled above with probabilities, the contestant can improve his chances of getting the car. But by (intuitive) symmetry arguments alone, the modelling assumption $P(H_2 | X_1, C_1) = P(H_3 | X_1, C_1) = 0.5$ is the way to go. – CopyPasteIt Jan 23 '23 at 22:05
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There are already a few answers, but none of them really stressed the point (in my opinion) which made me understand the Monty Hall Problem.

Once you chose a door, the doors you are offered to change to are never chosen at random since you are offered to change to another door after you made your initial decision. This seems like a minor detail, but this fact makes new information available, which results in a higher probability of winning if the doors are changed. Bayes' Theorem just formalizes this fact.

If the host selected a door prior to your choice, a potential offer to change the doors wouldn't alter the probabilities of winning. Hence the fact that the host offers you a new door is only relevant if this decision is made a posteriori; only in this case new information becomes available which can be used to increase the probability of winning.

lmaosome
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Let's start with very minimal assumptions -- nothing about how the host chooses a door:$$ P(C1)=P(C2)=P(C3)=\tfrac{1}{3}\\ P(C1,C2)=P(C1,C3)=P(C2,C3)=0\\ P(Xi|Cj)=P(Xi) \forall i, j\\ P(Hi|Xi)=0 \\ P(Hi|Ci)= 0 $$

Given the situation in the article, we want $P(C1|X1, H3)$ and $P(C2|X1, H3)$.

$$P(C1|X1, H3)=\frac{P(H3, X1|C1) P(C1) }{P(X1, H3)}\\ P(H3, X1|C1)=P(H3|X1,C1)P(X1|C1)\\ P(X1, H3)=P(H3|X1)P(X1)\\ P(C1|X1, H3)=\frac{P(H3|X1,C1)P(X1|C1)P(C1)}{P(H3|X1)P(X1)} $$

The same formulas apply for $C2$. By our initial assumptions, we get some cancellation: $$ P(C1|X1, H3)=\frac{P(H3|X1,C1)\tfrac{1}{3}}{P(H3|X1)}\\ P(C2|X1, H3)=\frac{P(H3|X1,C2)\tfrac{1}{3}}{P(H3|X1)} $$ $P(H3|X1,C2)=1$ (host can't open the door with the car behind it), so:$$ P(C2|X1, H3)=\tfrac{1}{3}\frac{1}{P(H3|X1)} $$

Since $C1$, $C2$, and $C3$ are mutually exclusive and sum to 1, $P(H3|X1)=P(H3|X1,C1)P(C1)+P(H3|X1,C2)P(C2)+P(H3|X1,C3)P(C3)$

$P(H3|X1,C2)=1$ and $P(H3|X1,C3)=0$ so $P(H3|X1)=P(H3|X1,C1)P(C1)+P(C2)=\tfrac{1}{3}(P(H3|X1,C1)+1)$

If $P(H3|X1,C1)=\tfrac{1}{2}$, then $P(H3|X1)=\tfrac{1}{2}$ -- but only in that case. If the host hates door 3, and never chooses it unless they have to, $P(H3|X1,C1)=0$ and $P(H3|X1)=\tfrac{1}{3}$.

This makes for:$$ P(C2|X1, H3)=\frac{1}{(P(H3|X1,C1)+1)}\\ P(C1|X1, H3)=\tfrac{1}{3}\frac{P(H3|X1,C1)}{P(H3|X1)}=\frac{P(H3|X1,C1)}{(P(H3|X1,C1)+1)} $$

This looks like a pretty clear dependence on the host's strategy -- why?

Let's look at it from another angle: if the host and player were in cahoots, but couldn't communicate except through the game (but could coordinate in advance) -- could the host give the player a hint, through their choice of door?

Suppose the host always opens the lowest numbered door they can. In that case, we have:

C1 C2 C3
X1 H2 H3 H2
X2 H3 H1 H1
X3 H2 H1 H1

If player picks 1, and the host responds 2 (X1, H3), the player would know that the car is behind 2 (C2) for sure.

Knowing nothing about the host's strategy, the player should use $1/2$ -- but that's not entirely satisfying -- because it seems like the host can influence the player's chances!

We can take a step back in the game, and look at the odds of the player winning, from the host's perspective. Call their final choice $Y$; take $X1$ as a given.

