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

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Posted by Joel on October 14, 2002 at 10:36:06:

In Reply to: Re: Probability (orig. posted by rob77)(small correction) posted by Joel on October 13, 2002 at 15:37:06:

: : 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).

: :
: : : : 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.

: : :
: : :

`: : :                                             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: : : : : : `

: 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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