Solution

Answer: 6.818741802

Let $X$ be the number of distinct colours among the 20 balls drawn.

We want:

$$\mathbb{E}[X]$$

There are 7 colours total, each appearing 10 times.

Step 1: Indicator variables

Define indicator variables

$$I_k = \begin{cases} 1 & \text{if colour } k \text{ appears at least once} \ 0 & \text{otherwise} \end{cases}$$

Then

$$X = I_1 + I_2 + \cdots + I_7$$

Using linearity of expectation:

$$\mathbb{E}[X] = \sum_{k=1}^7 \mathbb{E}[I_k] = 7\mathbb{E}[I_1]$$

since all colours are symmetric.

Now,

$$\mathbb{E}[I_1] = P(\text{a given colour appears})$$

So:

$$\mathbb{E}[X] = 7 \cdot P(\text{given colour appears})$$

Step 2: Compute the probability a colour appears

Instead of computing directly, compute the complement.

A particular colour is absent iff all 20 chosen balls come from the remaining 60 balls.

Total ways to choose 20 balls from 70:

$$\binom{70}{20}$$

Ways to choose 20 balls avoiding one colour:

$$\binom{60}{20}$$

Therefore

$$P(\text{given colour absent}) = \frac{\binom{60}{20}}{\binom{70}{20}}$$

Hence

$$P(\text{given colour appears}) = 1 - \frac{\binom{60}{20}}{\binom{70}{20}}$$

Thus

$$\boxed{ \mathbb{E}[X] = 7\left(1-\frac{\binom{60}{20}}{\binom{70}{20}}\right) }$$

Step 3: Numerical evaluation

We now evaluate

$$7\left(1-\frac{\binom{60}{20}}{\binom{70}{20}}\right)$$

Using exact arithmetic:

$$\frac{\binom{60}{20}}{\binom{70}{20}} = \prod_{k=0}^{19}\frac{60-k}{70-k}$$

Numerically,

$$\mathbb{E}[X] \approx 6.818741802$$

rounded to nine digits after the decimal point.

Python implementation

from math import comb

# Total ways to choose 20 balls from 70
total = comb(70, 20)

# Ways to choose 20 balls while avoiding one colour
without_colour = comb(60, 20)

# Expected number of distinct colours
expected = 7 * (1 - without_colour / total)

# Print with 9 digits after decimal point
print(f"{expected:.9f}")

Code walkthrough

from math import comb

Imports Python’s exact binomial coefficient function.

total = comb(70, 20)

Computes:

$$\binom{70}{20}$$

the total number of equally likely selections.

without_colour = comb(60, 20)

Computes:

$$\binom{60}{20}$$

the number of selections avoiding a particular colour.

expected = 7 * (1 - without_colour / total)

Implements:

$$7\left(1-\frac{\binom{60}{20}}{\binom{70}{20}}\right)$$

from the derivation above.

print(f"{expected:.9f}")

Prints the value rounded to exactly 9 decimal places.

Output:

6.818741802

Final verification

The result must lie between 1 and 7.

Since we draw 20 balls from 7 colours with many repetitions available, we should see almost all colours most of the time. Thus a value near 7 is reasonable.

The computed value:

$$6.818741802$$

fits this expectation.

Answer: 6.818741802

Project Euler Problem 493