$$ P(Y1, X1, C1)+P(Y2, X1, C2)+P(Y3, X1, C3)=P(Y1, H2, X1, C1)+P(Y1, H3, X1, C1)+P(Y2, H3, X1, C2)+P(Y3, H2, X1, C3)=P(Y1| H2, X1, C1)P(H2, X1, C1)+P(Y1|H3, X1, C1)P(H3, X1, C1)+P(Y2|H3, X1, C2)P(H3, X1, C2)+P(Y3| H2, X1, C3)P(H2, X1, C3)\\ =P(Y1| H2, X1, C1)P(H2|X1, C1)P(X1,C1)+P(Y1|H3, X1, C1)P(H3| X1, C1)P(X1,C1)+P(Y2|H3, X1, C2)P(H3| X1, C2)P(X1,C2)+P(Y3| H2, X1, C3)P(H2| X1, C3)P(X1,C3) $$ Using $P(X1)=1$ and simplifying the cases where the host has no choice:$$ P(Y1, X1, C1)+P(Y2, X1, C2)+P(Y3, X1, C3)=P(Y1| H2, X1, C1)P(H2|X1, C1)\tfrac{1}{3}+P(Y1|H3, X1, C1)P(H3| X1, C1)\tfrac{1}{3}+P(Y2|H3, X1, C2)\tfrac{1}{3}+P(Y3| H2, X1, C3)\tfrac{1}{3} $$ $P(Y1| H2, X1, C1)$ and $P(Y1|H3, X1, C1)$ are both "keep the same door" for the player. Say the player does that with probability $x$. Then the other two terms are "remaining door," chosen with probability $1-x$.

Then: $$ P(Y1, X1, C1)+P(Y2, X1, C2)+P(Y3, X1, C3)=\tfrac{1}{3}(x (P(H2|X1, C1)+P(H3| X1, C1))+2(1-x)) $$

$P(H2|X1, C1)+P(H3| X1, C1)=1$ because the host has to choose one of H2 or H3. This leaves $\tfrac{1}{3}(2-x)$ as the odds of winning. This is maximized by $x=0$, regardless of how the host chooses a door.

David K
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Alex K
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Use Bayes' theorem on it.

$$P(H_3|X_1) = \frac{P(X_1|H_3)P(H_3)}{P(X_1)}$$

$$= \frac{\frac12\frac13}{\frac13}$$

$$= \frac12$$

JMP
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Here is what turned the lightbulb on inside my head: Monte Hall is NOT going to open a door at random, all with equal probability. Instead, he's going to pick one of the doors you did not choose. Among those two, he will strategically take the one that's empty. Even if the probabilities were equal ex-ante, the subsequent non-random behavior of Hall makes the posterior probabilities unequal.

ABC
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$P(H3|C1,X1)$ is the probability that the host opens door number three GIVEN that the guest picked door one (and the car was behind door one). The host never opens the door in which the car lies—otherwise it wouldn't be a very interesting game show—and never opens the door the guest picks.

Hence, since the door that the guest picks in this situation (door one) is the door to the car, the host eliminates only that door as an option, thus giving them two doors to choose from: Doors two and three. Assuming that the host has no inherent bias towards picking one number over the other, they would be expected to pick door three half of the time.

Arjun Vyavaharkar
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$P(H_3|X_1)$ does not have to be one half. The important thing about the Monty Hall problem is that the host knows which door holds the prize. If the contestant guesses wrong, which will happen two thirds of the time, Monty is left with only one door to open.

When the contestant guesses correctly, it doesn't Matter which door Monty opens. He could choose randomly, in which case $P(H_3|X_1)=\frac{1}{2}$ or he may have a policy that says he will choose the unselected door with the highest number in which case $P(H_3|X_1)=1$.

Worrying about how Monty chooses which door to open in case of a correct guess does not change the problem. The reason we have a two thirds chance of winning by adopting the switching strategy is because we win when we were originally wrong. Since we are originally wrong two thirds of the time we win two thirds of the time by switching.

To see this more clearly, imagine that the contestant chooses from ten doors. The contestant guesses door one and Monty opens doors three though ten. In this case there is only a one tenth chance of guessing correctly. Either way, Monty will choose all doors but one. If the contestant guesses wrong, he will open all doors except the one with the prize. If the contestant guesses correctly, he will be faced with the same problem as in the game with three doors. Once again, his strategy for choosing the open doors is irrelevant. The contestant will still win nine tenths of the time if he switches.

