Two players take turns removing balls from a box containing three white and four red balls.

Two players take turns removing balls from a box containing three white and four red balls. The winner is the one who first draws the white ball. Find the probability of winning for each player. Consider two cases: a) the removed ball is returned to the box; b) the ball is not returned to the box.

1. Given:

n1 = 3 white balls;
n2 = 4 red balls;
n = 7 balls in total.
2. Let’s find the probabilities of events A and B – the payoff of each player, respectively:

a) the removed ball is returned to the box:

P (A) = 3/7 + 4/7 * 3/6 * (3/5 + 2/5 * 1/4);
P (A) = 3/7 + 4/7 * 1/2 * (6/10 + 1/10);
P (A) = 3/7 + 4/7 * 1/2 * 7/10 = 3/7 + (4 * 1 * 7) / (7 * 2 * 10) = 15/35 + 7/35 = 22 / 35;
P (B) = 1 – P (A) = 13/35.
b) the ball is not returned to the box:

P (A) [1] = 3/7;
P (A) [2] = (4/7) ^ 2 * P (A) [1];
P (A) [i + 1] = (4/7) ^ 2 * P (A) [i];
b1 = 3/7; q = (4/7) ^ 2;
P (A) = b1 / (1 – q ^ 2) = 3/7: (1 – (4/7) ^ 2) = 3/7: (1 – 16/49) = 3/7: 33/49 = 3/7 * 49/33 = (3 * 49) / (7 * 33) = 7/11;
P (B) = 4/11.
Answer:

a) P (A) = 22/35; P (B) = 13/35;
b) P (A) = 7/11; P (B) = 4/11.