Friday puzzle -- loaded dice

Re: Friday puzzle -- loaded dice

Bill Earl

The brute-force solution should not be hard to code. But it would be pretty ugly. There must be a more elegant solution.

Re: Friday puzzle -- loaded dice

Alex Y

"Harder than you think", "yes", and "yes and no".

Re: Friday puzzle -- loaded dice

Larry Barrett

I think I see a way to get to the answer, though it looks tedious.

Start with just the two and three combinations (I will use the word, rather than the number, for the combination, so that the roll of each dice can be represented by the numbers 1,2,3,4,5,6) (and I will use dice to mean either one or both dies).

There is just one way to get a two: Roll a 1 with each dice (1,1).

There are two ways to get a three: (1,2) and (2,1).

So the problem is how to adjust the probabilities of the 1 and 2 on each dice so that the probability of a two and a three are equal.

Let the prob of rolling a 1 = a and the prob of rolling a 2 = b.

So a*2 = (ab + ba) and from this, a =2b.

So if prob 1=2/6 and prob 2=1/6, then a*2=2ab is (2/6)(2/6) = 2(2/6)(1/6), which is the result we want.

Now consider the two, three, and four combinations:

There is one way to get a two: (1,1)

There are two ways to get a three: (1,2) and (2,1)

There are three ways to get a four: (1,3), (2,2), (3,1)

So we now want to adjust the probabilities of the 1, 2, and 3 on each dice so that the probabilities of each combinations are equal.

Let the prob of a 1=a, the prob of a 2=b, and the prob of a 3=c.

So we want a*2 = 2ab = (2ac +b*2).

From the first equation we can see that a=2b.

Substituting for a in the second equation, we have 4b*2 = 4bc +b*2.

Thus, 3b*2 = 4bc, or 3b=4c.

So we now have b=(4/3)c and a=2b = (8/3)c.

If we let c = 1/6, then a = 8/18 and b = 4/18.

Substitute these back into the ways of getting each combination and we have

(8/18)(8/18) = ((8/18)(4/18)+(4/18)(8/18)) = ((8/18)(1/6)+(4/18)(4/18)+(1/6)(8/18)), and we can see that this results in the probabilities of each combination being equal.

It seems to me that this can be extended to the remaining combinations. Since there is symmetry, I think it only necessary to consider the combinations up to seven (rest left for later).

Alex Y

Larry,

I'm afraid that you are about at a dead end here. You have assumed that both dice have the same weights, which is not a given. With that assumption, I think your reasoning is right as far as the relative sizes of a, b, and c. But you can also see from symmetry that the probability of 6 is also a, of 5 is b, and of 4 is c.

Now two additional constraints will lead to a contradiction.

• Since there are 11 possible values for the sum of two dice (2-12), for them to be equally likely, they each have to have a 1/11 probability. That means that a = sqrt(1/11), which gives the values of b and c.


• But we also know that the probabilities of all possible values on a die have to add to 1, so 2*(a+b+c) =1.

A similar problem may help

Alex Y

If you toss a fair penny and dime, the probability of getting zero heads is 1/4, the probability of getting exactly one head is 1/2, and the probability of getting two heads is 1/4.

Your mission here, should you decide to accept it, is to determine weights for the two coins such that the three possible outcomes for numbers of heads is the same.

Re: 

Larry Barrett

I came to the same conclusion overnight (actually, in the middle of the night when I do my best thinking).

Answer

Alex Y

The statement of each of these challenges contained an allusion to the answer -- these were indeed missions impossible. I've seen a clever algebraic proof that I never would have been able to come up with, even though it just uses 9th grade algebra:

Assume there is a solution. Label the dice A and B, and let the probability of throwing a 1 on die A be a1, a 3 on die B= b3, etc.

The only way to get a 2 is to throw a 1 with each die, so the probability is a1*b1 = 1/11 (since all eleven totals are equally likely).

And the only way to roll a 12 gives us a6*b6 = 1/11.

Now consider the probability of rolling a seven, which is a1*b6 + a2*b5 + ... + a6*b1. The middle four terms are non-negative, so the probability of rolling a 7, P(7) >= a1*b6 + a6*b1.

Multiply each term by one (in different forms):

P(7) >= (a1*b6)*(b1/b1) + (a6*b1)*(b6/b6)

Rearranging, we have

P(7) >= (a1*b1)*(b6/b1) + (a6*b6)*(b1/b6)

We know that a1*b1 and a6*b6 are both equal to 1/11, so this becomes

P(7) >= 1/11 * (b6/b1 * b1/b6)

And the parenthetical term is >=2, so P(7) cannot be equal to 1/11.

The solution for the coin problem is essentially the same.