Problem with a multivariable stochastic exercise

Question

Hey I have this problem where I want to check my solutions:

At the 2024 European Football Championship, 24 teams, each with a squad of 23 players, will compete against each other. Each player has a collection card. At the kiosk, a picture is sold in a privacy-protected package for ten cents. How much will a collector spend on average at the kiosk until he has (at least) one picture of each player? Hint: Let $N_i$ be the number of images acquired until you have one image from $i$ players. Determine the distribution of $D_i = N_i − N_{i−1}$.

What I have done is:

I have analyzed the acquisition of the pictures step by step.

Let $N$ be the total number of pictures a collector needs to buy in order to have at least one picture of each player. Our goal is to calculate the expected value $E(N)$.

Let's start with the acquisition of the first picture. Since the collector has no pictures initially, they will definitely have to buy one picture. Therefore, the distribution of $D_1$ is:

$P(D_1 = 1) = 1$ (the collector needs to buy one picture)

Next, we consider the acquisition of the second picture. There are two cases to consider:

The collector gets a new picture that they don't already have. This happens with a probability of $22/23$ because there are $23$ players remaining from whom they don't have a picture, out of a total of $24$ players.

The collector gets a picture they already have. This happens with a probability of $1/23$ because there is only one player left from whom they don't have a picture.

Therefore, the distribution of $D_2$ is:

$P(D_2 = 1) = 23/24$ (the collector gets a new picture) $P(D_2 = 0) = 1/24$ (the collector gets a picture they already have)

We can continue this approach for the remaining pictures by taking into account the number of players still missing.

In general, for $D_i, i > 1$:

$P(D_i = 1) = \frac{24-i+1}{24}$ (the collector gets a new picture) $P(D_i = 0) = \frac{i-1}{24}$ (the collector gets a picture they already have)

To calculate the expected value $E(N)$, we sum up the expected values of the individual $D_i$:

$E(N) = E(D_1) + E(D_2) + ... + E(D_23)$

$= 1 + (23/24) + (22/24) + ... + (1/24)$

By calculating this sum, we can determine the average amount a collector is expected to spend at the kiosk until they have at least one picture of each player.

I am really not sure about the results so far.

My biggest doubt is that my calculations do not take into account the fact that perhaps card $i$ has already been acquired by the collector in the previous packages to obtain card $j, (j<i)$. However, I dont know if I am right and how to do it. Can someone help me?