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Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #31704 Fun: (2.52) Difficulty: (2.34) Category: Probability Submitted By: Cogmatic Corrected By: Winner4600

One day, two friends named Al and Bob found a shiny quarter on the ground. To decide who would get it, Al would flip the coin. If it came up heads, then Al would keep it, but if it was tails, Bob would have to flip for it. If Bob got heads, Bob would keep the coin, but if he got tails, then Al would flip it and the entire process would repeat. Assuming that Al goes first and that the coin has an equal chance of landing on either heads or tail, what is the probability that Al would keep the coin?

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 grungy49 Jul 16, 2006 Very clever. Writer_fighter Aug 08, 2006 Nice job!!! Dedrik Aug 09, 2006 Wahoo now I can tell all the tutees a use for the summation of a geometric series ulan Sep 05, 2006 There is a simpler solution: Let P be the probability of keeping the coin, then with prob. 1/2 Al gets heads ans keeps the coin, and with prob. 1/2 he has to give the coin to Bob. In the latter case, Bob is in the same situation as Al, so probability that he will keep the coin is P. Using this, we can write P = 1/2 + 1/2 * (1 - P). Which gives P = 2/3. Jimbo Jan 21, 2007 Neat problem. Even neater solution Ulan! spinnercat Jul 20, 2007 Fun! InfernalDis Jun 27, 2009 Since the problem is recursive, I solved it via the following: Let p=the chance of Al winning when he flips the coin first; q=chance of Al winning when Bob flips the coin. We then have p=1/2+1/2*q and q=1/2*p. Solving for two equations and two unknows, we have p=2/3. Fun teaser!

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