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Two urns contain n balls each, numbered from 1 to n. We pick a ball from the first one and then a ball from the second. What is the probability that the number of the second ball is

a) smaller

b) equal to the number of the first ball?

My humble attempt:

b) $\frac{1}{n}$

a) Suppose the number picked is $0< k\leq n$

So we have $k-1$ numbers $< k$ and $n-k$ numbers $> k$. Isn't the probability $\frac{n-k}{n}$ ?

Community
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Samuel
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2 Answers2

2

You are right for part b. For part a, note that the probability that the two balls are the same number is $1/n$, so the probability that they are a different number is $(n-1)/n$. Now both balls are equally likely to be the smaller one of the two, so the probability that the second ball is smaller than the first equals $$\frac{n-1}{2n}.$$

If you do not see this directly, you can indeed condition the probabilities on the outcome of the first draw. We than have \begin{align*}P(\text{ball 2 is smaller})&=\sum_{k=1}^nP(\text{ball 2 is smaller than ball 1}|\text{ball 1 equals }k)P(\text{ball 1 equals }k)\\&=\sum_{k=1}^nP(\text{ball 1 is }k)\cdot P(\text{bal }2\text{ is smaller than }k)\\&=\sum_{k=1}^n\frac1n\cdot\frac{n-k}{n}\\&=\sum_{k=1}^n\frac1n-\sum_{k=1}^n\frac{k}{n^2}\\&=1-\frac{1}{n^2}\frac{n(n+1)}{2}\\&=\frac{n-1}{2n},\end{align*} as found before.

Václav Mordvinov
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  • Why both balls are equally likely to be the smaller one of the two?Let's say that we first pick a number n-1. Then most probably the second number will be smaller. So the probability is not equal. – Samuel Oct 08 '18 at 18:16
  • @Samuel I made an edit to clearify that the probability is the same if you use conditional probabilities. – Václav Mordvinov Oct 08 '18 at 18:29
  • Samuel, suppose you win if ball 1 is bigger and I win if ball 2 is bigger, and it's a tie if the balls are equal. Which one of us will win more often? If you can see that we'll each win equally often without doing any counting of outcomes, then that's the answer to your question about why the symmetry argument works. – Ned Oct 09 '18 at 09:58
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A symmetry argument is fine. Though in general it may not always be possible. Here is a more detailed approach.

The number of ordered pairs $(i,j)$ with $i<j$ can be counted as follows:

$$\begin{align} i=1&: j\in \{2,...,n\} &... &\ \ \ \ \ n-1 \text{ pairs }\\ i=2&: j\in \{3,...,n\} &... &\ \ \ \ \ n-2 \text{ pairs }\\ i=3&: j\in \{4,...,n\} &... &\ \ \ \ \ n-3 \text{ pairs }\\ \vdots\\ i=n-1&: j\in \{n\} &...& \ \ \ \ \ 1 \text{ pair } \end{align}$$

Adding these up, we find the number of simple and equally-likely events satisfying your compound event:

$$\sum_{i=1}^{n-1} (n-i) = \dfrac{n(n-1)}{2}$$

The total number of ordered pairs is $n^2$, thus the probability is

$$\dfrac{n(n-1)}{2}\div n^2 = \dfrac{n-1}{2n}$$

By a similar argument, there are exactly $n$ pairs of the form $(i,i)$. Therefore the probability that the are equal is

$$\dfrac{n}{n^2} = \dfrac{1}{n}$$

David P
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