Is this correct? Is there a better way to solve this?

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: : : <b>There was a mistake in the lower right-hand corner of the chart on the original post which is corrected on this one (although it didn't affect the result).</b> : : : : : : : : A,B,C are evenly matched tennis players. Initially A and B play a set, and the winner then plays C. This continues, with the winner always playing the waiting player, until one of the players has won two sets in a row. That player is then declared the overall winner. What is the probability that A is the overall winner? : : : : I guess it IS a trick question, and the trick is that although all are evenly matched, A and B, by being in the first round, have an advantage over C. : : : : I don't know how to compute the probability but I believe it is a sum of the form 1/4 + 1/16 + 1/32 + 1/128 + 1/256 + 1/1024 + 1/2048 ... (1/64 and 1/512 are intentionally omitted). : : : : Here is my "attempt" at an upside-down tree diagram of the possible outcomes, which (if the board displays it as I am hoping) may show how I am getting to that. : : : : : : : : <pre><font size=2> : : : : AB : : : : AC BC : : : : : : : : **A** CB --BA-- CA 1/4 : : : : --C-- BA -CB-- AB : : : : --B-- AC **A** BC 1/16 : : : : **A** CB --B-- CA 1/32 : : : : --C-- BA --C-- AB : : : : --B-- AC **A** BC 1/128 : : : : **A** CB --B-- CA 1/256 : : : : : : : : </font></pre> : : Looks like A and B each have a probability of slightly over 35.71%. and C's probability is a little over 28.57%. : (Assuming this is correct) Is there a way to solve this without going through all the diagramming? : Is this a correct way to represent the sum 1/4 + 1/16 + 1/32 + 1/128 + 1/256 + 1/1024 + 1/2048 ... : : Sum(from n=1 to infinity)1/(2^(n+1) - Sum(from n=1 to infinity) 1/(2^3n) : What should I look up (subject, topic, etc.) to find out about techniques for solving sums like this?

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