Re: Monty at it again
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First , let me make sure we're talking about the same question . I'm assuming now that the question is equivalent to picking numbers 1-10 and that only card 1 is the winner but a player cannot win twice . If by chance a player picks #1 twice , or possibly 3 times , then there is a re-deal and the players choose new cards .
The probability Albert wins one prize is, by the inclusion/exclusion principle : 1/10+1/10 -1/100 . Or , 1/10*9/10 +9/10*1/10 +1/10*1/10 =0.19
The probability Dennis wins one prize is : 3*1/10 - 3c2*(1/10*1/10) +3c3*1/10*1/10*1/10 = 0.271
Make sure you understand how I got these numbers before we move on .
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Are you assuming that if the second draw is invalid, it invalidates the first draw as well?
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