MadeEasy Workbook: Probability

Question

A pair of dice is rolled again and again till a total of 5 or 7 is obtained. The chance that a total of 5 comes before a total of 7 is??

Comments

https://gateoverflow.in/1116/rolled-again-again-total-obtained-chance-total-before-total

http://math.stackexchange.com/questions/537935/a-pair-of-unbiased-dice-are-rolled-together-till-a-sum-of-either-5-or-7-is-obtai

Answer 1

Favourable cases for 5: {(1,4), (2,3)}
Favourable cases for 7: {(1,6), (2,5), (3,4)}

(I ignored the reverse cases like (4,1) for (1,4) as this happens for both 5 and 7 and hence won't affect the probability)

So, chance that total of 5 comes before 7 = 2/5

(We can ignore all other cases, as questions asks for probability of sum = 5 compared to sum = 7)

Answer 2

1/9 is the probability of sum of 5 and 1/6 is the probability of sum of 7. Probability that neither happens = 26/36 = 13/18. 

$$S=1/9 + 13/18  \times 1/9 + \dots $$

In words win₅ +loss_(5,7)win₅ +....

It is a GP series with first term $1/9$ and common ratio $13/18$.

$$ S=\frac{1/9}{(1-(13/18)} =\frac{2}{5}$$

Comments

Your GP series formation is correct way of doing. But the common ratio you got is wrong. You cannot do (8/9) * (5/6) for loss_(5,7), as there is only one event happening here. So, probability of loss(_5,7) = 1- (10/36) (4 ways of sum becoming 5, and 6 ways of sum becoming 7)
= 13/18 and this would be the ratio of GP.