# 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. - Mathematics

Sum

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).

Concept: Probability Examples and Solutions
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