Question

Suppose there are three balls in a bag. One ball is black and two others are white. Three people, A, B and C, will pick a ball in this order. Instead of deciding the winner by the first black ball, the person who picks the black ball for the second time will be the winner. For example, if A picks the black ball for his first pick, A is not the winner. He just returns it to the bag. And then if B picks the black ball, B will be the winner because that is the second time the black ball is picked. A will start this game again, followed by B and C in this order. Obtain the formulas for the probability of winning for each person A, B, and C and then get their numerical estimates.

Answer 1

Probability for A to winning the game

During his first turn A can pick a black ball or not pick a black ball

Case 1: A picks a black ball in his 1^st attempt

Now for A to win, B and C should not pick up the black ball in their respective chances and A should pick up the black ball in the 2^nd attempt

Probability that A picks a black ball in his 1^st attempt = 1/3

Probability that B picks doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C picks doesn't pick a black ball in his 1^st attempt = 2/3

Probability that A picks a black ball in his 2^nd attempt = 1/3

So Probability that A will be the winner in Case 1 = 1/3 * 2/3 * 2/3 * 1/3

= 4 / 81

Case 2: A does not pick a black ball in his 1^st attempt

Since A will not pick up a black ball in his 1^st attempt, either B or C should pick up the black ball in their 1^st attempt so then A can pick the black ball in his 2^nd attempt

Case 2(a)

B picking a ball ball in his 1^st attempt and C not picking a black ball

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B picks a black ball in his 1^st attempt = 1/3

Probability that C doesn't pick a black ball in his 1^st attempt = 2/3

Probability that A picks a black ball in his 2^nd attempt = 1/3

So Probability that A will be the winner in Case 2a = 2/3 * 1/3 * 2/3 * 1/3

= 4 / 81

Case 2(b)

B picking a white ball in his 1^st attempt and C picking a black ball

Probability that A will be the winner in Case 2b = 4/ 81 (probabilities will be same as that of case 2(a) with probabilities of B and C exhanging)

So total Probability that A will be the winner = 4/81 + 4/81 + 4/81 = 12/ 81 = 0.1481

Probability for B to winning the game

Case 1: A picks a black ball in his 1^st attempt

B can pick up the ball in his 1^st attempt and can win the game

Probability that A picks a black ball in his 1^st attempt = 1/3

Probability that B picks a black ball in his 1^st attempt = 1/3

So Probability that B will be the winner in Case 1 = 1/3 * 1/3 = 1/9

Case 2: A does not pick a black ball in his 1^st attempt

Case 2a : B picks a black ball in his 1^st attempt

B will win the game C ans A doesn't pick a black ball in their 1^st and 2^nd attempt respectively, B picks the black ball in his 2^nd attempt

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B picks a black ball in his 1^st attempt = 1/3

Probability that C doesn't pick a black ball in his 1^st attempt = 2/3

Probability that A does not pick a black ball in his 2^nd attempt = 2/3

Probability that B picks a black ball in his 2^nd attempt = 1/3

So Probability that A will be the winner in Case 2a = 2/3 * 1/3 * 2/3 * 2/3 * 1/3

= 8/ 243

Case 2b : B doesn't picks a black ball in his 1^st attempt

Since B will not pick up a black ball in his 1^st attempt, either C or A should pick up the black ball in their 2^nd and 1^st attempt respectively so then B can pick the black ball in his 2^nd attempt

Case 2b(i) C picking a black ball and A not picking black ball in his both attempts

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C picks a black ball in his 1^st attempt = 1/3

Probability that A does not pick a black ball in his 2^nd attempt = 2/3

Probability that B picks a black ball in his 2^nd attempt = 1/3

So Probability that B wins = 8/ 243

Case 2b(ii) A picking a black ball in the 2^nd attempt and C not picking black ball in his 1^st attempt

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C doesn't pick a black ball in his 1^st attempt = 2/3

Probability that A picks a black ball in his 2^nd attempt = 1/3

Probability that B picks a black ball in his 2^nd attempt = 1/3

So Probability that B wins = 8/ 243

So total probability that B wins the game = 1/9 + 3 * (8/243)

= 1/9 + 8/81

= 17/81 = 0.2099

Probability for C to winning the game

Case 1 : Either A or B picking a black ball in their 1^st attempt and C picking the black ball in the 1^st attempt

Case 1 (a)

A picks the black ball and B doesn't pick the black ball

Probability that A picks a black ball in his 1^st attempt = 1/3

Probability that B doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C picks a black ball in his 1^st attempt = 1/3

Probability that C wins the game = 2/27

Case 2 : Both A and B not picking a black ball in their 1^st attempt and C picking the black ball in the 1^st attempt

So Both A and B should not pick a black ball in their 2^nd attempt too so C can pick the black in 2^nd attempt and win the game

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C picks a black ball in his 1^st attempt = 1/3

Probability that A does not pick a black ball in his 2^nd attempt = 2/3

Probability that B doesn't pick a black ball in his 2^nd attempt = 2/3

Probability that C picks a black ball in his 2^nd attempt = 1/3

Probability that C wins the game = 2/3 * 2/3 * 1/3 * 2/3 * 2/3 * 1/3

= 16 / 729

Case 3 : A, B and C doesn't pick a black ball in their 1^st attempt and either of A or B pick a black ball in 2^nd attempt

Case 3a ; A picks a black ball in 2^nd attempt and B doesn't pick a black ball in 2^nd attempt

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C doesn't pick a black ball in his 1^st attempt = 2/3

Probability that A picks a black ball in his 2^nd attempt = 1/3

Probability that B doesn't pick a black ball in his 2^nd attempt = 2/3

Probability that C picks a black ball in his 2^nd attempt = 1/3

Probability that C wins the game = 2/3 * 2/3 * 2/3 * 1/3 * 2/3 * 1/3

= 16 / 729

Case 3b ; A doesn't picks a black ball in 2^nd attempt and B picks a black ball in 2^nd attempt

Probability that A doesn't pick a black ball in his 1^st attempt = 2/3

Probability that B doesn't pick a black ball in his 1^st attempt = 2/3

Probability that C doesn't pick a black ball in his 1^st attempt = 2/3

Probability that A does't pick a black ball in his 2^nd attempt = 2/3

Probability that B picks a black ball in his 2^nd attempt = 1/3

Probability that C picks a black ball in his 2^nd attempt = 1/3

Probability that C wins the game = 2/3 * 2/3 * 2/3 * 2/3 * 1/3 * 1/3

= 16 / 729

So total Probability that C wins the game = 2/27 + 3 * (16/729)

= 2/27 + 16/243

= 34/ 243

So total Probability that A will be the winner = 12/ 81 = 0.1481

So total probability that B wins the game = 17/81 = 0.2099

So total probability that C wins the game = 34/243 = 0.1399