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My friend posed me the following question:

We have a bowl with 70 balls, 7 colors, and 10 balls in each color.

You draw 20 balls simultaneously from the bowl. What is the expected number of distinct colors?

I tried working out this problem for $2$, and $3$ colours the rest seemed difficult.

2: Here we need all the balls from 2 distinct colours. This can be done in $7P2 = 7 \cdot 6$ ways.

**3: ** For three balls this got more interesting we have the restraints $i + j + k = 20$, $i \in [1,10]$, $j = [1,10]$, $k \in [1,10]$. Also we have $i + j \geq 10$. This gives me

$$ \sum_{i = 1}^{10} \left( \sum_{j = \min(10-i,1)}^{9} 1\right) = 54 $$ Multiplying this with the number of ways to pick $3$ colors from $7$ I get $$ P(\text{3 distinct colors}) = 54\cdot (7\text{Pn} 3) = 11340 $$

I tried to solve this problem numerically, and it seems that the expected number of different balls is about $$ E(b) \approx 6.818835818835818685101912706159055233001708984375 $$

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Each colour has probability

$$ 1-\frac{\binom{60}{20}}{\binom{70}{20}} $$

of being drawn, so by linearity of expectation the expected number of colours drawn is

$$ 7\left(1-\frac{\binom{60}{20}}{\binom{70}{20}}\right)=\frac{763700091}{112000148}\approx6.81874\;. $$

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