Re: Monte hall problem
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This monte think is starting to get to me [img]/images/graemlins/smile.gif[/img]
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The way that I think it is best explained is by looking at all the possible outcomes.
Assume the prize is behind door 3
If you never switch the outcomes are:
You pick door 1, the host shows you door 2, you stay with door 1. Outcome: lose
You pick door 2, the host shows you door 1, you stay with door 2. Outcome: lose
You pick door 3, the host shows you another door, you stay with door 3. Outcome: win
If you always switch the outcomes are:
You pick door 1, the host shows you door 2, you switch to door 3. Outcome: win
You pick door 2, the host shows you door 1, you switch to door 3. Outcome: win
You pick door 3, the host shows you another door, you switch to the other door. Outcome: lose
Therefore by not switching you win 1/3 times. By switiching you win 2/3.
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FYP
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FYP.
You pick door 3 one time out of three. Half of those times, you are shown door one, half you are shown door 2. This means that when you always stay, you win [(0/3)+(0/3)+(1/6)+(1/6)]=1/3 of the time. When you always switch, you win [(1/3)+(1/3)+(0/6)+(0/6)]=2/3 of the time. Therefore, you should always switch.
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