Calculating Probability for a Game of Picking White Balls: Three Players

[stephenranger]

Homework Statement

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?

Homework Equations

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 P_A = 1/5
The probability that B picks the white ball is P_B = (4/5)x(1/5)
The probability that C picks the white ball is P_C = (4/5)x(4/5)x(1/5)

[/B]

 

[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?

 

[LCKurtz]

This is duplicate thread to https://www.physicsforums.com/threads/probability-problem.807091

 

[stephenranger]

This is duplicate thread to https://www.physicsforums.com/threads/probability-problem.807091

Oh yeah. I'm sorry because posting in the precalculus mathematics, no one gives me a decent answer and otherwise this problem is supposed to be beyond pre-calculus mathematics, right?

 

[LCKurtz]

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

 

[haruspex]

Oh yeah. I'm sorry because posting in the precalculus mathematics, no one gives me a decent answer and otherwise this problem is supposed to be beyond pre-calculus mathematics, right?

You could try responding to my post there.

 

[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)

 

[stephenranger]

You could try responding to my post there.

Ok

 

[stephenranger]

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)

Thanks for your suggestion.
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

 

[stephenranger]

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

≈0.4