Posted by: phedrick | November 24, 2008

More Gambling

So I have one more probability problem for everyone, this time on gambling.


A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6. 

The dealer says, “You can choose your bet on a number–any number from 1 to 6. I will then roll the 3 dice. If none show the number you bet, you’ll lose $1. If one shows the number you bet, you’ll win $1. If two or three dice show the number you bet, you’ll win $3 or $5, respectively.”

Is this game better for the dealer or the gambler?



It turns out that it is a fair game–both dealer and gambler have expected winnings of zero. I chose this problem because it seems to be skewed in the direction of the gambler–after all, he can only win or lose $1 depending on the existence of one die that he chose, but he’ll win $3 or $5 if there are more than one.

It’s probably easier to think of the solution in terms of 6 separate gamblers all playing at once, each one choosing their own number 1 through 6.

If the dealer rolls three different numbers, three players will win $1 each and the other three will lose $1 apiece (no one can gain $3 or $5 if the numbers are all different).

If two of the three dice are the same, then one person will win $3 and one person will win $1, but this time there will be four losers, losing a total of $4 (they lose $1 apiece).

If all three dice are the same, then only one person will win any money ($5), and the other five will lose a total of $5.

The cool part about the problem is that you actually don’t have to look all the permutations (i.e. whether it’s more likely for all three to be the same or only two to be the same, etc.) because in every type of situation the table would be even–the total amount won equals the total amount lost.



  1. I think its wrong. Game is not fair.

    The probability of winning is equal to the probability of losing.

    case 1 weight: when only one die has the number chosen = [1/6 + 1/6 + 1/6 – (1/6)(1/6) – (1/6)(1/6) – (1/6)(1/6)]x1 = 0.42$
    case 2 weight: when only two of the dies has the number = [(1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) – (1/6)(1/6)(1/6)]x3 = 0.24$
    case 3 weight: when all dies have the number chosen = (1/6)(1/6)(1/6)x5 =0.023$

    adding all equals = 0.68$ (winning)

    while losing = 0.5 x -1$ = – 0.5$

    So pure probability reveals that game is skewed towards gambler (not fair)

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