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
more results by Daniel)
11-sided die:
Probabilities:
{0.687834,0.212301,0.068419,0.0218153,0.00677347,0.00203704,0.000592617,0.000166863,0.0000455342,0.000012062,3.10704*10^-6}
Ratios:
{3.2399,3.10295,3.13628,3.2207,3.32515,3.43737,3.55151,3.66457,3.77502,3.88215}
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}
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Posted by Daniel
on 2015-07-04 09:45:05 |
re: more results (11 and 12)
