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You are playing a game in which there are four doors, behind one of them there is a car and behind the other 3, there are goats. The car is equally likely to be behind any door. The game show host knows where the car is.

You first choose a door. From the other 3 doors, the host opens (with equal likelihood) one door with a goat behind.

You must then make a decision: either you stay with your original door or you change for one of the remaining doors.

(a) suppose your strategy is to switch each time the host opens a door to show a goat, independently of which door the host opens. Whats your probability of winning the car?

(b) Now assume the following has happened:

  • You initially choose door 1
  • Then the host opened door 2
  • You switched to door 3
  • The host opened the door 4

You must now make a decision what's your probability of winning the car if you stay with the door 3? What if you switch back to door 1?


I try this

(b) xn = the car is behind the door n

suppose that the participant chooses door 1, this (and the other three) have the same possibility of being the winning door, that is, p (xn) = 1/4.

in this way, the probability that the car is in one of the doors that was not chosen is 3/4.

In a first event, the host opens door 2 revealing a goat, therefore p (x2) = 0 but the probability that the car is in one of the non-chosen doors must remain in 3/4, and therefore the old value of p (x2) is redistributed between p (x3) and p (x4), thus, the new probabilities are

p (x1) = 1/4 p (x2) = 0 p (x3) = 3/8 p (x4) = 3/8

and the participant changes to door 3.

in a second event, the host opens door 4 revealing a goat, therefore p (x4) = 0, again the old value of p (x4) must be redistributed between the non-chosen doors (only door 1 remains) Finally, the probabilities are

p (x1) = 5/8 p (x2) = 0 p (x3) = 3/8 p (x4) = 0

Then, if the participant decides to stay with door 3, his probability of winning is 3/8 while if he decides to change to door 1 it will be 5/8.

(a)

Note that the case of always changing the door coincides with that of the previous exercise when the participant finally chooses door 1, this because it does not matter which door the host opens in the second event, the probability that the car is in the door that the host leaves closed will always be 5/8 and if the strategy of the participant is always to change the door, in the end, it will be changed to this one.

But my professor told me this is true for (a) but not for (b), and I don't understand why.

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The solution to the Monty Hall problem is critically dependent on the rules of the game. In particular, you must know in advance that the host always opens a door after your first guess, that he never reveals the car at that time, and that he always gives you a second chance.

The wording for (a) kinda-sorta implies these rules in a subtle way, mainly in the use of the words "each time" as if the sequence of events up to that point is a repeating pattern.

I don't see any hint of such an implication for part (b), so I think the problem is ill-defined. We don't know what circumstances caused the host to open door 4. It is just something that "has happened", apparently for the first time ever, and we don't know why.

Perhaps the host is following a new secret rule: Open the one remaining unchosen, unopened door provided that the player switched and that the car is behind the player's first-choice door, but if the player switched to the door with the car, offer them $\$500$ to switch again; if the car is behind the remaining door, just open it so the player loses.

Or perhaps the host's new secret rule is to open the remaining unchosen, unopened door if the player switched and the car is behind the player's second-choice door, otherwise (if the player switched) just reveal where the car is, so the player loses.

We can assume the host's new rule is not "if the player switches, always open a door not already open or chosen, but don't open the door with the car," because in some games the car will be behind the only door not already open or chosen and it will be impossible to follow the rules.

But let's suppose that the new rule is, "After the player either stays or switches, open another door, but not the player's latest choice and not the door with the car. If there are two options, pick either with equal probability." And let's consider just the games in which the first three moves are you choose door 1, the host opens door 2, and you switch to door 3: "choose-1-open-2-switch-3."

  • As you have already shown, in $1/4$ of these games the car is behind door 1; and in all the cases where that is true, the host is forced to open door 4. This is the "car-1-open-4" variation of the choose-1-open-2-switch-3 games.
  • As you have already shown, in $3/8$ of the choose-1-open-2-switch-3 games the car is behind door 3. When it is, the host has a choice between doors 1 and 4, so $3/16$ of the choose-1-open-2-switch-3 games belong to the "car-3-open-1" variation and $3/16$ to the "car-3-open-4" variation.
  • As you have already shown, in the remaining $3/8$ of the choose-1-open-2-switch-3 games the car is behind door 4, and the host is forced to open door 1. This is the "car-4-open-1" variation.

In summary, in $1/4$ of all games with those first three moves, the host opens door 4 and the car is behind door 1 (car-1-open-4 variation); in $3/16$ of all games with those first three moves, the host opens door 4 and the car is behind door 3 (car-3-open-4 variation). In the other $3/8 + 3/16 = 9/16$ of all such games, including the car-3-open-1 and car-4-open-1 variations, the host opens door 1.

Bayes' Theorem then says that $$ P(\text{car is behind door 1}\mid\text{choose-1-open-2-switch-3-open-4}) = \frac{1/4}{1/4 + 3/16} = \frac47. $$

On the other hand, suppose the new rule is that the host will always open another door, not the player's latest choice and not the door with the car, but if the player has switched then the host will not open the first-choice door either unless forced to. Under those rules, after the first three moves described in part (b), the host will open door 4 in all cases where the car is behind door 3 as well as all cases where the car is behind door 1, and Bayes' Theorem says $$ P(\text{car is behind door 1}\mid\text{choose-1-open-2-switch-3-open-4}) = \frac{1/4}{1/4 + 3/8} = \frac25. $$

One more variation: suppose the new rule is that if the player switched, the host then opens the first-choice door if it has a goat, otherwise he opens the remaining unchosen door. Then the host opens door 4 only in the case where the car is behind door 1, so you are guaranteed to win if you switch.


I think this exercise points out a weakness in one of the usual explanations of the Monty Hall paradox, and (I'm sorry to say) it is an explanation that I have used myself without adequate safeguards. One difference between part (a) and part (b) is that in part (a), there is no a priori reason to think that the car is more likely to be behind one of the unchosen doors (2, 3, or 4) than any other, so the host's choice of door to open tells us nothing about the probability that the car was behind at least one of those doors in the first place. In part (b), however, in the given example the car is more likely to be behind door 4 than door 1 based on the first three moves; assuming the host can choose with equal probability if goats are behind both doors, then, the fact that the host opens door 4 is evidence against the guess that the car is behind either of the unchosen doors.

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  • $\begingroup$ Thanks for taking the time to answer me! Can you explain a little bit more how you calculate this: "So in 1/4 of all games with those first three moves, the host opens door 4 and the car is behind door 1; in 3/16 of all games with those first three moves, the host opens door 4 and the car is behind door 3. (In the other 7/16 of all such games, the host opens door 1.)" $\endgroup$ – Luis GC Mar 5 '18 at 13:51
  • $\begingroup$ "The other 7/16" should have been "the other 9/16." I have fixed that error, although we never actually use that fraction in any of the subsequent reasoning. I've also expanded the answer to explain the 1/4 and 3/16 figures in more detail. $\endgroup$ – David K Mar 5 '18 at 14:42

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