# Probability that the 10th drawn ball is black.

An urn contains $$7$$ white and $$13$$ black balls. We draw a ball from an urn, put it back and add $$2$$ additional balls of the same color (as the one we have just drawn and put back). We repeat this $$10$$ times (so the numbers of balls change throughout the whole experiment). What is the probability that the $$10$$th drawn ball is black?

I think it should be $$13/20$$. Basically, I think that no matter how many times the experiment is repeated, the probability of picking up the black ball at the end is $$13/20$$. I was trying induction but got stuck.

• I believe you can prove this by simply looking at the expected number of white and black balls. With the linearity of the expected value, it should be easy to prove by induction. Dec 20, 2018 at 18:16
• The probability of getting black ball the second time is $13/20$. $P(B_2) = P(B_2 | B_1) P(B_1) + P(B_2 | W_1) P(W_1) = \frac{15}{22} \cdot \frac{13}{20} + \frac{13}{22} \cdot \frac{7}{20} = \frac{286}{440} = \frac{13}{20}$. Dec 20, 2018 at 18:17
• @DonThousand What makes you say that? The probability that the second ball is black is 13/20 as well. Dec 20, 2018 at 18:20
• Dec 20, 2018 at 18:20
• Thank you, @saulspatz Dec 20, 2018 at 18:23

Direct computation shows that this is true for one round of the game, regardless of the starting configuration. (and for any "shift" where, in your problem, the "shift" is $$3$$).

Inductively, suppose we have shown that the probability is unchanged for $$n-1$$ rounds of the game (regardless of the starting configuration).

Now suppose we start with $$w$$ White and $$b$$ Black (denoting that by $$(w,b)$$). On round two we either have $$(w+3,b)$$ with probability $$\frac w{b+w}$$ or $$(w,b+3)$$ with probability $$\frac b{b+w}$$. By the inductive hypothesis the probability is then unchanged for at least the next $$n-1$$ rounds.

Of course, if you are in state $$(w+3,b)$$ the (new) probability of drawing a black ball is $$\frac b{b+w+3}$$ and if you are in state $$(w,b+3)$$ it is $$\frac {b+3}{b+w+3}$$.

Thus the probability on the $$n^{th}$$ round of the original game is $$\frac w{b+w}\times \frac b{b+w+3}+\frac b{b+w}\times \frac {b+3}{w+b+3}=\frac {b(b+w+3)}{(b+w)(b+w+3)}=\frac b{b+w}$$

as desired.

(To be sure, the $$3$$ is incidental as well. As the computation makes clear, we could replace it with any natural number).

The comments suggest that the logic of this argument is unclear, so let me elaborate. I am inductively supposing that we know the desired claim for any starting configuration of White and Black balls for $$n-1$$ rounds. I then take the desired configuration and look at the second round. To be sure, the probability will have changed depending on the first draw. Inductively, though, I know that the new probability will be constant for the next $$n-1$$ rounds. So it suffices to consider the two possible branch paths, weighted by the probability of being on one or the other. Of course, the probability of being on the "White" branch is $$\frac w{b+w}$$ and the new probability of drawing black along that branch is $$\frac b{b+w+3}$$. That explains the first term of my sum. The second term, corresponding to the "Black" branch, is entirely similar.

• you want a $b+w$ in the final denominator, I think
– Ned
Dec 20, 2018 at 18:34
• You meant $b/(b+w)$ in the last line of the calculation. Dec 20, 2018 at 18:36
• I have some problems with this proof. It seems like you assume sth like: ok, at the $n-1$ round the probability of drawing a black ball is $\frac{b}{b+w}$. But then for calculating the probability for the $n$th round, you assume that you actually have $b+w$ balls in the urn which seems a bit weird for me. I am not telling your proof is incorrect. I am just saying I am not fully convinced. Dec 20, 2018 at 18:47
• I have doubts about this. There aren't $b$ and $w$ balls in the urn after $n$ draws. Even if we assume there are $kb$ black and $kw$ white balls, at the next draw there will be $kw+kb+3,$ and if I'm not mistaken, the algebra doesn't work out to $b/(b+w).$ Dec 20, 2018 at 18:48
• @Ned Yes, thank you.
– lulu
Dec 20, 2018 at 19:00

Here's a conceptual solution based on symmetry which also works regardless of how many extras you put back at each stage. First note that you can play the games with more than 2 colors and ask the same question.

