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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 $\frac{2}{3}$ and that of $P_2$ is $\frac{1}{3}$.The winner will be the one who is ahead by two games as compared to the other player and wins atleast $6$ games.Now if the player $P_2$ wins first 4 matches,prove that the probability of $P_1$ winning the championship is $\frac{1088}{3645}$

I have couple of doubts in this question after following the solution given by "Aretino" .

Find the probability of $P_1$ winning the championship

Its assumed that After a Parity occurs, Probability of winning $P1$ is $p_0$.

But there are different parity situations right?

For example we have

$P_2 P_2 P_2P_2 P_2 P_1P_1P_1P_1P_1$ also a parity situation.why cant we consider this case?

am i missing something here?

Ekaveera Gouribhatla
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  • @KeyFlex Note that there is a specific "doubts" regarding the solution of an existing duplicate linked in the post. – Lee David Chung Lin Oct 16 '18 at 17:30
  • @LeeDavidChungLin Actually the question has $2$ duplicates. The one which I linked and the other which the OP has linked in the question. – Key Flex Oct 16 '18 at 17:31
  • @KeyFlex That's exactly my point. The asker has a specific doubt about one of the duplicates, thus this post (as intended) is not really a duplicate to the one you linked. – Lee David Chung Lin Oct 16 '18 at 17:33
  • @KeyFlex Having said that, I agree that one should still link that "other" duplicate (and perhaps more out there) as you did. Just that when addressing the asker's issues, one should keep in mind it is not about solving this math problem in general but about a specific argument. – Lee David Chung Lin Oct 16 '18 at 17:39
  • i do not know why irrelevant comments coming up here – Ekaveera Gouribhatla Oct 16 '18 at 17:45

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