5 Heads and Tails

Dan, Sam, Dimitri, Angie and Joe are playing a game in which each one has a fair coin and everyone tosses it at the same time. If, in the result, there are more heads than tails, the owners of the tail coins lose. If there are more tails than heads, the owners of the head coins lose. If the 5 coins are all heads, or all tails, they repeat the process until someone loses (or some of them).

If the probability that Dan does not lose can be expressed as a b \dfrac { a }{ b } , where a a and b b are coprime positive integers, find a + b a+b

The answer is 5.

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1 solution

Mark C
Apr 27, 2016

There are 2 5 2^5 or 32 equiprobable ways the coins can land, 30 of them determining winners. If Dan wins with 3 others, the loser can be any of four people and the winning side can be heads or tails, so there are 8 ways for that to happen. If he wins with 2 others, the other two winners can be any pair of the other 4 (there are ( 4 2 ) 4\choose 2 , or 6 pairs) and the winning side can be heads or tails, for 12 ways. So the probability Dan wins is 20/30 or 2/3, yielding 5 \boxed{5} .

And what if they all get tails or heads? Dan does not lose, but they have to throw the coins again.

Mateo Matijasevick - 5 years, 1 month ago

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Right. Eventually one of the 30 possible winner-determining combinations will come up, and all are equally likely. That's why we divided by 30 at the end, not 32.

Mark C - 5 years, 1 month ago

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