Thread: Rolling a die
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Old November 2nd, 2009, 07:41 AM
Aquafina Aquafina is offline
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Quote:
Originally Posted by Robb View Post
This question is similiar to a thread I created before... see the second post, change homer and marge to A and B, and the probability from \frac{2}{6} of winning to\frac{1}{6}

So assuming that A goes first;


P(A)=\frac{1}{6} \cdot \frac{1}{1-\frac{25}{36} } =\frac{1}{6}\cdot \frac{36}{11}=\frac{6}{11}
How do you get a geometric sum? The variable n is not defined, so Homer/A could win the game in the first game or the 301st game.. etc?
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