"SnakeEyes" 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 "SnakeEyes" 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 mouseover 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(n1) + 1/6  5/36*P(tn==10).
The transition matrix has characteristic polynomial
f(x) ~ (t1)(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) = (t1)*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(n1) + ... + 48828125 E(n12) = 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)/(t1)^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 20080801 22:16:07 