Two players `P_(1)and P_(2)` are playing the final of a chess championship, which consists of a series of matches. Probability of `P_(1)` winning a match is 2/3 and that of `P_(2) ` is 1/3. Thus winner will be the one who is ahead by 2 games as compared to the other player and wins at least 6 games. Now, if the player `P_(2)` wins the first four matches, find the probability of `P_(1)` wining the championship.

Explanation

`P_(1)` can win in the following matually exclusive ways:
(a) `P_(1)` wins the next dix matches.
(b) `P_(1)` wins five out of next six matches, so that after new six matches scored of `P_(1)and P_(2)` are tied up. This tie continues up to next 2n matches `(nge0)` and finally `P_(1)` wins 2 consective matches. Now, for case (a), probability is given by `(2//3)^(6)` and probability of tie after 6 matches [in case(b)] is
`""^(6)C_(5)((2)/(3))^(5)((1)/(3))=6xx(2^(5))/(3^(6))=(2^(6))/(3^(5))`
Now probability that scores are still tied up after another next two matches is
`2/3xx1/3+1/3xx2/3=4/9`
[First match won by `P_(1)` and second by `P_(2)` or first by `P_(2)` and second by `P_(1).]`
Similarly, probability that scores are still tied up after another 2n matches is `(4//9)^(n).`
Therefore, the total probability of `P_(1)` winning the championship is
`((2)/(3))^(6)+(2^(6))/(3^(5))(underset(n=0)overset(oo)sum((4)/(9))^(n)((2)/(3))^(2))`
`=((2)/(3))^(6)+(2^(6))/(3^(5))((2)/(3))^(2)((1)/(1-(4)/(9)))`
`=17/5((2)/(3))^(6)=1088/3645`