4
$\begingroup$

A box contains $n$ balls numbered from 1 to $n$. Suppose you take a ball at a time, putting it back on the box, until you pick a ball twice. How many balls are you expected to take from the box?

Let $X$ be the r. v. of interest. Its support is every natural number from $2$ to $n+1$. If $k$ is one such number, to get the first repetition on the $k$-th pick, the $k-1$ previous ones must be distinct and $k$-th equal to some of those, therefore, $$\begin{align} \forall k\in\mathbb N\cap[2,\,n\!+\!1]\qquad \mathbb P(X=k) &= \frac nn\frac {n-1}n\ldots\frac {n-(k-2)}n \;\cdot\; \frac{k-1}n =\\ &=\frac{(n-1)!}{(n-k+1)!}\frac{k-1}{n^{k-1}} =\\ &= \frac{k-1}n\,\prod_{l=0}^{k-2}\left(1-\frac{l}n\right)\quad, \end{align}$$ so that $$\begin{align} E(X) &= \sum_{k=2}^{n+1}\,k\cdot\frac{k-1}n\,\prod_{l=0}^{k-2}\left(1-\frac{l}n\right) =\\ &=\frac1n\sum_{k=2}^{n+1}\,k(k-1)\prod_{l=0}^{k-2}\left(1-\frac{l}n\right)\quad. \end{align}$$

However, the book answer is $$E(X) = 2 + \sum_{k=1}^{n-1}\prod_{l=1}^k\left(1-\frac ln\right)\quad.$$

What am I doing wrong? Or, if the answers are actually the same, how to show that?

$\endgroup$
4
$\begingroup$

Both answers are the same. The solution in the question is clearly presented and holds up. For reference, the answers it gives for $n=2, 3, \ldots, 7$ are

$$E(X) = 2,\frac{5}{2},\frac{26}{9},\frac{103}{32},\frac{2194}{625},\frac{1 223}{324},\frac{472730}{117649}, \ldots$$

A simpler way to obtain the expectation is to sum the survival function $S_n(k) = \mathbb{P}(X\gt k), k\ge 0.$ This is the probability that the first $k$ balls will be unique:

$$S_n(k) = \prod_{l=1}^{k-1} \left(1 - \frac{l}{n}\right).$$

Clearly $S_n(0)=S_n(1)=1$ (the value, by definition, of an empty product). Accounting for their sum separately yields

$$E(X) = S_n(0) + S_n(1) + \sum_{j=2}^n S_n(j) = 2 + \sum_{k=1}^{n-1} S_n(k+1).$$

(writing $j=k+1$ for the last step). That is the textbook answer and it gives the same sequence of values. Incidentally, a useful closed form expression for these values is

$$E(X) = 1 + n \int_0^{\infty } e^{-n t} (1+t)^{n-1} \, dt.$$

For instance, expanding the log of the integrand through second order at $0$ and integrating that MacLaurin series gives the approximation $$E(X)\approx \frac{2}{3} + \sqrt{\frac{\pi n}{2}}$$ for large $n$; it turns out to have two significant figures of accuracy even for $n\ge 5$ and three for $n \ge 75.$

$\endgroup$
2
  • $\begingroup$ Indeed, when asking the distribution of $X$ on the previous item, the book suggested investigating $\mathbb{P}(X\gt k)$, but I found it via the rationale I showed on the question. $\endgroup$ – Luke May 20 '14 at 1:37
  • 1
    $\begingroup$ I rolled back an edit because it reflected a misunderstanding of the (conventional) notation: an empty product is, by definition, equal to $1$. Thus when $k\le 1$, $\prod_{l=1}^{k-1}f(l) = 1$ for any function $f$. Although it might seem like a technical nicety, including the possibility of such empty products in the formula reflects the process of computing the survival function for $k=0$ and $k=1$, which--because such values of $k$ are less than the support of $X$--is (axiomatically) equal to $1$. $\endgroup$ – whuber May 20 '14 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.