Three coins are lying on a table: a quarter, a half dollar, and a silver dollar. You claim one coin, I’ll claim the other two, and then we’ll toss all three.
A coin that lands tails counts zero, and a coin that lands heads wins its value
(in cents, 25, 50, or 100) for its owner.
Whichever of us has the larger score wins all three coins. If all three coins land tails then we call it a draw and toss again.
Which coin should you claim to make the game fair — that is, so that each of us has an expected win of zero?
Source: Martin Gardner, “Charles Addams’ Skier and Other Problems,” in Wheels, Life and Other Mathematical Amusements, 1983.
(In reply to
Fair Game (spoiler) by Steve Herman)
This works because the winner of all the coins is the largest coin to be face up. And because they are powers of 2, the probability of winning everything is proportional to the "entry price".
This can be extended without limit to more coins and more players. For instance, add a $2 coin and a $4 coin and an $8 coin and a $16 coin. The 7 coins can be distributed in any way among 2 to 7 players, and it is still a fair game, as long as all of the coins are in play.
re: Fair Game (spoiler)