Second note that if two colors have the same number of initial balls, then the probabilities of obtaining those colors on the $$j$$-th draw are the same (by symmetry, nothing favors one of the two colors over the other).

Now given the initial configuration of $$w$$ white and $$b$$ black, imagine that the balls are all different colors (maybe, $$w$$ shades of white and $$b$$ shades of black). On draw $$j$$ each of the $$w+b$$ shades is equally likely to be drawn, since we started with the same number (namely $$1$$) of each. But that says the probability of drawing a shade of Black is $$b/(w+b)$$.

Here is I what I have found based on a problem on Polya's urn scheme but I elaborate this formally below

Proof for the specific case

We start with $$7$$ white balls and $$13$$ black balls. On each round we add additional $$2$$ balls according to the rules described in the question.

Think that for each round you either draw a ball that was there in the previous round or you draw a ball that was added in the previous round.

Assume inductively that on round $$n$$ the probability of drawing a black ball is $$p$$ ($$:= \frac{b}{b+w}$$). Let's see what happens on round $$n+1$$. Let $$k$$ denote the number of balls in the $$n+1$$ round. So with probability $$\frac{k-2}{k}$$ we draw an old ball (that was there in the $$n$$th round) and with probability $$\frac{2}{k}$$ we draw a new ball (one of the two that were added on the $$n$$th round). By the inductive hypothesis, the probability of drawing a black ball from the old ones is $$p$$. The probability of drawing a black ball from the new ones is simply $$p$$ (because you either have drawn a black ball with probability $$p$$ on the $$n$$th round and hence necessarily a black one on the $$n+1$$th round, or you have drawn a white one on the $$n$$th round and hence a white one on the $$n+1$$th round).

All in all, the probability of drawing a black ball on the $$n+1$$th round is:

$$p \cdot \frac{k-2}{k} + p \cdot \frac{2}{k} = p \cdot ( \frac{k-2}{k} + \frac{2}{k}) = p$$

Proof for the general case

You can use this idea to give a general inductive proof for the number $$k$$ of balls that you add each time to the urn and arbitrary initial configuration $$b,w$$ of black and white balls, respectively.

Let $$b,w$$ be our starting configuration. On the first round the probability of drawing a black ball is $$\frac{b}{b+w}$$.

Now, assume inductively that on the $$n$$th round the probability of drawing a black ball is $$\frac{b}{b+w}$$. The probability of drawing a black ball on the $$(n+1)$$th round is:

$$P(B_{n+1}) = P(B_n) P(B_{n+1} | B_n) + P(W_n) P(B_{n+1} | W_n) ,$$

where $$B_i$$, $$W_i$$ are the events of drawing a black ball and the white ball on the $$i$$th round, respectively.

We have:

$$P(B_{n+1}) = \frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{w}{b+w}\frac{(b+w+(n-1)k)\frac{b}{b+w}}{b+w+nk}=$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{w}{b+w}\frac{b}{w}\frac{(b+w+(n-1)k)\frac{b}{b+w}\frac{w}{b}}{b+w+nk}=$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{b}{b+w}\frac{(b+w+(n-1)k)\frac{w}{b+w}}{b+w+nk}=$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k + (b+w+(n-1)k)\frac{w}{b+w}}{b+w+nk}$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)(\frac{b}{b+w} + \frac{w}{b+w}) + k}{b+w+nk}$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k) + k}{b+w+nk}$$ $$\frac{b}{b+w} \frac{b+w + nk}{b+w+nk} = \frac{b}{b+w}$$

This ends the proof.