# Problem on probability

1. Apr 7, 2015

### stephenranger

1. The problem statement, all variables and given/known data
Three friends play a game in which one picks blind–folded from a bag containing white and
black balls. In the bag there are four black balls and one white ball. The player
whose turn it is picks one ball. If the ball is white the player has won; otherwise
the ball is returned to the bag and the next player gets the turn. The turn rotates
until the white ball is picked.
a) What is the probability that the game ends before any of the players has
picked twice?
b) Let the players be A, B, and C, in this order. What is each player’s probability
of winning the game?

2. Relevant equations

3. The attempt at a solution
The following is my solution. But I'm not sure if it is correct. Please correct me if there's any mistake. Thanks.

a)
The probability of the 1st player picks the white ball is P1 = 1/5
The probability of the 1st player picks a black ball and then the 2nd player picks the white ball is P2 = (4/5)x(1/5)
The probability of the 1st player picks a black ball and then the 2nd player picks a black ball and then the 3rd picks the white ball is P3 = (4/5)x(4/5)x(1/5)

So the probability that the game ends before any of the players has picked twice is: P = P1+P2+P3 = (1/5) + (4/5)x(1/5) + (4/5)x(4/5)x(1/5) = 61/125 = 0.488

b)
The probability that A picks the white ball is PA = 1/5
The probability that B picks the white ball is PB = (4/5)x(1/5)
The probability that C picks the white ball is PC = (4/5)x(4/5)x(1/5)

2. Apr 7, 2015

### paisiello2

I don't think your answer is correct for part b). You seem to have tried to calculate the individual probabilities of each player winning after one pick only. However, can't the game go on for multiple rounds before a winner happens?

Last edited: Apr 7, 2015
3. Apr 7, 2015

4. Apr 8, 2015

5. Apr 8, 2015

### LCKurtz

Instead of duplicating a thread, you can always ask a moderator to move it to a more appropriate forum if needed.

6. Apr 9, 2015

### haruspex

You could try responding to my post there.

7. Apr 9, 2015

### Delta²

Just calculate the probability of the player A winning on the n-th round $P_A(n)$. Then the probability for A to win the game is $P_A=\sum\limits_{n=1}^{\infty}P_A(n)$

Last edited: Apr 9, 2015
8. Apr 9, 2015

Ok

9. Apr 9, 2015

### stephenranger

So, the probability that A picks the white ball at:
the 1st round: 1/5
the 2nd round: (4/5)3.(1/5)
the 3rd round: (4/5)6.(1/5)
the 4th round: (4/5)9.(1/5)
.............................................................
.............................................................
the n-th round: (4/5)3n.(1/5)

Therefore: ∑(4/5)3n.(1/5) when n runs from 0 → ∞ ≈ 0.32

10. Apr 9, 2015

≈0.4