Viewing the Monty Hall Dilemma through Bayes' Theorem 📂Mathematical Statistics

Viewing the Monty Hall Dilemma through Bayes' Theorem

Explanation

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As you know, in the Monty Hall game, it is beneficial to change your choice regardless of where the actual prize is located. Apart from whether one accepts this fact or not, there are people who either cannot intuitively understand the Monty Hall game or are not skilled in expressing it mathematically.

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For the sake of argument, assume that you are the player and you have chosen door number 1.

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At this point, since we have no information about the prize, we must assume that the probability of choosing any door is the same. But after making this choice, let’s say the host opens door number 3. The host selects a door to open based on the player’s choice and the actual location of the prize:

If the actual prize was behind door 3, then door 3 cannot be opened, so the probability of door 3 being opened is $0$.
If the actual prize was behind door 2, since the player chose door 1 and door 2 cannot be opened, the probability of door 3 being opened is $1$.
If the actual prize was behind door 1, then the host is free to choose between door 2 or 3, so the probability of door 3 being opened is $1/2$.

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Let’s reconsider that the event of the host opening a door is not a random occurrence. If the host, without any information, had randomly picked a door and had revealed no prize, it would be no different than if the game had started with only two doors from the beginning. The event of opening a door is not a coincidence but a hint that eliminates a definite incorrect option after observing the player’s action. Naturally, the player, being more informed than just making a random choice, is in a position where either keeping the first choice or changing it becomes favorable.

Since the host’s choice ‘updates’ the game, let’s calculate the posterior probability using Bayes’ theorem. Let’s say the probability of the prize being behind door 1 after keeping the initial choice is $p_1$, and the probability of the prize being behind door 2 after changing the choice is $p_2$. $$ p_{1} = {{ 1/3 \cdot 1/2 } \over {1/3 \cdot 1/2 + 1/3 \cdot 1 + 1/3 \cdot 0}} = {{1} \over {3}} $$

$$ p_{2} = {{ 1/3 \cdot 1 } \over {1/3 \cdot 1/2 + 1/3 \cdot 1 + 1/3 \cdot 0}} = {{2} \over {3}} $$ $20190430\_080007.png$ Sometimes, some people prefer specific calculations over ‘simply seeing it as concentrating probability’ as an intuitive explanation. Using Bayes’ theorem allows one to come to the same conclusion ‘without having to think hard’ by just calculating, dividing the cases.