Sanderson M. Smith

DOUBLE UP (Casino game constructed by Cate students Beth, Jamah, and Billy)

Here is the game description as written by the inventors:

Each player is allowed to bet however much he/she wants to. He/she rolls 2 dice and makes a note of the total shown. If the dice show two of the same number, for instance a four and a four, the player loses. If the dice show two different numbers, the player rolls again. If the total the second time is greater than the total the first time, he/she gets twice the amount of money bet. If the number is the same as or smaller than the first, he/she loses.

What follows is an analysis of this student-constructed game.

Note that the player can win only by rolling a total of 3, 4 (but not with 2,2), 5, 6 (but not with 3,3), 7, 8 (but not with 4,4), 9, 10 (but not with 5,5), or 11 on the first roll. The player can't win with a total of 2 or 12 on the first roll, since these totals can only be obtained by rolling doubles.

Probability (win with 3 on first roll) = Prob(3 on first roll)xProb(4 or more on second roll) = (2/36)(33/36) = .0509.

Probability (win with 4 on first roll) = Prob(4 on first roll without doubles)xProb(5 or more on second roll) = (2/36)(30/36) = .0463.

Probability (win with 5 on first roll) = Prob(5 on first roll)xProb(6 or more on second roll) = (4/36)(26/36) = .0802.

Probability (win with 6 on first roll) = Prob(6 on first roll without doubles)xProb(7 or more on second roll) = (4/36)(21/36) = .0648.

Probability (win with 7 on first roll) = Prob(7 on first roll)xProb(8 or more on second roll) = (6/36)(15/36) = .0694.

Probability (win with 8 on first roll) = Prob(8 on first roll without doubles)xProb(9 or more on second roll) = (4/36)(10/36) = .0309.

Probability (win with 9 on first roll) = Prob(9 on first roll)xProb(10 or more on second roll) = (4/36)(6/36) = .0185.

Probability (win with 10 on first roll) = Prob(10 on first roll without doubles)xProb(11 or more on second roll) = (2/36)(3/36) = .0046.

Probability (win with 11 on first roll) = Prob(11 on first roll)xProb(12 on second roll) = (2/36)(1/36) = .0015.

 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

The probability of a player winning this game is thus

.0509+.0463+.0802+.0648+.0694+.0309+.0185+.0046+.0015 = .3671, or 37.71%

Hence, the probability of losing a game is 1 - .3671 = .6329, or 63.29%.

Let's assume that the player pays \$1 to play, and that this is not returned. That is, the player loses the dollar upon losing, and gets a payoff of \$2 upon winning. Under this interpretation of the rules, a player's expected payoff is

(\$2)(.3731) + (\$0)(.6329) = \$0.7462

In other words, for each \$1 spent, the player expects a return of \$0.7462. This is approximately equivalent to losing 25 cents each time a dollar is spent to play Double Up. Hence, the game is a money-maker for the casino.

Note however that if the \$2 won when a player wins is (routlette-style) simply put on top of the \$1 spent to play, then the player's expected payoff is

(\$3)(.3731) + (\$0)(.6329) = \$1.1193

In this situation, the player expects a return of \$1.12 for each \$1 spent to play Double Up. It goes without saying that a casino would not offer up the game under these circumstances.