An urn contains 10 kinds of pebbles, and 100 pebbles of each kind. We draw 100 pebbles (without replacement). What is the probability that we get between 8 and 12 pebbles of each kind?
Combinatorics: likelihood of a uniform draw
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0without replacement (updated) – 2012-10-12
1 Answers
The most likely of the admissible combinations is the completely uniform one, with a probability of
$ \frac{\binom{100}{10}^{10}}{\binom{1000}{100}^{\hphantom{10}}}\approx3.8\cdot10^{-8} $
(computation). Presumably the most unlikely of the admissible combinations is the most non-uniform one with $12$ pebbles of five kinds and $8$ of the others, with a probability of
$ \frac{\binom{100}{12}^5\binom{100}8^5}{\binom{1000}{100}}\approx4.5\cdot10^{-9} $
(computation). The number of admissible combinations can be calculated using the formula at the bottom of this page as the number of ways of distributing $k=20$ excess pebbles over $m=10$ kinds with a capacity of $R=4$ each, which yields
$ \sum_{t=0}^4(-1)^t\binom{10}t\binom{29-5t}9=856945 $
(computation). Thus the desired probability $p$ satisfies
$ 0.03\approx 856945\cdot\frac{\binom{100}{10}^{10}}{\binom{1000}{100}^{\hphantom{10}}}\gt p\gt 856945\cdot\frac{\binom{100}{12}^5\binom{100}8^5}{\binom{1000}{100}}\approx0.004\;. $
The exact answer is
226031412377730730814344253428220298277915460779610728832457924491489212422618433457300376001429754322127222112213012269223936000000000 ————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————— 18724490300969246403723903560710344364006715745998044509175019419801870086437214647777411833371759499800416660391039940703106220925825529
or about $0.012$, as computed by this code, which enumerates all admissible combinations and checks the result with a simulation (and also checks the number of admissible combinations).
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0@Alexis: You're welcome! – 2012-10-13