Question

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 ratio12 : 11.

Answer 1

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:

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   s_∞ = \(\cfrac{a}{1-r}\).

⇒ P(B_wins ) = 1 - P(A_wins )

⇒ P(A_wins ) : P(B_wins ) = 12 : 11

∴ Thus proved