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