# Probability of Winning Game for A, B, and C

• stephenranger
In summary, the probability that the game ends before any of the players has picked twice is 0.488. The probability that A picks the white ball is 0.4, the probability that B picks the white ball is 0.32, and the probability that C picks the white ball is 0.
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?

## 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)

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

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

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

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

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).

haruspex said:
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.

stephenranger said:
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.

## 1. What is the probability of player A winning the game?

The probability of player A winning the game depends on various factors, such as their skill level, the rules of the game, and the strategies used by both players. It is not possible to determine the exact probability without more information.

## 2. Is it possible for player B to have a higher probability of winning than player A?

Yes, it is possible for player B to have a higher probability of winning than player A. This could be due to a number of reasons, such as player B having more experience, using better strategies, or having a stronger skill set.

## 3. How does the probability of player C winning change if player A and B are evenly matched?

If player A and B are evenly matched, the probability of player C winning would depend on their own skill level and strategies. However, it is likely that the probability would increase as the game may become more unpredictable and any player could potentially win.

## 4. What is the relationship between the probabilities of winning for players A, B, and C?

The relationship between the probabilities of winning for players A, B, and C would depend on the specific game and the factors that influence their chances of winning. In some cases, the probabilities may be equal, while in others they may vary significantly.

## 5. How can the probability of winning be calculated for players A, B, and C?

The probability of winning for players A, B, and C can be calculated using mathematical formulas and statistical analysis. However, the accuracy of these calculations would depend on the availability of data and the complexity of the game. In some cases, it may not be possible to accurately determine the probability of winning for each player.

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