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Aryabhatta
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Loaded pair of dice
« on: Mar 15th, 2007, 12:46am » |
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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}.
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| « Last Edit: Mar 19th, 2007, 10:48am by Aryabhatta » |
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towr
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Re: Loaded pair of dice
« Reply #1 on: Mar 15th, 2007, 1:30am » |
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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.
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Aryabhatta
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Re: Loaded pair of dice
« Reply #2 on: Mar 15th, 2007, 8:59am » |
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Yes. Assume you can assign any probabilities to each face you like. Also, you can load each die differently.
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Aryabhatta
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Re: Loaded pair of dice
« Reply #3 on: Mar 15th, 2007, 9:55am » |
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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.
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| « Last Edit: Mar 15th, 2007, 10:10am by Aryabhatta » |
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towr
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Re: Loaded pair of dice
« Reply #4 on: Mar 15th, 2007, 10:11am » |
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on Mar 15th, 2007, 9:55am, Aryabhatta wrote:| In fact the sum of face probabilites for a die need not even total 1. |
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Really?
Then I pick the probability zero for all faces. 
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| « Last Edit: Mar 15th, 2007, 1:54pm by towr » |
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Aryabhatta
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Re: Loaded pair of dice
« Reply #5 on: Mar 15th, 2007, 10:12am » |
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on Mar 15th, 2007, 10:11am, towr wrote:
Really?
Then I pick the probability zero for all faces.
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Shoot! You are fast. You got this in while I was editing that post...
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Aryabhatta
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Re: Loaded pair of dice
« Reply #6 on: Mar 18th, 2007, 1:15pm » |
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Hint1: It is not possible.
Hint2: In fact it is not possible that Pbt(Sum=2) = Pbt(Sum=12) = Pbt(Sum=7)
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Grimbal
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Re: Loaded pair of dice
« Reply #7 on: Mar 19th, 2007, 3:52am » |
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P(sum=7) should be 6 times P(sum=2)
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Aryabhatta
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Re: Loaded pair of dice
« Reply #8 on: Mar 19th, 2007, 9:33am » |
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on Mar 19th, 2007, 3:52am, Grimbal wrote:| P(sum=7) should be 6 times P(sum=2) |
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Can you prove this?
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Grimbal
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Re: Loaded pair of dice
« Reply #9 on: Mar 19th, 2007, 9:42am » |
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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.
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rmsgrey
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Re: Loaded pair of dice
« Reply #10 on: Mar 19th, 2007, 10:15am » |
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on Mar 19th, 2007, 9:42am, 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. |
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I read the original problem statement as:
P(sum=a)=P(sum=b) for all a,b in [2,12]
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Grimbal
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Re: Loaded pair of dice
« Reply #11 on: Mar 19th, 2007, 10:38am » |
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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.
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Aryabhatta
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Re: Loaded pair of dice
« Reply #12 on: Mar 19th, 2007, 10:39am » |
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on Mar 19th, 2007, 10:15am, rmsgrey wrote:
I read the original problem statement as:
P(sum=a)=P(sum=b) for all a,b in [2,12] |
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Yes. That was the intent.
I will modify the original post to make it more clear.
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Aryabhatta
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Re: Loaded pair of dice
« Reply #13 on: Mar 19th, 2007, 10:50am » |
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on Mar 19th, 2007, 10:38am, Grimbal wrote: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. |
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Hmm.. This is interesting too.
Seem like it should be possible by slightly changing the probabilities... Just guessing now, having tried proving it.
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towr
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Re: Loaded pair of dice
« Reply #14 on: Mar 19th, 2007, 11:01am » |
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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..
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Aryabhatta
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Re: Loaded pair of dice
« Reply #15 on: Mar 19th, 2007, 2:01pm » |
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on Mar 19th, 2007, 10:38am, 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. |
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This one turned out to be more interesting than the original one!
It looks like it is not possible!
Suppose Dice i has probability pij/6 of getting the number j.
Consider the polynomial
P(x) = Sum_{j=1 to 6} pij xj
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.
Thus it is not possible.
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| « Last Edit: Mar 19th, 2007, 2:20pm by Aryabhatta » |
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Aryabhatta
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Re: Loaded pair of dice
« Reply #16 on: Mar 19th, 2007, 2:20pm » |
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I have hidden the above as the approach used there can probably be used for the original problem.
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Grimbal
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Re: Loaded pair of dice
« Reply #17 on: Mar 20th, 2007, 1:46am » |
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Interesting approach. The polynomial is a neat way to express all constraints. I started to write all the equations on the pij 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 Mar 19th, 2007, 11:01am, 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.. |
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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.
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Aryabhatta
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Re: Loaded pair of dice
« Reply #18 on: Mar 20th, 2007, 8:32am » |
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on Mar 20th, 2007, 1:46am, Grimbal wrote:Interesting approach. The polynomial is a neat way to express all constraints. I started to write all the equations on the pij and tried to solve them to show they must all be 1/6. It is a bit tedious and I stopped before being through.
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Yup. Power of Generating Functions!
on Mar 20th, 2007, 1:46am, Grimbal wrote:
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. |
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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)!).
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| « Last Edit: Mar 20th, 2007, 8:34am by Aryabhatta » |
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Grimbal
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Re: Loaded pair of dice
« Reply #19 on: Mar 20th, 2007, 9:59am » |
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A constraint is that the coefficients of the polynomials must all be integers >=0 and only 6 of them >0.
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| « Last Edit: Mar 20th, 2007, 10:01am by Grimbal » |
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Aryabhatta
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Re: Loaded pair of dice
« Reply #20 on: Mar 20th, 2007, 10:08am » |
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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)
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