Given that A,B and C throws a die.

The first who throw 6 wins the game.

⇒ P(S₆) = P(getting 6)

⇒

⇒ P(S_N) = P(not getting 6)

⇒

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Let us assume A starts the game, A wins the game only when he gets 6 while throwing dice in 1^st,4^th,7^th,…… times

Here the probability of getting sum 6 on throwing a dice is same for the players A, B and C

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(A_wins) = P(S₉) + P(S_N)P(S_N)P(S_N)P(S₉) + P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S₉) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

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B wins the game only when he gets 6 while throwing dice in 2^nd,5^th,8^th,…… times and others doesn’t get 6.

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(B_wins) = (P(S_N)P(S₉)) + (P(S_N)P(S_N)P(S_N)P(S_N)P(S₉)) + (P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S₉)) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

⇒ P(C_wins) = 1-P(A_wins)-P(B_wins)

⇒

⇒

∴ The probabilities of winning of A, B and C is .