The Monty Hall Problem: This classic probability puzzle involves a game show where a contestant must choose between three doors, one of which has a prize behind it. After the initial choice, the host (Monty Hall) opens one of the other two doors that doesn't contain the prize. The contestant is then given the option to switch their choice. What is the optimal strategy, and what is the probability of winning if the contestant switches?
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The optimal strategy in the Monty Hall Problem is for the contestant to switch their choice after the host opens one of the other doors.
Initially, when the contestant chooses one door out of three, the probability of choosing the door with the prize is 1/3. This means the probability of not choosing the door with the prize is 2/3.
When the host, Monty Hall, opens one of the remaining two doors to reveal that it does not contain the prize, the probabilities change. The door the contestant originally chose still has a probability of 1/3 of having the prize behind it. However, the other door that wasn't chosen initially (but is still closed) now has a probability of 2/3 of having the prize behind it. This is because Monty Hall deliberately revealed a door without the prize, increasing the likelihood of the remaining unchosen door having the prize.
Therefore, when given the option to switch, the optimal strategy is to switch doors. This maximizes the probability of winning.
If the contestant switches their choice, the probability of winning becomes 2/3.
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