The game of Chuck-a-Luck

probability

The game of Chuck-a-Luck is described as follows here. 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.)

My approach is as follows:-
Since the problem is symmetric, I took a bet on one single number, say $4$. Now, the expected amount I win (including the original stake) is:

Occurance     Loss
   0            1 
   1            0
   2           -1
   3           -2

And getting a $4$ in n dies is:

   n          Prob
   0        125/216 
   1         75/216
   2         15/216
   3          1/216

which would make my expected loss

$$\frac{1\times 125 + 0\times 75 – 1 \times 15 – 2 \times 1}{216} = 0.5$$

The answer can be found in the book provided, which comes out to be ~8%. In the author's approach, he has considered expected winnings instead of the expected loss. While calculating the loss, why is he not including the original stake?

Best Answer

The way you are computing is incorrect. The book is computing the net win. Whether you compute net win or net loss, the figure will come the same, only sign will change. By keeping the stake out of the computation of wins, the computation of net win/loss is simpler.

$E(W) = \dfrac{-1*125 +1*75+2*15+3*1}{216}= \dfrac{-17}{216}$

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