Probability of Winning Game for A, B, and C

[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

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)

 

[cpscdave]

What are you asking exactly??

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

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

 

[stephenranger]

What are you asking exactly??

I'm asking people to check if my solutions is correct, especially my solution for the first question.

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

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.

 

[cpscdave]

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

 

[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).

 

[stephenranger]

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).

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.

 

[haruspex]

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.

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.