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Question
A, B and C throw a pair of dice in that order alternatively till one of them gets a total of 9 and wins the game. Find their respective probabilities of winning, if A starts first.
Solution
Let E: getting a total of 9
∴ `"E" = {(3,6),(4,5),(5,4),(6,3)}`
So, clearly 4 1 probability of winning = P(E) = `(4)/(36) = (1)/(9), "P"(bar"E") = (8)/(9)`
As A starts the game so, he may win in 1st, 4th, 7th, ... trials.
∴ `"P"("A wins") = "P"(bar"E") + "P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")+ "P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E") + ...`
⇒ `"P"("A wins") = (1)/(9) + (8/9)^3 xx (1)/(9) (8/9)^6 xx (1)/(9)+...`
∴ `"P"("A wins") = ((1)/(9))/((1- 512)/(729))`
= `(81)/(217)`.
Now B may win in 2nd, 5th, 8th,...trials.
∴ `"P"("A wins") = "P"(bar"E")"P"(bar"E") + "P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")+ "P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")"P"(bar"E")+...`
⇒ `"P"("A wins") = (8)/(9) xx (1)/(9) + (8/9)^4 xx (1)/(9) + (8/9)^7 xx (1)/(9) +...`
∴ `"P"("A wins") = ((8)/(81))/(1 - (821)/(729))`
= `(72)/(217)`
And, finally
`"P"("C wins") = 1-["P"("A wins")+"P"("B wins")]`
= `1 - [(81)/(217 + (72)/(217)]]`
= `(64)/(217)`.
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