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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*de****
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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Comments

grungy49Aau*
Jul 16, 2006

Very clever.
Writer_fighterAus*
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*au*
Jan 21, 2007

Neat problem. Even neater solution Ulan!
spinnercatAus*
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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