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Consider a single player game with a nugget hidden under one of the with 3 upside-down cups and suppose (without loss of generality) that the player chooses cup 1. Now the game maker lifts cup 2 and shows the player that it is empty and gives him the option to either change his earlier choice or stick to it.
We want to know if the player can gain from switching his choice to cup 3. So we have to compare the probability of choosing the nugget by
staying with cup 1, and the probability of choosing the nugget by switching to cup 3.
Let E1 be the event “nugget is under is cup 1,” let E2 be the event “nugget is under cup 3,” and let E3 be the event “Game maker lifts cup 2 and its empty.”
Since from the player’s initial reasoning, the nugget is equally likely to be under any of the three doors, Pr(E1) = Pr(E2) = 1/3
We want to know Pr(E1|E3) and Pr(E2|E3).
According to bayes rule:
Pr(E1|E3) = Pr(E3|E1).Pr(E1)/ Pr(E3) = 1/3 × Pr(E3|E1) / Pr(E3)
Pr(E2|E3) = Pr(E3|E2).Pr(E2)/ Pr(E3) = 1/3 × Pr(E3|E2) / Pr(E3)
First, suppose Game maker randomly picks one of the remaining cups. The probability that he picks cup 2 is then 1/2. The probability that there’s a no nugget under any given cup is 1 − 1/3 = 2/3, so the probability that game maker picks cup 2 and it is empty is Pr(E3) = 1/2× 2/3 = 1/3
because the two events are independent by our assumption that game maker picks randomly.
We now need to determine Pr(E3|E1) and Pr(E3|E2). If the nugget is under cup 1, then cup 2 certainly is empty, and hence the probability that game maker picks cup2 and cup 2 is empty, is equal the probability that game maker picks cup2 : Pr(E3|E1) = 1/2.
Similarly, if the nugget is under cup 3, then also cup 2 is empty for sure, yielding Pr(E3|E2) = 1/2.
Putting all this together yields:
Pr(E1|E3) =1/3 ×1/2 1/3=1/2
Pr(E2|E3) = 1/3 × 1/2 1/3 = 1/2
Hence, the player cannot gain from switching. This is not surprising: if game maker makes a random choice, his action reveals nothing.
we realize that there is another case we should deal with and which is an example of strategic thinking.
Suppose now that game maker would never pick a cup with a nugget under it (much more likely in a real monty hall show). As before, Pr(E3 | E1) = 1/2 because if the nugget is under cup 1, then both cup 2 and 3 certainly are empty, and so game maker will pick randomly between them. However, Pr(E3|E2) = 1 because if the nugget is under cup 3, then cup 2 certainly is empty, but sincegame maker never reveals the nugget (and cannot pick cup 1 because the player already picked it), he will certainly pick cup 2. Finally, because game maker will never pick a cup with nugget under it, Pr(E3) = 1/2.
You can use the total probability theorem to obtain this number.
Let “2” denote the event “the nugget is under cup 2.” Then:
Pr(E3) = Pr(E3|E1)Pr(E1) + Pr(E3|2) Pr(2) + Pr(E3|E2) Pr(E2) = 1/2×1/3 + 0×1/3 + 1 × 1/3 = 1/6 + 1/3 = 1/2.
We used the fact that the probability of the event “Game maker picks cup 2 and it is empty” is zero if there is a nugget under that cup: Pr(E3|2) = 0.
This now yields:
Pr(E1|E3) = (1/2×1/3)/( 1/2) = 1/3
Pr(E2|E3) = (1×1/3 )/(1/2) = 2/3.
Since Pr(E2|E3) > Pr(E1|E3), the player should definitely switch.
The reason is that game maker’s informed choice is a public announcement that reveals additional information which the player should incorporate in his beliefs.
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