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        riddles >> medium >> Loaded pair of dice
        
(Message started by: Aryabhatta on Mar 15^th, 2007, 12:46am)

Title: Loaded pair of dice
Post by Aryabhatta on Mar 15^th, 2007, 12:46am

Is it possible to load a pair of dice such that every sum 2,3,...,12 is equally likely? Assume the dice are distinguishable (so (1,3) and (3,1) are different outcomes)

To clarify:

Is it possible to assign probabilities to each face of the dice such that when we throw the two dice the chances that the sum of the top faces is X is same as the chances that the sum is Y for any X, Y in {2,3,...,12}.

Title: Re: Loaded pair of dice
Post by towr on Mar 15^th, 2007, 1:30am

Depends how loading works. It will be harder in practice than theory, because making 1 more likely decreases the likelyhood of 6. If you can just assign probabilities to each face as you see fit (but still adding to one obviously), then possibly.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 15^th, 2007, 8:59am

Yes. Assume you can assign any probabilities to each face you like. Also, you can load each die differently.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 15^th, 2007, 9:55am

~~In fact the sum of face probabilites for a die need not even total 1.~~

Scratch that. We need the sum of probabilities of the faces of a die to total 1. The split among the 6 could be anything of your choice.

Title: Re: Loaded pair of dice
Post by towr on Mar 15^th, 2007, 10:11am

on 03/15/07 at 09:55:49, Aryabhatta wrote:
In fact the sum of face probabilites for a die need not even total 1.
Really?
Then I pick the probability zero for all faces. :P

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 15^th, 2007, 10:12am

on 03/15/07 at 10:11:12, towr wrote:

Really?
Then I pick the probability zero for all faces.

Shoot! You are fast. You got this in while I was editing that post...

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 18^th, 2007, 1:15pm

Hint1: [hide]It is not possible.[/hide]
Hint2: [hide]In fact it is not possible that Pbt(Sum=2) = Pbt(Sum=12) = Pbt(Sum=7)[/hide]

Title: Re: Loaded pair of dice
Post by Grimbal on Mar 19^th, 2007, 3:52am

P(sum=7) should be 6 times P(sum=2)

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 19^th, 2007, 9:33am

on 03/19/07 at 03:52:26, Grimbal wrote:

P(sum=7) should be 6 times P(sum=2)

Can you prove this?

Title: Re: Loaded pair of dice
Post by Grimbal on Mar 19^th, 2007, 9:42am

Well, that is how it is on regular dice (there is one way to get 2 and 6 ways to get 7). And I understand the aim is to manage to get the same probabilities for the loaded dice.

Title: Re: Loaded pair of dice
Post by rmsgrey on Mar 19^th, 2007, 10:15am

on 03/19/07 at 09:42:12, Grimbal wrote:

Well, that is how it is on regular dice (there is one way to get 2 and 6 ways to get 7). And I understand the aim is to manage to get the same probabilities for the loaded dice.

I read the original problem statement as:

P(sum=a)=P(sum=b) for all a,b in [2,12]

Title: Re: Loaded pair of dice
Post by Grimbal on Mar 19^th, 2007, 10:38am

I see. I understood "equally likely ... as for a pair of regular dice".

But it happens that the problem as I understood it:
"Is there a pair of loaded dice that still have the same probabilities as a regular set of dice to get sums 2 to 12?"
is equally interesting, and probably related.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 19^th, 2007, 10:39am

on 03/19/07 at 10:15:23, rmsgrey wrote:

I read the original problem statement as:

P(sum=a)=P(sum=b) for all a,b in [2,12]

Yes. That was the intent.

I will modify the original post to make it more clear.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 19^th, 2007, 10:50am

on 03/19/07 at 10:38:56, Grimbal wrote:

Hmm.. This is interesting too.
Seem like it should be possible by slightly changing the probabilities... Just guessing now, having tried proving it.

Title: Re: Loaded pair of dice
Post by towr on Mar 19^th, 2007, 11:01am

I do know you can change the numbers on the dice and keep the same probability for each sum, but that's yet again another problem..

