Question

Starting with 10 blue balls, in each of 10 sequential rounds, we remove a random ball and replace it with a new red ball. For example, after the first round we have 9 blue balls and one red ball, after the second round, with probability 9/10 we have 8 blue balls and 2 red balls, and with probability 1/10 we have 9 blue balls and one red ball, etc. What is the probability that the ball we remove at the 11th round is blue?

Answer 1

Answer:

Suppose the fraction of blue balls at any given state n is $f_{n}$. At the start of the process we have: $f_{0}$=1.

Now consider the situation after having drawn n times.

The current fraction of blue balls is $f_{n}$, the number of blue balls is therefore 10$f_{n}$.

We either draw a red ball, with probability 1−$f_{n}$.  and the number of red balls, blue balls stay the same (we swap red for red):

$f_{n+1}$(red)=$f_{n}$

or we draw a blue ball, with probability $f_{n}$  and we obtain 10$f_{n}$−1 blue balls on a total of 10 balls, thus:

$f_{n+1}$(blue) = (10$f_{n}$− 1)/10

The total probability of (fraction of) blue balls after this n'th draw is therefore:

$f_{n+1}$=$f_{n+1}$(red)(1−$f_{n}$)+$f_{n+1}$(blue)$f_{n}$

which can be simplified to:

$f_{n+1}$=0.9$f_{n}$

which leads to:

fn=0.9$f_{n}$

At the end of the 10'th draw the fraction of blue balls is equal to:

$f_{10}$ = $0.9^{10}$ ≈ 0.348678