Sanderson M. Smith
Home  About Sanderson Smith  Writings and Reflections  Algebra 2  AP Statistics  Statistics/Finance  Forum
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 studentconstructed 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. 

.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 moneymaker for the casino.
Note however that if the $2 won when a player wins is (routlettestyle) 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.
Home  About Sanderson Smith  Writings and Reflections  Algebra 2  AP Statistics  Statistics/Finance  Forum
Previous Page  Print This Page
Copyright © 20032009 Sanderson Smith