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Dec
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comment Monty hall problem extended.
@DwB you may consider the answer ridiculous, but that doesn't make it wrong. The 50% answer is simply not correct.
Dec
18
comment Monty hall problem extended.
Gerrat is not correct. Another argument to intuit the right answer is this; say you and a friend play this game. You never switch, and your friend always does. You both always pick the same initial door. You have a 1/3 chance of winning, always. Since Monty ALWAYS picks a losing door, and your friend always switches, you now have collectively picked both remaining doors, and Monty has shown you a losing door. So ONE of you is ALWAYS going to win. Since you are always going to win 1/3 of the time, your friend is going to win the other 2/3 of the time.