A jar contains $m=90$ white balls and $n=10$ red balls, the balls are drawn under the following constraints:
- the ball is thrown away if it is white;
- the ball is put back if it is red and another ball is drawn; this time, the ball is thrown away no matter what color it is.
The question is, what is the probability to exhaust all balls and have the last one in white color? My guess is $\frac{1}{2}$ but I might be wrong. Please show how you deduce the answer. Thanks!
EDIT: Thanks for posting the solution and simulation, which are all appreciated. B. E. Oakley and R. L. Perry discussed a very similar problem in their A Sampling Process paper published on The Mathematical Gazette, Vol. 49, No. 367 (Feb., 1965), pp. 42-44.
The problem presented in the paper is:
A bag contains m > 0 black balls and n > 0 white balls. A sequence of balls from the bag is discarded in the following manner:
(i) A ball is chosen at random and discarded. (ii) Another ball is chosen at random from the remainder. If its colour is different from the last it is replaced in the bag and the process repeated from the beginning (i.e. (i)). If the second ball is the same colour as the first it is discarded and we proceed from (ii). Thus the balls are sampled and discarded until a change in colour occurs, at which point the last ball is replaced and the process starts afresh.
The question is: what is the probability that the final ball should be black?
Their induction gives $\frac{1}{2}$ which is what I have here. Apparently having no constraint on colors like what's in the paper changes the situation significantly.
But the story hasn't ended. At the end of their paper, they proposed a seemingly more interesting problem: What is the solution if there are balls of 3 different colours and the sampling process is as before?