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: computer aided solution by Jer)
Concerning the ratio of the probability of a given number being the result, to that of the next number, hypothesized as sqrt(10) ~= 3.16227766016838, note the ratio of the probability of 1 to that of 2, as the number of sides progresses from 2 to 6 is:
3
3.166666666666667
3.217934165720772
3.233459959104922
3.238064338219506
getting farther from sqrt(10).
Ratio of p(2) to p(3) as m progresses from 3 to 6 follows a different trajectory:
3
3.069686411149826
3.092602637934237
3.099855399609466
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Posted by Charlie
on 2015-07-02 15:34:07 |
re(2): computer aided solution
