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 1^{st}, 4^{th}, 7^{th}, ... 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 2^{nd}, 5^{th}, 8^{th},...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)`.