A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12:11.

Question

Philoid · Accepted Answer

Given that A and B throws two dice.

The first who throw 10 awarded a prize.

The possibilities of getting 10 on throwing two dice are:

⇒ P(S₁₀) = P(getting sum 10)

⇒

⇒ P(S_N) = P(not getting sum 10)

⇒

Let us assume A starts the game, A wins the game only when he gets 10 while throwing dice in 1^st,3^rd,5^th,…… times

Here the probability of getting sum 10 on throwing a dice is same for both the players A and B

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

⇒ P(A_wins) = P(S₁₀) + P(S_N)P(S_N)P(S₁₀) + P(S_N)P(S_N)P(S_N)P(S_N)P(S₁₀) + ……………

⇒

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(B_wins) = 1-P(A_wins)

⇒

⇒ P(A_wins):P(B_wins) = 12:11

∴ Thus proved