GeekDad Puzzle of the Week Solution: Dice Games

Geek Culture

The puzzle as presented last Monday:

Chuck-a-luck is a dice game played with three dice that are rolled within a closed container. After wagers on the numbers 1-6 (from each face on a standard die) are placed, the dice are rolled, and payouts are made per the following schedule:

1-1 if your number appears on just one die
2-1 if your number appears on two of the dice, and
3-1 if your number appears on all three dice.

What is the expected loss per roll for a five-die game, with the same payouts as above, and is it better than a three-die game? Would the payoff be better or worse if 4-sided die were used? What about d8? Additionally, propose a set of payoffs that would even the odds, i.e., instead of 5-1 return if your number appears on all five dice, posit a 6-1 or higher return.

The solution takes us back to a Statistics 101 course, learning about basic probability. To determine the payout (or loss) for a given type, we can just add up all of the ways you can win, and multiply each by its odds of happening. Simple, right? With a standard d6, rolled three times (aka “3d6”), your odds of winning nothing are (5/6)3 = 125/216 or about 58%. Your odds of winning your dollar back are a little more challenging to calculate. The odds of winning on the first die are (1/6)^1*(5/6)2 = 25/216 or about 11 1/2%. Since there are three dice in this game, there are 3 ways to win on a single die; the odds of winning your dollar back are (1/6)1*(5/6)2*3 = 75/216 or just under 35%. To be very specific, and so that this extends to all cases, the number of ways to win on a single die in this game are 3C1, read as “3 choose 1” and presented in equations as the number 3 over 1 in larger parenthesis, like this:

Similar calculations can lead you to the odds of winning on two dice (15/216, just under 7%) and on all three (1/216, or less than 1/2 of 1%). Multiplying each by its winnings, we get ([0*125)+(1*75)+(2*15)+(3*1)]/216 = 108/216 or exactly 1/2. This makes the expected loss also 1/2 — for every dollar wagered, over time you would expect to lose $0.50.

Using the above, we can generalize the payout as follows:

It turns out that with the simple $1 for one die, $2 for two dice, etc., this simplifies to something really straightforward: n/f

That is, if we have n dice each with f sides, the expected payout at 1 face = 1$ back, 2 faces = $2 back, etc., is [the number of dice] / [the number of faces per die]. The case stated above is 3d6, 3 dice with 6 faces, so the payout is 3/6 dollars, or $0.50. The payout for 5d6 would be $0.83, 3d4 would be $0.75, and 3d8 would pay out at $0.38. If you wanted to run the math on your favorite computational knowledge engine, enter “n=3;f=6;sum(binomial(n,i)*(1/f)^i*((f-1)/f)^(n-i)*i,{i,0,n})” for a specific solution. As with fractions in general, the smaller denominator (die size) or larger the numerator (number of dice), the better the payout.

In terms of altering the payout to make the odds more, well, even, you could examine and change the “i” term in the equation above for one or more terms. For example, in the 3d6 situation, we could change the winning payout on “three dice” from 3$ returned to $111. This would get us ([0*125)+(1*75)+(2*15)+(111*1)]/216 = 216/216 or even money on your bet.

Congratulations to Clay Brehm on winning this week’s $50.00 ThinkGeek Gift Certificate. If you submitted a solution, or are just interested in puzzles and all things geeky in general, use the code GEEKDAD44AF and get $10 off your next $50.00 ThinkGeek order.

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