A standard six-sided die is to be rolled repeatedly until a side appears a number of times equal to its number. In other words until the n-th n appears.
Let P(n)=the probability the game terminates with the n-th n.
Find the distribution of n.
Feel free to generalize for m sides.
Warning: I have not managed this past m=4.
(In reply to
re: more results (11 and 12) by Daniel)
12-sided die:
Probabilities:
{0.687834,0.212301,0.0684188,0.0218152,0.00677342,0.00203701,0.0005926,0.000166855,0.0000455305,0.0000120604,3.10643*10^-6,7.79319*10^-7}
Ratios:
{3.2399,3.10296,3.13629,3.22071,3.32517,3.43741,3.55158,3.66469,3.7752,3.88241,3.98608}
I didn't really expect those ratios to be quite √10, but I did expect them to have a more regular behavior. That down and up is a bit of a shock.
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Posted by Jer
on 2015-07-05 16:07:34 |
re(2): more results (11 and 12)
