16. Extra Crocodiles (from 9th grade, 6 points). A chocolate egg manufacturer with a toy inside announced that a new collection is being released, featuring ten different crocodiles. The crocodiles are uniformly and randomly distributed among the chocolate eggs, meaning that in a randomly chosen egg, each crocodile can appear with a probability of 0.1. Leshа wants to collect the entire collection. Every day, his mother buys him one chocolate egg with a crocodile.

At first, a crocodile with glasses appeared in Leshа's collection, then a crocodile with a newspaper. The third unique specimen was a crocodile with a cane. Find the mathematical expectation of the random variable "the number of crocodiles with a cane that Leshа had by the time he completed his collection."
Answer: approximately 3.59.

Solution. Let's assume there are a total of $n$ crocodiles in the collection. We will number the exhibits in the order in which they appeared in the collection. For example, the crocodile with glasses was the first, so it is a crocodile of type 1. After some number of attempts, a crocodile of type 2 appeared, which had not been seen before (with a newspaper), and so on.

Consider the moment when the $k$-th crocodile appeared in the collection. We are interested in the expected number of crocodiles of this type that will be purchased before the collection is completed.

Starting from the moment when the $k$-th crocodile appeared, it will take $\varphi_{k}$ purchases for the next missing exhibit (number $k+1$) to appear. The probability that it will appear in each separate attempt is $\frac{n-k}{n}$. Therefore, $\mathrm{E} \varphi_{k}=\frac{n}{n-k}$, and the number of "extra" crocodiles that will appear on average is

$$
\mathrm{E} \varphi_{k}-1=\frac{n}{n-k}-1=\frac{k}{n-k}
$$

Of these extra $\frac{k}{n-k}$ crocodiles, some (let $\xi_{k}$) belong to type $k$, and since in each attempt each of the $k$ types of extra crocodiles appears with equal probability

$$
\mathrm{E} \xi_{k}=\frac{1}{k} \cdot \frac{k}{n-k}=\frac{1}{n-k}
$$

Similarly, between exhibits number $k+1$ and $k+2$, on average

$$
\mathrm{E} \varphi_{k+1}-1=\frac{n}{n-k-1}-1=\frac{k+1}{n-k-1}
$$

extra crocodiles will appear, of which on average $\mathrm{E} \xi_{k+1}=\frac{1}{n-k-1}$ belong to type $k$, and so on.

Therefore, the total number $X_{k}$ of extra crocodiles of type $k$ is

$$
X_{k}=\xi_{k}+\xi_{k+1}+\xi_{k+2}+\ldots+\xi_{n-1}
$$

and the expected value of this quantity is

$$
\mathrm{E} X_{k}=\frac{1}{n-k}+\frac{1}{n-k-1}+\ldots+1=H_{n-k}
$$

where $H$ denotes the harmonic number: $H_{m}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{m}$.

Adding to all these extra crocodiles of type $k$ the first such crocodile (not extra), we find that the average number of such crocodiles is $H_{n-k}+1$.

In the conditions of the problem, $n=10, k=3$, so the expected number of crocodiles with a cane, that is, crocodiles of type 3, is $H_{7}+1 \approx 2.59+1=3.59$.

Remark. It is known that $H_{m} \approx \ln m+\gamma$, where $\gamma=0.577215664 \ldots$ is the Euler-Mascheroni constant. Using this formula, we find:

$$
\mathrm{E} X_{k}+1 \approx \ln 7+0.578+1 \approx 3.524
$$

The accuracy is acceptable.

![](https://cdn.mathpix.com/cropped/2024_05_06_142b35c881fc974fcd25g-13.jpg?height=132&width=1339&top_left_y=134&top_left_x=381)

## Alexandra Nesterenko

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