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        riddles >> putnam exam (pure math) >> Dice sum divisibility
        
(Message started by: Eigenray on Aug 7^th, 2008, 9:41am)

Title: Dice sum divisibility
Post by Eigenray on Aug 7^th, 2008, 9:41am

Let S_n denote the sum of n standard dice. In increasing order of difficulty:

0) If m|6, show that for all n, S_n is uniformly distributed mod m.

1) Find all m,n such that S_n is uniformly distributed mod m.

2) For each m http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 12, find all n such that the probability that m divides S_n is 1/m.

3) Show that for any m > 12, there are at most finitely many n for which P( m | S_n ) = 1/m.

4) Are there any pairs m,n, m > 12, for which P( m | S_n ) = 1/m?

Title: Re: Dice sum divisibility
Post by l4z3r on Aug 30^th, 2008, 5:44am

on 08/07/08 at 09:41:33, Eigenray wrote:

S_n is uniformly distributed mod m.

i dont get this. uniformly distributed?

Title: Re: Dice sum divisibility
Post by towr on Aug 30^th, 2008, 11:37am

on 08/30/08 at 05:44:07, l4z3r wrote:

i dont get this. uniformly distributed?

It means any residue modulo m is equally probable.

So take, for example: n=2 m=6
We can get sums
2 (x1), 3 (x2), 4 (x3), 5 (x4), 6 (x5), 7 (x6), 8 (x5), 9 (x4), 10 (x3), 11 (x2), 12 (x1)

modulo 6, we get
2 = 2 (mod 6) x1
3 = 3 (mod 6) x2
4 = 4 (mod 6) x3
5 = 5 (mod 6) x4
6 = 0 (mod 6) x5
7 = 1 (mod 6) x6
8 = 2 (mod 6) x5
9 = 3 (mod 6) x4
10 = 4 (mod 6) x3
11 = 5 (mod 6) x2
12 = 0 (mod 6) x1

So we have 6 out of 36 of each of 0..5 (mod 6), they are all equally probable residues.

Title: Re: Dice sum divisibility
Post by Eigenray on Aug 30^th, 2008, 12:14pm

How about starting with something easier then:

0) If m|6, show that for all n, S_n is uniformly distributed mod m.

That is, S₁ is uniformly distributed mod m if and only if (for all n, S_n is uniformly distributed mod m).

Hint for the rest: [hide]generating functions[/hide] and [hide]roots of unity[/hide].

Title: Re: Dice sum divisibility
Post by Michael Dagg on Aug 31^st, 2008, 5:50pm

Nice problem set.

In Towr's example, the probability mass function of the sum
is a triangle, so the sum of the slopes at the two sides is
a constant. This does not seem to be true for n>2 . (Generalizing,
it seems to work also with n=2 and any m , if you use m-sided
rather than 6-sided dice.)

Title: Re: Dice sum divisibility
Post by towr on Sep 1^st, 2008, 12:34am

Mod 6, any extra die just adds a random "shift", uniformly. So it doesn't change the distribution mod 6.
And if the distribution is uniform modulo k, then it's uniform modulo m for any m|k, because you take an equal number of groups together (k/m).

Title: Re: Dice sum divisibility
Post by william wu on Oct 17^th, 2008, 9:46am

Problem 0)

Here's a long way of saying the same thing towr said ... just my own way of seeing it. Have not got to the other problems yet.

To show that S_n is uniformly distributed mod m, it suffices to show that S_2 is uniformly distributed mod m. We can then deduce a uniform distribution for any larger sums by induction.

Let X be a random variable describing the outcome of the first roll of the die, and Y denote the outcome of the second roll. Then S_2 = X+Y. Now imagine making a 6x6 matrix that enumerates all the possible ways sums can be made from X and Y. The rows are decorated with the possible values of X, and the columns are decorates with the possible values of Y. Then for i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif{1,2,3,4,5,6} and j http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif{1,2,3,4,5,6}, the ij-th entry of the matrix contains i + j:

http://www.ocf.berkeley.edu/~wwu/images/riddles/diceResidues_1.png

We can make the matrix is Toeplitz by reversing the ordering of the columns, so that instead of the jth column representing Y=j, it will now represent Y = 6-j+1. Then

http://www.ocf.berkeley.edu/~wwu/images/riddles/diceResidues_2.png

Now, the ij-th entry contains (6 - j + 1) + i = (i-j) + (6+1), a function of (i-j), which indicates that the matrix is Toeplitz.

Lastly, let m be any factor of 6, and take residues of our Toeplitz matrix mod m. For example, if m=6, the matrix of residues is

http://www.ocf.berkeley.edu/~wwu/images/riddles/diceResidues_3.png

The ij-th entry is now

( (i-j) + (6+1) ) mod 6 = ((i-j) + 1) mod 6

a function of (i-j) mod 6, which makes this a circulant matrix. Thus, to assure that every entry in a circulant matrix appears the same number of times, it suffices to show that every entry in the first column appears an equal number of times. Setting j=1, we get that the ith entry of the first column is:

i mod 6 : i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif{1,2,3,4,5,6}

and thus all numbers in {0,1,2,3,4,5} appear exactly once, proving that we have an uniform distribution mod 6. Similarly, if m is any factor of 6, and we take residues mod m, then the ith entry of the first column is

i mod m : i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif{1,2,3,4,5,6}

and since m | 6, every possible residue will occur the same number of times; namely, (6/m) times.

To show that the result holds as well for a die with k sides, just replace all instances of 6 above with k.