# Combinatorics: likelihood of a uniform draw

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?

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Find the probability that, say, Type A is picked 8 times, 9 times, 10 times, 11 times and 12 times. Add those up. Put the sum to the power of 10 and there's your answer. –  Imray Oct 12 '12 at 15:30
wouldn't this mean that there is a probability to draw 12 pebbles of each kind, while we only draw 100 pebbles? –  Alexis Oct 12 '12 at 15:43
@Alexis: Yes, that, and it would assume that these events are independent, which they aren't. –  joriki Oct 12 '12 at 15:58
Do we draw with or without replacement? –  joriki Oct 12 '12 at 16:01
without replacement (updated) –  Alexis Oct 12 '12 at 17:11

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

$$0.03\approx 856945\cdot\frac{\binom{100}{10}^{10}}{\binom{1000}{100}^{\hphantom{10}}}\gt p\gt 856945\cdot\frac{\binom{100}{12}^5\binom{100}8^5}{\binom{1000}{100}}\approx0.004\;.$$

 226031412377730730814344253428220298277915460779610728832457924491489212422618433457300376001429754322127222112213012269223936000000000

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).
So the geometric mean of the bounds $0.03$ and $0.004$, which is about $0.011$, isn’t a bad estimate. –  Brian M. Scott Oct 13 '12 at 4:04