Gambling games are fascinating examples that illustrate human irrationality because of their straightforward mathematics. We have spent several times on roulette wheels in the past. Now, it’s the game Chuck-a-Luck.
A player can bet on one of the numbers 1, 2, 3, 4, 5, 6. Three dice are rolled. If the player’s number comes up in one, two or three of the dice, she gets, respectively, one, two or three times the original stake (in addition to her original wager); else loses the money.
So what is the house advantage of Chuck-a-Luck?
Imagine the player chooses X (a number between 1 to 6) and places 1 dollar bet. The expected value of the casino then becomes,
E(X) = 1 x P(X=0) – 1 x P(X=1) – 2 x P(X=2) – 3 x P(X=3)
E(X) is the expected value for the casino for X
P(X=0) = probability of no appearance of X (in three dice rolling)
P(X=1) = probability of one appearance of X (in three dice rolling)
P(X=2) = probability of two appearances of X (in three dice rolling)
P(X=3) = probability of three appearances of X (in three dice rolling)
If you forgot how to calculate the expected value of a die, read this post; it is the payoff of an event x its probability. And the probabilities can be calculated by applying the binomial theorem.
E(X) = 1 x [3C0 x (1/6)0 x (5/6)3] – 1 x [3C1 x (1/6)1 x (5/6)2] – 2 x [3C2 x (1/6)2 x (5/6)] – 3 x [3C3 x (1/6)3 x (5/6)0]
E(X) = [(5/6)3] – [3 x (1/6) x (5/6)2] – 2 x [3 x (1/6)2 x (5/6)] – 3 x [(1/6)3]
0.0787 or 7.87%; at par with the European style Roulette!
Reference
Fifty Challenging Problems In Probability: Frederick Mosteller
