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 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?

(Source: Mathematical Puzzles of Sam Loyd, Volume 1, edited by Martin Gardner)

This is an example of the binomial distribution in action. The probability that exactly one die comes up with the chosen number is $\binom 31 \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^2$ or about 0.347. Computed in much the same way, the probabilities that exactly two or three die come up with the chosen number are, respectively, about 0.069 and 0.005. One notices there is actually a more or less geometric decay in the probabilities of these latter outcomes, and that should be seen as foreshadowing. The probability of none of the dice coming up with the chosen number is then one minus the sum of the three other probabilities, or about 0.579. The expected value of the game is then about $0.579 \times -1 + 0.347 \times 1 + 0.069 \times 2 + 0.005 \times 3$ or a little more than negative 8¢. The game is favorable to the operator, but only by a bit. This is also how the standard game of craps goes, although the edge in favor of the house is even smaller, a bit more than 1¢. Perhaps this can help explain why craps remains popular while—to my knowledge—this game is no longer played at carnivals. (Samuel Loyd lived from 1841 to 1911.)