Posted by Joel on October 13, 2002 at 15:37:06:
In Reply to: Re: Probability (orig. posted by rob77)(small correction) posted by Joel on October 13, 2002 at 14:50:36:
: 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 ... which seems to be approaching .375. Not quite (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%.
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