"Snake-Eyes" Joe introduced a die of his own into a game of chance.

He was subsequently challenged that the die was biased.

Very rigorously test to see if there are grounds to substantiate this claim; don't accept just two or three trial runs. Are you able to offer a theoretical model consistent with your findings?

Test "Snake-Eyes" Joe's Die with this simulator which has a run of 60,000 at a time:

No:	1	2	3	4	5	6	Total
Scores	0	0	0	0	0	0	0

Note: the data changes with each subsequent mouse-over visitation to the link.

(In reply to re(2): error in solution - fallacy? by Eigenray)

The expected number of 1s in 60000 rolls is actually 9231.159763...:

We can form an 11x11 transition matrix on the state of tn after each roll, and then the expected number of 1s after n rolls is E(n) = E(n-1) + 1/6 - 5/36*P(tn==10).

The transition matrix has characteristic polynomial

f(x) ~ (t-1)(48828125 - 595703124t + 3271484376t^2
- 10650390624t^3 + 22762109376t^4 - 33370890624t^5 + 33988709376t^6
- 23748090624t^7 + 10893989376t^8 - 2962842624t^9 + 362797056t^10),

and then the recurrence for E(n) has characteristic polynomial

g(t) = (t-1)*f(t)
~ 48828125 - 693359374t + 4511718749t^2 - 17789062500t^3
+ 47334375000t^4 - 89545500000t^5 + 123492600000t^6
- 125096400000t^7 + 92378880000t^8 - 48498912000t^9
+ 17182471680t^10 - 3688436736t^11 + 362797056t^12.

That is, E(n) satisfies the recurrence
362797056 E(n) - 3688436736 E(n-1) + ... + 48828125 E(n-12) = 0,
with E(n)=n/6 for n=0..10, and
E(11) = 11/6 - 5/6^12 = 3990767611/2176782336.

Then for some constants c,c_1,...,c_10, we have
E(n) = 2n/13 + c + sum c_i r_i^n,
where r_1,...,r_10 are the roots of g(t)/(t-1)^2. The constant c ~ 0.390533, and the other terms decay exponentially since |r_i| < 0.976732 for all i.

So in conclusion, for large n, E(n) ~ 2n/13 + 0.390533, with an error that's O(0.976732^n).
Edited on August 1, 2008, 10:57 pm

Posted by Eigenray on 2008-08-01 22:16:07