11. The First Ace. In a certain card game, four players participate. To determine who starts the game, they usually draw lots "until the first ace." There are 32 cards in the deck, from sevens to aces. One of the players deals the cards face up to each player in turn, clockwise (to himself - last), until someone gets an ace. This player will start the game.

a) (8th grade. 3 points) For each player, find the probability that they will get the first ace.

b) (6th grade. 1 point). How can the rules be changed to make the draw fair?
Solution. a) It is naturally assumed that the deck is well shuffled, meaning each card can end up in any of the 32 possible positions with equal probability.

Let's assign the players numbers $1, 2, 3$, and 4. The dealer (4) starts with player (1). Player (1) will get the ace first if the first ace is in position $1, 5, 9, 13, 17, 21, 25$, or 29 in the deck.

An ace is the first card in the deck with a probability of $\frac{1}{8}$, since there are 8 different ranks in the deck from seven to ace.

The first ace will be in the 5th position if the first four cards are not aces. The probability of this is

$$
\frac{28}{32} \cdot \frac{27}{31} \cdot \frac{26}{30} \cdot \frac{25}{29} \cdot \frac{4}{28}=\frac{1}{8} \cdot \frac{28 \cdot 27 \cdot 26 \cdot 25}{31 \cdot 30 \cdot 29 \cdot 28}=\frac{1}{8} \cdot \frac{C_{28}^{4}}{C_{31}^{4}}
$$

Similarly, the first ace will be in the 9th position if the first eight cards are not aces. The probability of this is $\frac{1}{8} \cdot \frac{C_{28}^{8}}{C_{31}^{8}}$. And so on. Summing these probabilities, we find the probability that player (1) will start the game:

$$
p_{1}=\frac{1}{8}+\frac{C_{28}^{4}}{8 C_{31}^{4}}+\frac{C_{28}^{8}}{8 C_{31}^{8}}+\ldots+\frac{C_{28}^{28}}{8 C_{31}^{28}}=\frac{1}{8} \sum_{k=1}^{7} \frac{C_{28}^{4 k}}{C_{31}^{4 k}}
$$

For player (2), a similar calculation gives:

$$
p_{2}=\frac{C_{28}^{1}}{8 C_{31}^{1}}+\frac{C_{28}^{5}}{8 C_{31}^{5}}+\frac{C_{28}^{9}}{8 C_{31}^{9}}+\ldots+\frac{C_{28}^{25}}{8 C_{31}^{25}}=\frac{1}{8} \sum_{k=1}^{6} \frac{C_{28}^{4 k+1}}{C_{31}^{4 k+1}}
$$

Notice that this sum has one fewer term than the first. For players (3) and (4), we similarly get:

$$
p_{3}=\frac{1}{8} \sum_{k=1}^{6} \frac{C_{28}^{4 k+2}}{C_{31}^{4 k+2}} \text { and } p_{4}=\frac{1}{8} \sum_{k=1}^{6} \frac{C_{28}^{4 k+3}}{C_{31}^{4 k+3}}
$$

For the calculation, any suitable tool can be used. For example, paper and pencil, although this is tedious. Calculation in Excel gives:

$$
p_{1} \approx 0.3008, p_{2} \approx 0.2647, \quad p_{3} \approx 0.2320 \text { and } p_{4} \approx 0.2024
$$

The draw is unfair—the chances of the player who deals the cards "to the first ace" are about one and a half times lower than those of the first player.

Answer: approximately $0.3008, 0.2647, 0.2320$, and 0.2024.

Comment. We can almost understand without calculation that the described draw is unfair. Imagine that there is 1 ace in the deck. Then the chances for everyone are the same and equal to 0.25. Now imagine that there are 32 aces in the deck. Then with probability 1, the first player immediately gets an ace. Thus, as the number of aces increases, the chances of the first player seem to increase. Therefore, with four aces, the chances of the first player should be greater than 0.25, and those of the others should be less. However, this is not a very convincing argument. The calculation performed gives an exact solution to the problem.

b) The simplest change, as it seems to us, is to deal cards until the ace of spades. Obviously, if the "lucky" card in the deck is unique (not four), then the chances for all players are equal.