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 19^th, 2007, 2:01pm

on 03/19/07 at 10:38:56, Grimbal wrote:

"Is there a pair of loaded dice that still have the same probabilities as a regular set of dice to get sums 2 to 12?"
is equally interesting, and probably related.

This one turned out to be more interesting than the original one!

It looks like it is not possible!

[hide]
Suppose Dice i has probability p_ij/6 of getting the number j.

Consider the polynomial
P(x) = Sum_{j=1 to 6} p_ij x^j

Thus if P and Q are two polynomials for different dice, the product PQ gives us the probabilities of the corresponding sums by taking each coefficient in the product and dividing by 36. For instance P(Sum = 12) would be the coefficient of x^12 in PQ divided by 36.

For fair dice we see that the polynomial is

Fair(x) = sum_{j=1 to 6} xⁱ = x(x⁶ - 1)/(x-1) = x(x³-1)(x³+1)/(x-1) = x(x+1)(x²+x +1)(x²-x +1)

Thus the Sum polynomial is
Fair(x) * Fair (x) = [x(x+1)(x²+x +1)(x²-x +1)]² = Sum(x) say

Now the problem is to factorize this Sum(x) into two other polynomials T(x) and U(x) such that
0) T(x) and U(x) are both 6th degree polynomials
1) x divides both T(x) and U(x)
2) each coeffiecient of T(x) and U(x) is positive
3) Sum(x) = T(x) * U(x)

Now the roots of Sum(x) are 0, -1, w, w², -w and -w² (multiplicity of each is 2) where w is the cube root of unity.

Since T(x)*U(x) = Sum(x), the union of roots of T and U must be roots of S and vice versa.

Let us now try to assign T(x) and U(x) the roots.

The roots in question are
0, 0, -1, -1, w, w, w²,w², -w, -w, -w² and -w²

Since x divides both T and U, one zero goes to T and one goes to U.

Consider the -1's. Suppose both go to T. Then roots of T would be 0,1,-1 and three others.
But the complex roots always come in conjugate pairs. Hence the -1's should also be split.

So the only other possible split for roots is:

T: 0, -1, w, w, w², w²
and
U:0, -1, -w, -w, -w², -w²

With this split, we can easily verify that not all coefficients of U are positive.

[/hide]
Thus it is not possible.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 19^th, 2007, 2:20pm

I have hidden the above as the approach used there can probably be used for the original problem.

Title: Re: Loaded pair of dice
Post by Grimbal on Mar 20^th, 2007, 1:46am

Interesting approach. The polynomial is a neat way to express all constraints. I started to write all the equations on the p_ij and tried to solve them to show they must all be 1/6. It is a bit tedious and I stopped before being through.

on 03/19/07 at 11:01:14, towr wrote:

I do know you can change the numbers on the dice and keep the same probability for each sum, but that's yet again another problem..

They are called Sicherman dice. One of the die has numbers in range 1-8, the other 1-4. If they had been in range 1-6, it would have been like changing the probabilities of each number, and it would solve the problem.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 20^th, 2007, 8:32am

on 03/20/07 at 01:46:27, Grimbal wrote:

Yup. Power of Generating Functions!

on 03/20/07 at 01:46:27, Grimbal wrote:

The generating function approach used above can also be used to derive the Sicherman dice numbers I think. (Look at the split for T and U in the above proof, if degrees are allowed to be other than 6, we can redistribute the roots, one to a 4th degree and one to an 8th degree polynomial). Need to work it out though, but I am pretty sure that would be the case (We _have_ to get two factors of Sum(x)!).

Title: Re: Loaded pair of dice
Post by Grimbal on Mar 20^th, 2007, 9:59am

A constraint is that the coefficients of the polynomials must all be integers >=0 and only 6 of them >0.

Title: Re: Loaded pair of dice
Post by Aryabhatta on Mar 20^th, 2007, 10:08am

Yes.

But once we get the factorizing polynomials, we have to just check if they match the constraints. The number of such polynomials is not high in this case. At max 3 I believe. (we just have to consider degrees 6-6, 8-4, 2-10, the degrees have to be even, we can show that)