riddle of the day

Sometimes a single visible piece of information is enough to change the weight of an entire situation. This riddle is deeper than it seems.

The urn that reveals itself

Strategist
Chance and uncertainty

Riddle statement

An urn contains 100 balls. Some are red and the rest are green.

You don't know how many red there are, except this: the number of red balls was chosen at random with the same probability between 0 and 100.

You draw a ball. It comes out red and you put it aside.

Now you are going to draw a second ball from the same urn.

What is the probability that this second ball is also red?

Show solution

Solution

Let $k$ be the number of red balls. Before any draw, all values

$k=0,1,2,\dots,100$

are equally likely.

After observing that the first ball is red, larger values of $k$ become more plausible: an urn with many red balls was more likely to produce that observation.

If there are $k$ red balls, the probability of drawing a red ball first and another red ball second is

$\frac{k}{100}\cdot\frac{k-1}{99},$

while the probability of drawing a red ball on the first draw is

$\frac{k}{100}.$

By conditional probability,

$ P(\text{second red}\mid\text{first red}) = \frac{\displaystyle\sum_{k=0}^{100}\frac{k}{100}\cdot\frac{k-1}{99}} {\displaystyle\sum_{k=0}^{100}\frac{k}{100}} = \frac{\displaystyle\sum_{k=0}^{100}k(k-1)} {\displaystyle 99\sum_{k=0}^{100}k}. $

Now,

$\sum_{k=0}^{100}k=\frac{100\cdot101}{2}=5050,$

and

$\sum_{k=0}^{100}k(k-1) = \sum_{k=0}^{100}k^2-\sum_{k=0}^{100}k = \frac{100\cdot101\cdot201}{6}-5050 = 333300.$

Therefore,

$P=\frac{333300}{99\times5050}=\frac{2}{3}.$

So the probability that the second ball is also red is $\frac{2}{3}$.