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A step by step thinking approach to these types of problem.

Q: Given Player A,B,C, and D randomly lined up. first two players in the line play games, winner of the game plays with the person third in line, further the winner plays with the fourth in the line. Finding probability that A wins the tournament, given A wins each game it plays with probability p.

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If $A$ is number $1$ on the line (probability on that is $\frac14$) then he must win $3$ consecutive games (probability on that $p^3$).

If $A$ is number $2$ on the line (probability on that is $\frac14$) then he must win $3$ consecutive games (probability on that $p^3$).

If $A$ is number $3$ on the line (probability on that is $\frac14$) then he must win $2$ consecutive games (probability on that $p^2$).

If $A$ is number $4$ on the line (probability on that is $\frac14$) then he must win $1$ game (probability on that $p$).

Can you take it from here?

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  • $\begingroup$ Yes. You solved the problem. I went blank when I saw problem, my intention is to know how should I approach whenever I encounter such problems. Your explanation made this problem so simple, and I wonder why I haven't thought it this way. Wanting to develop this thinking process. $\endgroup$ – Roopesh Singh Jun 24 at 8:43
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    $\begingroup$ Just keep going in your tries. Then mathematical maturity will surely grow. It needs time though ;). $\endgroup$ – drhab Jun 24 at 8:47
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First calculate the probability that $A$ is in each position, i.e. first second third and last. Then, for each position, calculate the probability that $A$ wins the tournament. Then, multiply and sum these probabilities, i.e.

$$P(A\mbox{ wins}) = P(A\mbox{ wins |} A\mbox{ is first})P(A\mbox{ is first}) + \ldots + P(A\mbox{ wins |} A\mbox{ is last})P(A\mbox{ is last})$$

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