Chuck-A-Luck is a carny game in which 3 dice are tossed. You pick a number from 1 to 6. If one of the dice shows your number you gain the value of your bet; if two of the dice show your number you gain twice the value of your bet and if three of your number show, you gain three times the value of your bet. Otherwise you lose the amount you bet.
What is your expected loss for each game?
At first blush, it looks like this might be a fair game. You win the value of your bet for each time your number appears on the dice.
Let's say that you bet $1.
There are three dice, and your number will appear on each with probability 1/6, so your expected winnings are 3*(1/6)*$1 = $.50.
If you lost the value of your bet half the time, then this would actually be a fair game. Unfortunately, the carneys "never give the sucker an even break".
You lose if your number does not come up on any of the three dice, so the probability of losing = (5/6)*(5/6)*(5/6)= 125/216, which is approximately 58%. So your expected loss on a $1 bet is $.58
Net expected loss = 50 cents - 58 cents = 8 cents on a $1 bet. Exact figure is 1/2 - 125/216 = -7/216
Fair Game?