John Douma
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  • This is incorrect, as we're not specifying how Monty chooses a door (this truly doesn't matter); rather, we're specifying the player's uncertainty/knowledge of how Monty chooses the doors. Then $P(H_3|X_1,C_1) = 1$ can only be interpreted, in this context, as the player having insider information on how the selection of doors is made. In fact, if we have such insider information, then $P(C_1|X_1,H_3) = 1/2$ (and yes, this analysis even extends to the many-more-doors example) – Brian Moehring Jan 22 '23 at 06:51
  • @BrianMoehring How do we know that Monty chooses randomly? The point of my answer is that it doesn't matter. If you originally guess right, the remaining door will not have the prize and if you originally guess wrong, the remaining door will have the prize. We cannot say that $P(H_3|X_1, C_1)=\frac{1}{2}$ because that assumes Monty chooses randomly. He may have a system for choosing. Consider the case with ten doors where he must choose one of nine doors to leave open. He may choose the closest prime number door to the price. We cannot say the probability is $\frac{1}{9}$. – John Douma Jan 22 '23 at 06:57
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    Again, $P(H_3|X_1,C_1) = 1/2$ doesn't mean Monty chooses randomly. It means that the player has no knowledge of how Monty chooses the doors. How Monty chooses the doors doesn't matter whatsoever as long as the choice appears random to the player. – Brian Moehring Jan 22 '23 at 07:04
  • @BrianMoehring Even without the player knowing, we could run an experiment where a judge keeps track of the player selections, the locations of the prize and Monty's selections. If Monty doesn't choose randomly it may not turn out that the selection of remaining doors is uniformly distributed. That's really my point. That and the fact that computing that probability is irrelevant to the outcome of the problem. In fact we can show that, given any probability $p$ for picking a door to leave closed after the contestant has guessed correctly, the outcomes of winning and losing are the same. – John Douma Jan 22 '23 at 07:16
  • If $P(H_3|X_1,C_1) = p$, then $$\begin{align}P(C_1|X_1,H_3) &= \frac{P(H_3,X_1,C_1)}{P(H_3,X_1,C_1) + P(H_3,X_1,C_2)} \ &= \frac{pP(X_1,C_1)}{pP(X_1,C_1) + P(X_1,C_2)} \ &= \frac{p/9}{p/9+1/9} \ &= \frac{p}{p+1}\end{align}$$ Are you saying that this doesn't depend on $p$? (that's the only interpretation I can give your most recent comment) Note in particular that for $p=1$, we have the probability that we initially chose correctly is $1/2$ while if $p=1/2$ then we get the expected probability of $1/3$. – Brian Moehring Jan 22 '23 at 07:25
  • @BrianMoehring No. I am saying that the outcome of winning doesn't depend upon how Monty chooses which door to leave closed when the contestant guesses correctly. The outcome depends upon the probability of guessing correctly and the strategy, i.e. stay or switch. No matter how Monty chooses the other door, you lose if you guess correctly and switch and you win if you guess correctly and stay. – John Douma Jan 22 '23 at 07:45
  • If replace "the probability of guessing correctly" with "whether the player initially guesses correctly", then I agree, as this just describes the essential parts of the rules of the game. The evaluation of the probability that the initial guess is correct, calculated from the player's perspective, is further determined by the prior probability that the initial choice is correct along with all the information given by which door Monty chooses to open. In the usual statement of the problem, the door doesn't change the prior, which symbolically represents $P(H_3|X_1,C_1) = 1/2$. – Brian Moehring Jan 22 '23 at 08:26
  • I can't stress hard enough that my argument with your answer, since the beginning, has been that $P(H_3|X_1,C_1)$ doesn't represent how Monty chooses the door but rather the player's knowledge/uncertainty of how Monty chooses the door. You have written it as a representation of how Monty chooses the door, but this is nonsense as a filter of sorts that the player applies to their prior. In any case, if my concern is not yet apparent, then it's unlikely to become apparent with more comments. I'll stop. – Brian Moehring Jan 22 '23 at 08:29
  • My answer explains in detail why this conditional probability matters (if the player knows what it is) and doesn't matter (if the player is using one of the two strategies). This answer is incorrect. – Alex K Jan 23 '23 at 00:04
  • @AlexK This answer is absolutely correct. The probability of winning depends upon the number of doors and the strategy. You answer is long and convoluted. The Monty Hall problem boils down to one simple fact: If you guess correctly, the door offered doesn't contain the prize and if you don't, the offered door contains the prize. Monty's method for picking the door in the former case is irrelevant to the final outcome and need not be random. – John Douma Jan 23 '23 at 01:12
  • My answer was thorough, and explicitly shows the dependence on the probability in question. The question was about a Bayesian approach to the problem. That requires consideration of what the player does and doesn't know about the process. If the player knows how Monty picks the door, and it's not random, the player has a better chance of winning. Read just the part with the table, starting with "Suppose the host always opens the lowest numbered door they can. " – Alex K Jan 23 '23 at 01:35
  • @AlexK Monty can use an arbitrary process that is unknown to the player and not random. How Monty selects which door to offer when the player guesses correctly is irrelevant to the problem. If there are $n$ doors and you stay, you win $\frac{1}{n}$ times and if you switch you win $\frac{n-1}{n}$ times. The probabilities of interest are the probability that you win given that you stay and the probability that you win given that you switch. Anything else is a distraction. – John Douma Jan 23 '23 at 01:42
  • that is unknown to the player -- exactly right. Perhaps you should update your answer to mention that, because $P(H3|X1)$ i in a Bayesian decision making context is all about what is known to the player! – Alex K Jan 23 '23 at 02:03
  • @AlexK $P(H_3|X_1)$ can be unknown, random and still have a probability that is not equal to $\frac{1}{2}$. Suppose Monty uses the following process. If the player guesses correctly he glances at his watch. If the minute hand is on a whole number he shifts left one space to decide which door to offer and if not, he shifts right. Then $P(H_3|X_1)=\frac{1}{5}$. But that is completely irrelevant because if the player guesses correctly, he will be offered a box with no prize. – John Douma Jan 23 '23 at 02:22
  • "can be unknown" no, it must be either unknown or 1/2 for the answer to be valid. – Alex K Jan 23 '23 at 03:13