#1
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Simple (hopefully) probability question
Here are the prerequisites:
Hand A stands to win 1/3 of showdowns. Hand B stands to win 2/3 of showdowns. Ties are impossible. Hands A and B do battle precisely 4 times. What is the probability of: A) Hand A defeating Hand B exactly 2 times out of 4. B) Hand A defeating Hand B 2 or more times out of 4. Any comments on how to do this problem are appreciated, as well as answers! |
#2
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Re: Simple (hopefully) probability question
probability a wins 4 =(1/3)^4 =0.01234567
probability a wins 3 =(1/3)^3*(2/3)^1*COMBIN(4,3) =0.098765432 probability a wins 2 =(1/3)^2*(2/3)^2*COMBIN(4,2)=0.296296296 probability a wins 1 =(1/3)^1*(2/3)^3*COMBIN(4,1)=0.39506172 probability a wins 0 =(2/3)^4 = 0.197530864 So the probability of A winning exactly 2 is ~.296 and 2 or more is ~.407 |
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