Probability of A winning the game

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SUMMARY

The discussion focuses on calculating the probability of player A winning a game against player B, where each game can result in a win for A, a win for B, or a draw. The probabilities are denoted as P(A), P(B), and P(D). The overall winner is determined by the first game that does not end in a draw. The formula for P(A wins at game n) is derived from the probabilities of draws in the first n-1 games multiplied by the probability of A winning in the nth game, leading to the conclusion that P(A wins) can be expressed as the sum of probabilities of A winning in each game.

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TomJerry
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Suppose A and B play over and over , independently a game which each time results in either a win for A , a win for B or draw (meaning no decision) with probabilities P(A), P(B) and P(D). Suppose they keep playing until the first game that does not result in a draw , and call the winner of that game the overall winner. Compute
i) P(A wins at game n)
ii) P(a wins)
 
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P(A,n) = P(draw in first n-1 games) * P( A wins in nth game | draw in first n-1 games)

P(A wins) = P(A wins first game) + P(A wins second game) + ...

This seems like it belongs to the homework guidance forum though
 
I think laying out clearly the sample space should take you around 90% towards
the answer.
 

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