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Re: (no subject)
Posted:
Aug 11, 2013 2:48 PM


The answer is not 6. It is greater than 6.0035228729. I computed success probabilities up through 24 flips. 24 flips accounts for 0.9999858141 of the pdf.
Interestingly, the solution to this problem is related to this one: http://mathforum.org/kb/thread.jspa?threadID=2579173
After computing the probability for success after 4 flips (4th row of Pascal's triangle: 6/16, not 1/4 as somebody else indicated), the entries representing success (C(4,2) in this case) are replaced with zeros because the process stops with them. This affects the numbers in the following rows. Then it's on to the 6th row, which is now 1 6 9 8 9 6 1. The probability for success after 6 flips is (1  6/16)*(9+8+9)/(1+6+9+8+9+6+1). Continue for as long as you like.
Message was edited by: Mark Rickert



