A bag contains 10 balls: 9 are red and 1 is black. There are 10 players. What is the probability of picking the black one, given the balls are not replaced each time a player picks one?

Question

(if you picked the black the game is over, you win)

1. How would change the result if the balls are replaced it?

Answer 1

1/10

then 1/9

then 1/8

then 1/7

....

Answer 2

Assume out of 10 you are the n^th player. The probability that the n^th player picks the black ball (and loses) is

P_n = (Probability that n-1 players picked red) * (Probability of you picking black)

If Probability that n-1 players picked red = R_(n-1)

At n^th turn (10-n+1) total balls are left and (9-n+1) red balls are left. (Also no black ball is picked or else the game would have ended) (Not valid for n = 1)

guy 1 guy 2....... guy n-1

R_(n-1) = (9/10)*(8/9)*....( [9-n+2)] / [10-n+2] )

Simplifying

R_(n-1) = [9-n+2]/10

No of balls remaining = 10-(n-1) = 9 - n + 2

Probability of you picking black = 1/[9-n+2]

P_n = R_(n-1)*{1/[9-n+2]} = {[9-n+2]/10}*{1/[9-n+2]} = 1/10

This is because before it is the n^th person's turn (n-1) people had to win. Even though the ratio Red:Black decreases, this is offset by the unlikeliness of the people before you to winning.

With Replacement P_n is same for all Players.

P_n = 1/10.

Sorry I couldn't use latex. I don't know how to use it on this platform.

Answer 3

It seems that each person picking a ball is an independent event. Therefore, you should be able to use Bayes theorem to calculate the probability that, for example, the fifth person in the series will pick the black ball. If the red balls are not replaced the probability of picking the black ball increases, in essence, the probability that the fifth person in the game will pick the black ball is the product of the individual probabilities for each of the first five people.

If the balls are replaced each time then each person has the same probability of picking the black ball. The key here is that the events are all independent.

Answer 4

If the balls are replaced each time, then the probability is the same for each player to win 1/10. This also would change the game since multiple players could win since the black ball is replaced if it is drawn.