# How to solve Chuck-a-Luck puzzle

### Story (spoilers!)

The puzzle:

Chuck-a-Luck is a gambling game often played at carnivals and gambling houses. A player may bet on any one of the numbers 1,2,3,4,5,6. Three dice are rolled. If the player's number appears on one, two, or three of the dice, he receives respectively one, two, or three times his original stake plus his own money back; otherwise he loses his stake. What is the player's expected loss per unit stake? (Actually the player may distribute stakes on several numbers, but each such stake can be regarded as a separate bet.)

You can find a solution in Fifty challenging problems in probability with solutions (F. Mosteller).

Before reading the solution of course, I worked on my own, and I thought "hey, there is nice symmetry here, so let's assume I bet on 1 and then calculate loss". The Author thought (just an assumption) "hey, there is nice symmetry here, so let's assume I bet at the same time on all numbers".

I calculated loss about 11%, the Author calculated loss at ~8%. I rechecked my calculations, they seem fine, I rechecked the Author's calculations and they also look fine. However, no matter how long I stare at those outcomes, 11% is not 8%.

Btw. you can find the same comparison one by one here: http://www2.washjeff.edu/users/mwoltermann/Mosteller/contents.htm -- number 6.

### Question

Are those approaches actually both incorrect? I mean, the question is plain and simple "What is the player's expected loss per unit stake?". You can model the scenarios, but adding something to the question like "... assuming the player bet on all the numbers" changes the question. So Author answers on similar, yet different question.

So, MY question is -- am I right with above suspicion, that in order to correctly solve this problem one should calculate expected loss per unit stake period. The calculation should be done really not with parameter (bet on 1) or (bet on all) but (bet A on 1, bet B on 2, bet ..., bet F on 6) and only with such parameters calculate the expected loss.

Am I right with my new approach?

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All three methods should produce the same result for the house edge. To check yours, you should find

Number 1s     Prob
0        125/216
1         75/216
2         15/216
3          1/216


which would make your expected winnings (for a stake of 1)

$$\frac{-1\times 125 + 1\times 75 + 2 \times 15 + 3 \times 1}{216} = - \frac{17}{216} \approx -0.0787.$$

You may need to check your calculations again.

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 Thank you, you wrote this in much cleaner way, than I did :-) – greenoldman Oct 3 '11 at 5:56