Q. 255.0( 4 Votes )

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.

Answer :

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


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