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Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability p of winning each game (independently). They play with a “win by two” rule: the first player to win two games more than his opponent wins the match. Find the probability that Calvin wins the match (in terms of p), by conditioning, using the law of total probability.

What I did is following:

• Let P(C) be the probability that Calvin wins the match
• Let W stands for Winning a game and L stands for loosing a game
  So, Possible ways of Games for winning the match = WW, WLW, LWW
     (i)   (WW => Calvin wins first two games)
     (ii)  (WLW => Calvin wins first game, looses second game and wins third game)
     (iii) (LWW => Calvin looses first game and wins next two games)

$=> P(C) = P(WW) + P(WLW) + P(LWW) $

$=> P(C) = p*p + p*(1-p)*p + (1-p)*p*p $

$=> P(C) = p^2(3-2p) $

But this is not the right answer, the right answer is: $P(C) = \frac{p^2}{p^2+q^2}$

What I am doing wrong here and how to do it right?

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  • $\begingroup$ LWW is not a win. If Hobbes has won one game, Calvin needs to win three games to win. $\endgroup$ – Thomas Andrews Sep 15 '19 at 16:28
  • $\begingroup$ You are right. I wrongly interpreted the question that to win the match one needs to win 2 games. It's actually two more than ones opponent. $\endgroup$ – Ashutosh Sep 16 '19 at 20:04
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You are underestimating the number of possible games that can happen. They could swap wins all afternoon and neither of them would ever be ahead by two.

Here's how I would set up the problem:

Let $T$ be the probability that Calvin wins the match from a tie game. Let $A$ be the probability that Calvin wins the match when he is ahead by one point, and let $B$ be the probability that Calvin wins the match when he is behind by one point. Then show that: $$T=pA+(1-p)B\\A=p+(1-p)T\\B=pT$$ and solve that for $T$.

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  • $\begingroup$ This is a very novel and elegant solution. $\endgroup$ – Ashutosh Sep 16 '19 at 20:07

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