Pólya’s urn is the without-replacement intuition turned upside-down: each draw reinforces its color, making the same color more likely next time. And yet, because the reinforcement rule treats the colors symmetrically, the sequence of draws is exchangeable — the probability of any history depends only on the color counts, not the order. The corollary that breaks the problem open: $P(\text{n-th draw is black}) = \tfrac{b}{b+w}$ for every $n$, identical to the very first draw. When you spot Pólya, don’t enumerate — invoke the symmetry and you’re done in one line.
✓ Intro · expand
Try first (productive failure)
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✓ Try first · expand
Worked example
An urn contains 1 black ball and 1 white ball. You draw one ball uniformly at random, observe its color, and return it to the urn along with one extra ball of the same color. Then you draw again from the new urn. What is the probability that the second draw is black?
✓ Worked example · expand
Practice 1 of 3Type a fraction, decimal, or expression — mathjs parses it.
✓ Practice · expand
Reflection
In your own words, why does exchangeability hold for Pólya’s urn even though each draw clearly changes the composition? And what’s the cue in a problem statement that tells you to reach for exchangeability rather than computing histories one by one?