We draw a ball, look at it, then return it to the urn and add another ball of the same color. Pr: "after $n$ draws the urn contains $j$ white balls"?

Question

The urn contains $1$ white ball and $1$ black ball. We make a sequence of $n$ draws according to the following scheme: we draw a ball, look at it, then return it to the urn and add another ball of the same color. For $j \in \{ 1, 2, . . . , n + 1 \}$, let $p_j$ denote the probability that after $n$ draws the urn contains exactly $j$ white balls. Prove that $p_j = \frac{1}{n + 1}$ for all $j$.

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I came up with a recoursive equation describing the number of white balls after n draws.

• $\mathbb{P}_1(j,n) =$ probability that after $n$ draws we have $j$ white balls in the urn,
• $\mathbb{P}_2(b, n-1) =$ probability, that after $n-1$ draws we will draw a black ball from the urn.

The equation goes like this:

$$\mathbb{P}_1(j,n) = \mathbb{P}_2(b, n-1) \cdot \mathbb{P}_1(j,n-1) + \mathbb{P}_2(w, n-1) \cdot \mathbb{P}_1(j-1,n-1)$$

$$\mathbb{P}_1(j,n) = \frac{n-1+2-j}{n-1+2} \mathbb{P}_1(j,n-1) + \frac{j-1}{n-1+2} \mathbb{P}_1(j-1,n-1)$$

$$\mathbb{P}(j,n)_1 = \frac{n-j+1}{n+1} \mathbb{P}_1(j,n-1) + \frac{j-1}{n+1} \mathbb{P}_1(j-1,n-1)$$

Now I would like to substitute the assumed formula: $\mathbb{P}_1(j,n) = \frac{1}{n + 1}$ and check the result, but I can't since we have $\mathbb{P}_1(j-1,n-1)$ in the equation and I don't know how to evaluate that.

Anybody knows how to do that?

Answer 1

There are basically two approaches to solve your problem. One approach is to formulate a $2D$ recursion. It is very tedious to solve it that way. I'll share another solution, which simplifies it and helps in appreciating the problem on a deeper level.

For any $j \in \{ 1, 2, . . . , n + 1 \}$, for any permutation $\pi$ of W and B balls to achieve $j$ number of white balls, denote $P(j, \pi)$ = Probability of the urn containing $j$ white balls by the permutation $\pi$, where $\pi$ is a permutation of $W$ and $B$ of size $n$ having $(j-1)$ number of $W's$ and $(n - j + 1)$ number of $B's$ .

Claim$1$: $P(j ,\pi)$ is independent of $\pi$

Consider $n = 100$, $j = 5$. Let's calculate $P(5, \pi_1)$ where $\pi_1 = WWWWBBBBB....B$. $$P(5, \pi_1) = \frac12 \cdot \frac23 \cdot \frac34 \cdot \frac45 \cdot \frac16 \cdot \frac27 \cdot \frac38 \cdot \frac49 \cdot ... \cdot \frac{96}{101} = \frac{4! \cdot 96!}{101!} = \big({101} \cdot {100 \choose 4}\big)^{-1}$$

Calculate $P(5, \pi)$ for any other possible $\pi$ leading to $j = 5$, notice that the only thing that varies is the order in which the numbers occur in the numerator. This argument can be extended to any $j$. Thus $$P(j,\pi_1) = P(j, \pi_2) = ... = P(j, \pi_k) = \big((n+1)\cdot{n \choose j - 1} \big)^{-1}$$

Claim$2$: $P(j)$ is independent of $j$, in particular, $P(j) = \frac{1}{n + 1}$

Consider the same example as in Claim$1$. How many different $\pi_i's$ are possible for $n = 100, j = 5$? It is of course, the number of ways to choose $4$ objects from $100$, i.e. $100 \choose 4$. From Claim$1$ each of these is equally likely. Hence, for $n = 100, P(5) = {100 \choose 4} \cdot \big({101} \cdot {100 \choose 4}\big)^{-1} = \frac{1}{101}$.

Generalising, the number of different $\pi_i's$ for $n, j = {n \choose j - 1}$, each of which is equally likely by Claim$1$, hence:

$$P(j) = {n \choose j-1 } \cdot P(j, \pi_i)$$ $$=> P(j) = {n \choose j-1 } \cdot \big((n+1)\cdot{n \choose j - 1} \big)^{-1}$$

$$=>\boxed{ P(j) = \frac{1}{n + 1}}$$

Answer 2

Check out this problem. Your problem is basically the same problem but reframed. The basketball being scored in the original problem is analogous to a white ball being drawn, and a miss is analogous to a black ball being drawn.

Note that in the original problem, the probability of scoring is proportional to the number of goals scored. In your problem as well, the probability of adding a ball of a particular colour is proportional to the number of balls of that colour already present in the urn as well.

Thus the two problems are equivalent.

The only minor difference is in the definition of n. In the original problem the value of n includes the first two shots hence the no of scores varies from $1, 2, 3, ..., n - 1$. Hence $P$(scoring k shots) = $\frac{1}{n-1}$. Whereas in your problem the n draws are made after the first two balls are put in, hence, the no of white balls that can be there in the urn after all the draws have been made varies from $1, 2, 3, ..., n + 1$.

Thus P(no of white balls = $j$) = $\frac1{n+1}$

Note: In order to understand how the $2D$ recurrence is formed you can read this answer to the original problem. In order to understand how it is to be solved (without induction), you can have look at this answer to my question regarding the same.