If we've got 10 coupons, what is expected number of different ones if there are 25 different types

Question

I can't figure out this problem :

There are 25 different types of coupon, all equally probable to get. If we have got 10 coupons, what is expected number of different coupons between them?

Solution is : E(x) = 8.38

Answer 1

Using Stirling numbers we get

$$\frac{1}{25^{10}} \sum_{q=1}^{10} q\times {25\choose q} \times {10\brace q} \times q!$$

which gives the expectation

$${\frac {31964050675249}{3814697265625}} \approx 8.379184100.$$

What we have done here is choose the $q$ different values from the $25$ different ones where we can obviously represent at most ten different values. We then partition the $10$ slots into $q$ sets and choose a permutation of the $q$ that gives the value assigned to each set. Finally multiply by $q$ because we are computing the expectation of the number of different values which is precisely $q.$

Observe also that the probability of a coupon number $p$ not appearing is $$\frac{24^{10}}{25^{10}}$$ which yields for the expectation

$$\sum_{p=1}^{25} \left(1-\frac{24^{10}}{25^{10}}\right) = {\frac {31964050675249}{3814697265625}}$$

the same as before.

Answer 2

Let $X_i$ be an indicator random variable $=1$ if the $i^{th}$ coupon is present, and $=0$ otherwise.

Then $P[i^{th}$ coupon is present$] = [1 - (\frac{24}{25})^{10}]$
Now the expectation of an indicator r.v. is just the probability of the event it indicates, so $E[X_i] = [1 - (\frac{24}{25})^{10}]$

By linearity of expectation we have expectation of sum = sum of expectations,
$E[\sum{(X_i)}] = \sum{E(X_i)} = 25[1 - (\frac{24}{25})^{10}] \approx 8.38$