# A and B take turns in throwing two dice, the first to throw 9 being awarded the prize. Show that their chance of winning are in the ration 9:8.

Given that A and B throws two dice.

The first who throw 9 awarded a prize.

The possibilities of getting 9 on throwing two dice are:

P(S9) = P(getting sum 9)

P(SN) = P(not getting sum 9)

Let us assume A starts the game, A wins the game only when he gets 9 while throwing dice in 1st,3rd,5th,…… times

Here the probability of getting sum 9 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(Awins) = P(S9) + P(SN)P(SN)P(S9) + P(SN)P(SN)P(SN)P(SN)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(Bwins) = 1-P(Awins)

P(Awins):P(Bwins) = 9:8

Thus proved

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses