Home > Riddles > The urn that reveals itself
A Bayesian miniature in classic puzzle clothing: a single visible piece of information is enough for the entire situation to change weight.
An urn contains 100 balls. Some are red and the rest are green.
You don't know how many red ones there are, except this: the number of red balls was chosen at random with the same probability between 0 and 100. You take out a ball.
It comes out red and you push it away. Now you are going to draw a second ball from the same urn.
What is the probability that this second ball is also red?
Let $k$ be the number of red balls. Before looking at anything, all the values
$$ k=0,1,2,\dots,100 $$
They are equiprobable. But when observing that the first ball comes out red, the large values of $k$ become more plausible than the small ones, because an urn with many reds had more chances of producing that observation. The clean account is this. If there are $k$ reds, the probability of drawing red first and red later is
$$ \frac{k}{100}\cdot\frac{k-1}{99}, $$
while the probability of drawing red first is
$$ \frac{k}{100}. $$ By averaging over all possible values of $k$ and conditioning on having seen red first, the result is reduced to comparing $$
\sum_{k=0}^{100} k(k-1)
$$ with $$
99\sum_{k=0}^{100} k.
$$ When you do the math you get exactly
$$ \frac23. $$ Therefore, the probability that the second ball is also red is $$
\frac23.
$$ $$
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