Carnival Dice Game


The following dice game is very popular at fairs and carnivals, but since two persons seldom agree on the chances of a player winning, I offer it as an elementary problem in the theory of probability.

On the counter are six squares marked 1, 2, 3, 4, 5, 6. Players are invited to place as much money as they wish on any one square. Three dice are then thrown. If your number appears on one die only, you get your money back plus the same amount. If two dice show your number, you get your money back plus twice the amount you placed on the square. If your number appears on all three dice, you get your money back plus three times the amount. Of course if the number is not on any of the dice, the operator gets your money.

To make this clearer with an example, suppose that you bet 1 dollar on No. 6. If one die shows a 6, you get your dollar back plus another dollar. If two dice show 6, you get back your dollar plus two dollars. If three dice show 6, you get your dollar back plus three dollars.

A player might reason: the chance of my number showing on one die is 1 / 6, but since there are three dice, the chances must be 3 / 6 or 1 / 2, therefore the game is a fair one. Of course this is the way the operator of the game wants everyone to reason, for it is quite fallacious.

Is the game favorable to the operator or the player, and in either case, just how favorable is it?

Solution



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