## Task 3B - 231233B

Two people A and B play the following game:

20 cards, each marked with one of the numbers $1, 2, 3, \ldots, 20$ (with each of these numbers appearing exactly once), are laid face up on the table so that the numbers are visible. A has mentally selected two of these cards without B knowing which cards they are. B now tries to identify these two cards as follows: In the first move, B takes two cards of his choice, and A tells him how many of these two cards are correct (0, 1, or 2 cards).

B then lays these cards back face up.

If these were not the two correct cards, B takes two more cards of his choice in the second move, and A tells him how many of these are correct; B then lays these cards back.

This procedure is continued with the 3rd, 4th, ... move until B takes the two correct cards in one of these moves.

B wins if he takes the two correct cards at the latest by the 12th move. In one execution of this game, B starts the game with the following strategy: He takes the cards 1, 2 in the 1st move, and if these are not the two correct cards, he takes the cards 3, 4 in the 2nd move, and so on, if in none of the previous moves the two correct cards (simultaneously in one and the same move) appeared, he takes the cards (17, 18) in the 9th move.

a) Provide a continuation strategy for the subsequent moves for this strategy started by B, with which B can enforce a win.

b) What is the probability that B, with the given strategy, will even take the two correct cards at the latest by the 11th move?
a) There are exactly the following two cases:

1. Case: Among the pairs $(1,2),(3,4), \ldots,(19,20)$, the correct pair is included.

In this case, $B$ either takes the pair with the correct cards in at most 9 moves, or he can deduce from the "Zero" answers to the first 9 moves that $(19,20)$ is the correct pair. He takes it on the 10th move and thus wins the game.

2. Case: Among the pairs $(1,2),(3,4), \ldots,(19,20)$, there are exactly two pairs, each containing one correct card.

In this case, $B$ knows at the latest after the 9th move which pairs these are.

Let these pairs be $\left(a_{1}, a_{2}\right)$ and $\left(a_{3}, a_{4}\right)$ (with $a_{1}<a_{2}<a_{3}<a_{4}$). The following four cases are possible (the notation $W$ under $a_{i}$ means that $a_{i}$ is one of the correct cards; the notation $F$ means that $a_{i}$ is not one of the correct cards):

| Case | $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ |
| :---: | :---: | :---: | :---: | :---: |
| 2.1. | $\mathrm{W}$ | $\mathrm{F}$ | $\mathrm{W}$ | $\mathrm{F}$ |
| 2.2. | $\mathrm{W}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{W}$ |
| 2.3. | $\mathrm{F}$ | $\mathrm{W}$ | $\mathrm{W}$ | $\mathrm{F}$ |
| 2.4. | $\mathrm{F}$ | $\mathrm{W}$ | $\mathrm{F}$ | $\mathrm{W}$ |

Now $B$ takes the cards $\left(a_{1}, a_{3}\right)$ on the 10th move.

In case 2.1., he has thus taken the correct cards and won the game.

In case 2.4., he learns from the "Zero" answer to the 10th move that $\left(a_{2}, a_{4}\right)$ is the correct pair. He takes it on the 11th move and thus wins.

In cases 2.2. and 2.3., which are characterized by the "One" answer to the 10th move, $B$ takes the cards $\left(a_{1}, a_{4}\right)$ on the 11th move. If case 2.2. was true, $B$ has thus won. But if case 2.3. was true, he learns this from the "Zero" answer to the 11th move. He takes the cards $\left(a_{2}, a_{3}\right)$ on the 12th move and thus wins.

By the given strategy, $B$ thus enforces the win.

b) For each pair $\left(z_{1}, z_{2}\right)$ of two different numbers from $1, \ldots, 20$, the probability that the number $z_{1}$ is one of the two correct numbers is $P_{1}=\frac{2}{20}$; the probability that then $z_{2}$ is the other correct number is $P_{2}=\frac{1}{19}$.

Thus, the probability that $\left(z_{1}, z_{2}\right)$ is the correct pair is $P_{1} \cdot P_{2}=\frac{1}{190}$.

Therefore, the probability that the correct pair is among the pairs $(1,2), \ldots,(19,20)$ (Case 1) is $10 \cdot \frac{1}{190}=\frac{1}{19}$.

Thus, the probability that Case 2 occurs is $1-\frac{1}{19}=\frac{18}{19}$. The four cases 2.1. to 2.4. have equal probability. $B$ has not yet achieved the win after 11 moves exactly when Case 2.3. is true. The probability for this is thus $\frac{1}{4} \cdot \frac{18}{19}=\frac{9}{38}$.

Therefore, the probability that $B$ achieves the win at the latest after 11 moves is $1-\frac{9}{38}=\frac{29}{38} \approx 0.763$.

Adapted from $[5]$