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Question
A and B throw a die alternatively till one of them gets a number greater than four and wins the game. If A starts the game, what is the probability of B winning?
Solution
Let S denote the success, i.e. getting a number greater than four and F denote the failure, i.e. getting a number less than four.
`∴ P(S)=2/6=1/3, P(F)=1−1/3=2/3`
Now, B gets the second throw, if A fails in the first throw.
∴ P(B wins in the second throw) =` P(FS) = P(F)P(S) = 2/3xx1/3`
Similarly, P(B wins in the fourth throw) = P(FFFS) = P(F)P(F)P(F)P(S) = ` (2/3)^3xx1/3`
P(B wins in the sixth throw) = P(FFFFFS) = P(F)P(F)P(F)P(F)P(F)P(S) `= (2/3)^5xx1/3` and so on.
Hence,
P(B wins) = `2/3xx1/3 + (2/3)3xx1/3 + (23)5xx1/3 + ...`
`=2/3xx1/3xx[1+(2/3)2+(2/3)4+......`
`= 2/3xx1/3xx(1/(1−4/9)) ` [∵a+ar+ar2+... =a/(1−r)]
`= 2/5`
Thus, the probability that B wins is 2/5.
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