Two people, A and B, each have a bag containing three cards with the numbers 1, 2, and 3 written on them. They simultaneously draw one card from their bag and compare the numbers. If the numbers are the same, it's a draw. If the numbers are different, the person with the larger number wins. (1) Find the probability of a draw in one round. (2) If this game is played 4 times, with the drawn card being returned to the bag each time, find: (i) The probability that A wins at least 3 times. (ii) The probability that A wins once, loses once, and there are two draws. (iii) The probability that A's number of wins equals B's number of wins, and the probability that A's number of wins is greater than B's number of wins.

1. Problem Description

Two people, A and B, each have a bag containing three cards with the numbers 1, 2, and 3 written on them. They simultaneously draw one card from their bag and compare the numbers. If the numbers are the same, it's a draw. If the numbers are different, the person with the larger number wins.

(1) Find the probability of a draw in one round.

(2) If this game is played 4 times, with the drawn card being returned to the bag each time, find:

(i) The probability that A wins at least 3 times.

(ii) The probability that A wins once, loses once, and there are two draws.

(iii) The probability that A's number of wins equals B's number of wins, and the probability that A's number of wins is greater than B's number of wins.

2. Solution Steps

(1) The possible outcomes of a single round are: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3). There are a total of $3 \times 3 = 9$ possible outcomes. The draws are (1,1), (2,2), (3,3). Therefore, the number of draws is

3. The probability of a draw is $\frac{3}{9} = \frac{1}{3}$.

$L = 1$ and $M = 3$.

(2)

In a single round, let the probability of A winning be $p$, the probability of A losing be $q$, and the probability of a draw be $r$.

Since the game is symmetric, $p = q$. We also know that $p + q + r = 1$.

We found $r = \frac{1}{3}$, so $p + q = \frac{2}{3}$. Since $p = q$, we have $2p = \frac{2}{3}$, which means $p = \frac{1}{3}$ and $q = \frac{1}{3}$.

(i) We want to find the probability that A wins at least 3 times in 4 rounds. This means A can win 3 times or 4 times.

The probability of A winning exactly 3 times is given by the binomial probability formula:

$P(A=3) = \binom{4}{3} p^3 (1-p)^{4-3} = \binom{4}{3} (\frac{1}{3})^3 (\frac{2}{3})^1 = 4 \times \frac{1}{27} \times \frac{2}{3} = \frac{8}{81}$

The probability of A winning exactly 4 times is:

$P(A=4) = \binom{4}{4} p^4 (1-p)^{4-4} = \binom{4}{4} (\frac{1}{3})^4 (\frac{2}{3})^0 = 1 \times \frac{1}{81} \times 1 = \frac{1}{81}$

So the probability of A winning at least 3 times is:

$P(A \ge 3) = P(A=3) + P(A=4) = \frac{8}{81} + \frac{1}{81} = \frac{9}{81} = \frac{1}{9}$

$N = 1$ and $O = 9$.

(ii) We want to find the probability that A wins once, loses once, and there are two draws.

The number of ways to arrange 1 win, 1 loss, and 2 draws is given by $\frac{4!}{1!1!2!} = \frac{24}{2} = 12$.

The probability of each arrangement is $p^1 q^1 r^2 = (\frac{1}{3})^1 (\frac{1}{3})^1 (\frac{1}{3})^2 = (\frac{1}{3})^4 = \frac{1}{81}$.

Therefore, the probability of A winning once, losing once, and having two draws is:

$12 \times \frac{1}{81} = \frac{12}{81} = \frac{4}{27}$.

$P = 4$, $Q = 2$, and $R = 7$. Thus the probability is $\frac{4}{27}$.

(iii) A's number of wins equals B's number of wins. Since there are 4 rounds, this means A and B both win 0, 1, or 2 times.

Case 1: A and B both win 0 times. This means all 4 rounds are draws. The probability is $r^4 = (\frac{1}{3})^4 = \frac{1}{81}$.

Case 2: A and B both win 1 time. This means there are 2 draws. The number of arrangements is $\frac{4!}{1!1!2!} = 12$. The probability is $12 \times (\frac{1}{3})^1 (\frac{1}{3})^1 (\frac{1}{3})^2 = 12 \times \frac{1}{81} = \frac{12}{81}$.

Case 3: A and B both win 2 times. This means there are 0 draws. The number of arrangements is $\frac{4!}{2!2!} = 6$. The probability is $6 \times (\frac{1}{3})^2 (\frac{1}{3})^2 = 6 \times \frac{1}{81} = \frac{6}{81}$.

The total probability is $\frac{1}{81} + \frac{12}{81} + \frac{6}{81} = \frac{19}{81}$.

Therefore, $ST = 19$ and $UV = 81$.

Since the only possibilities are A wins more than B, B wins more than A, or A and B win the same number of times, and the probabilities of A winning more and B winning more are the same, we have

$P(A > B) + P(B > A) + P(A = B) = 1$.

$P(A > B) = P(B > A)$.

$2 P(A > B) + P(A = B) = 1$.

$2 P(A > B) = 1 - \frac{19}{81} = \frac{62}{81}$.

$P(A > B) = \frac{31}{81}$.

Therefore, $WX = 31$.

3. Final Answer

L/M = 1/3

N/O = 1/9

P/QR = 4/27

ST/UV = 19/81

WX/UV = 31/81