Two people, A and B, each have a bag containing three cards numbered 1, 2, and 3. They each draw a 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 game. (2) The game is repeated 4 times, with the drawn cards replaced each time. (i) Find the probability that A wins at least 3 times. (ii) Find the probability that A wins once, loses once, and draws twice. (iii) Find the probability that A and B win the same number of times. Then, find the probability that A wins more times than B.

1. Problem Description

Two people, A and B, each have a bag containing three cards numbered 1, 2, and

3. They each draw a 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 game.

(2) The game is repeated 4 times, with the drawn cards replaced each time.

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

(ii) Find the probability that A wins once, loses once, and draws twice.

(iii) Find the probability that A and B win the same number of times. Then, find the probability that A wins more times than B.

2. Solution Steps

(1) Probability of a draw in one game:

There are a total of $3 \times 3 = 9$ possible outcomes when A and B each draw a card. The draws occur when both players draw the same number. This happens when both draw 1, both draw 2, or both draw

3. Thus, there are 3 possible draws. Therefore, the probability of a draw is $3/9 = 1/3$.

$L = 1$

$M = 3$

(2) The game is repeated 4 times. We need to calculate the probability of A winning, losing or a draw in one round.

The probability of a draw is $1/3$. The probability of not a draw is $1 - 1/3 = 2/3$.

For A to win, A's number must be greater than B's number. The total number of cases is

9. The cases when A wins are (A=2, B=1), (A=3, B=1), (A=3, B=2). So there are 3 cases where A wins. The probability of A winning is $3/9 = 1/3$. Similarly, the probability of B winning is $3/9 = 1/3$.

So, $P(A \ win) = 1/3$, $P(B \ win) = 1/3$, and $P(draw) = 1/3$.

(i) Probability that A wins at least 3 times:

This can happen in two ways: A wins 3 times or A wins 4 times.

P(A wins 3 times) = $C(4,3) * (1/3)^3 * (2/3)^1 = 4 * (1/27) * (2/3) = 8/81$

P(A wins 4 times) = $C(4,4) * (1/3)^4 * (2/3)^0 = 1 * (1/81) * 1 = 1/81$

P(A wins at least 3 times) = $8/81 + 1/81 = 9/81 = 1/9$

$N = 1$

$O = 9$

(ii) Probability that A wins 1 time, loses 1 time, and draws 2 times:

P(A wins 1, B wins 1, draw 2) = $C(4,1,1,2) * (1/3)^1 * (1/3)^1 * (1/3)^2 = \frac{4!}{1!1!2!} * (1/3)^4 = \frac{24}{2} * \frac{1}{81} = 12/81 = 4/27$

$P = 4$

$QR = 27$

(iii) Probability that A and B win the same number of times:

Since there are 4 games, A and B can each win 0, 1, or 2 games.

Case 1: A wins 0, B wins 0, draw 4: $C(4,0,0,4) * (1/3)^0 * (1/3)^0 * (1/3)^4 = 1/81$

Case 2: A wins 1, B wins 1, draw 2: $C(4,1,1,2) * (1/3)^1 * (1/3)^1 * (1/3)^2 = 12/81$

Case 3: A wins 2, B wins 2, draw 0: $C(4,2,2,0) * (1/3)^2 * (1/3)^2 * (1/3)^0 = \frac{4!}{2!2!0!} * \frac{1}{81} = 6/81$

P(A and B win the same number of times) = $1/81 + 12/81 + 6/81 = 19/81$

The probability that A wins more than B is the same as the probability that B wins more than A because $P(A \ win) = P(B \ win) = 1/3$.

Let x be the probability of A winning more than B and y be the probability of A and B winning equal times. Let z be the probability of B winning more than A.

Since these are all possibilities, $x + y + z = 1$. Since $x=z$, $2x + y = 1$.

$2x = 1 - y = 1 - 19/81 = 62/81$.

$x = 31/81$.

$ST = 19$

$UV = 81$

$WX = 31$

$UV = 81$

3. Final Answer

(1) $L/M = 1/3$

(2) (i) $N/O = 1/9$

(ii) $P/QR = 4/27$

(iii) $ST/UV = 19/81$, $WX/UV = 31/81$