9. Card Drawing. Thirty-six players play a game: from a deck of 36 playing cards, players take turns to randomly draw a card. If a player draws the ace of diamonds, the player wins; if another card is drawn, the player returns it to the deck, and the next player draws a random card. They draw cards in a circle: first the first player, then the second, and so on. If no one draws the ace of diamonds in the first round, the players draw cards in the same order in the second round. This continues until someone draws the ace of diamonds.

a) (From 7th grade, 2 points) Is this a fair game, that is, are the players' probabilities of winning equal?

b) (From 9th grade, 2 points) Find at least an approximate probability of the first player winning.
Solution. a) Let the probability of the first player winning be $p_{1}$. Suppose the first player did not draw the diamond ace on the first try. The probability of this is $35 / 36$. At this point, the second player becomes the first, and the probability that he wins becomes equal to $p_{1}$. Returning to the beginning of the game, we see that for the second player to win, the first player must miss first. Therefore, at the start of the game, the probability of the second player winning is $\frac{35}{36} p_{1}$, which is less than $p_{1}$.

Similarly, the probability of winning for each subsequent player is the same multiple less than the previous one: $p_{k+1}=\frac{35}{36} p_{k}$. The game is unfair.

b) For the sake of generality, let's assume there are $n$ players and $n$ cards in the deck. The probability of the first player winning is

$$
p_{1}=\frac{1}{n}+\left(\frac{n-1}{n}\right)^{n} \cdot \frac{1}{n}+\left(\frac{n-1}{n}\right)^{2 n} \cdot \frac{1}{n}+\ldots=\frac{1}{n} \cdot\left(1-\left(1-\frac{1}{n}\right)^{n}\right)^{-1}
$$

For $n=36$, we find that $p_{1} \approx 0.0436$.

Note. The obtained expression can be replaced by an approximate one, which gives a small error for large $n$. It is known that $(1-1 / n)^{n} \approx 1 / e$, where $e=2.718281...$ is the base of the natural logarithm ${ }^{4}$. The larger $n$ is, the more accurate the equality. Then

$$
p_{1} \approx \frac{1}{n}\left(1-e^{-1}\right)^{-1}=\frac{e}{n(e-1)}
$$

For $n=36$, we get: $p_{1} \approx \frac{2.72}{36 \cdot 1.72} \approx 0.0439$.

Answer: a) no; b) approx. 0.044.