How Not To Play Dice

The question is as follows: Can two loaded dice be constructed such that, rolling the pair gives the numbers 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 with equal probability?

Now, a couple of warnings going in. First and foremost, I don’t really know how you go about loading dice. I’m a mathematician, not a dicetition, so this takes a very abstract approach to loading dice. For each die, we’re basically choosing the probability with which each number occurs in a roll.

Second, as little as I know about loading dice, I’m even less sure about my abilities with probability. Honestly, I avoid doing probability as often as I possibly can. Well, I don’t go out of my way to find probability to avoid, but it does make me very uncomfortable. So if I make any mistakes, feel free to comment. The only reason I’m doing this problem at all is that it involves a neat algebraic jump.

So here we go.

First off, I’m going to say that by constructing loaded dice, what we’re doing is for each number on the die, we’re picking the probability that number occurs when rolled. So, say that P1(n) is the probability that the number n shows up when die 1 is rolled, and P2(n) is the probability that n shows up when die 2 is rolled.

So we are choosing values for P1(1), P1(2), P1(3), P1(4), P1(5), P1(6), and likewise for P2.

Next, since we’re considering the probabilities of rolling the numbers 2 through 12, we first must consider how we might achieve those numbers. So below I have written out all the possible ways of rolling each number in terms of (first die, second die).

2: (1, 1)
3: (2, 1) (1, 2)
4: (3, 1) (2, 2) (1, 3)
5: (4, 1) (3, 2) (2, 3) (1, 4)
6: (5, 1) (4, 2) (3, 3) (2, 4) (1, 5)
7: (6, 1) (5, 2) (4, 3) (3, 4) (2, 5) (1, 6)
8: (6, 2) (5, 3) (4, 4) (3, 5) (2, 6)
9: (6, 3) (5, 4) (4, 5) (3, 6)
10: (6, 4) (5, 5) (4, 6)
11: (6, 5) (5, 6)
12: (6, 6)

You might say, but Fox, isn’t rolling a 6 and a 4 the same as rolling a 4 and a 6? Why are you counting them both? The thing is that since the probabilities of rolling a 6 on the first die and on the second die are different, the two different, though symmetric, rolls must be counted separately, since the net probability of each roll will be different.

To refresh, the probability of any one roll is going to be the probability of the value on the first die times the probability of the value on the second die, P1(i)*P2(j). The probability of any one number between 2 and 12 is going to be the sum of the probabilities of all possible rolls that yield that sum. So the probability of rolling a 4 is actually P1(1)P2(3) + P1(2)P2(2) + P1(3)P2(1).

Now, we know we want the numbers to occur with equal probability. As there are 11 numbers, this means we want each number to occur with probability 1/11.

One approach to this problem would be to set up a massive system of equations and try to solve explicitly for the values of the P1’s and P2’s. But as that would give us 12 variables in 11 equations, off hand, it would be difficult to say the least. But there’s this really need algebraic trick we can use the simplify the whole shebang.

To begin with, look at the probabilities of rolling 2’s and 12’s. The probability of rolling a 2 is P1(1)P2(1), and the probability of rolling a 12 is P1(6)P2(6). But we know that both of these must equal 1/11, so we have the following.

Now, look at the probability of rolling a 7. We want it to be 1/11, so full out that is

To simplify, I’m going to group P1(5)P2(2) + P1(4)P2(3) + P1(3)P2(4) + P1(2)P2(5) together and just call it S, which is short for STUFF.

Now let’s square each side of the equation.

It looks messy, but we’re very near the end. Notice that probabilities are always positive. You can’t really have a negative probability of something happening. S is therefore the sum of a bunch of positive numbers, and is itself positive. The squared terms in the above equation are also positive, and we can group many things together under a variable I’ll call Z, and declare with certainty that Z is positive.

But notice, we’ve already calculated the value of the P1(1)P2(6)P1(6)P2(1) term to be 1/121. Thus, substituting in and simplifying the above equation,

Look at that equation. We have a sum of three necessarily non-negative terms coming out to be a negative number. This is impossible. Clearly a contradiction : )

As such, we can say that value for P1(1) to P1(6) and P2(1) to P2(6) cannot be chosen to yield the desired probabilities.

I always feel a little let down when the answer to an interesting question is something decidedly not interesting, but it’s really difficult to argue with the correct solution.