# Probability problem

1. Apr 6, 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
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 6, 2015

### cpscdave

Assuming you want to check to see if your logic is right.

A) *EDIT* misread BEFORE they picked twice yes your logic is sound

3. Apr 6, 2015

### stephenranger

I'm asking people to check if my solutions is correct, especially my solution for the first question.
Why do you have to edit? the first question is ''What is the probability that the game ends before any of the players has picked twice?". That means that you have to find the probability that one of 3 players picks the white ball in the first round.

4. Apr 7, 2015

### cpscdave

Prior the edit I thought it was before each play has picked twice :) (so 2 rounds of the game) I blame Monday

5. Apr 7, 2015

### haruspex

For a), your answeris right but there's a slightly easier way. Think about the converse condition, the probability that it does not end on the first round.

For b), I think you are supposed to assume that the game continues until somebody wins. I.e. this part is independent of a).

6. Apr 9, 2015

### stephenranger

Thanks. This is my attempt to solve the b question:

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, the probability that B wins the game is : ∑(4/5)3n.(1/5) when n runs from 0 → ∞ ≈ 0.4

The probability that B picks the white ball at:
the 1st round: (1/5).(4/5)
the 2nd round: (1/5).(4/5)4
the 3nd round: (1/5).(4/5)7
the 4nd round: (1/5).(4/5)10
.............................................................
.............................................................
the n-th round: (1/5).(4/5)3n+1

Therefore, the probability that B wins the game is: ∑(4/5)3n+1.(1/5) when n runs from 0 → ∞ ≈ 0.32

The same process with C.

7. Apr 9, 2015

### haruspex

That looks right. You can avoid having to perform the sum by letting x be the probability that A wins, and note that if all three fail once then it's back to x again: x = 1/5+(4/5)3x.