> Toss a fair coin until the number of heads exceeds the number of tails. This determines a stopping time. What is the probability that this stopping time is even?

Claude Code did pretty well:

> Zero. > > The stopping time always falls on an odd toss. Here's why: after >n tosses, heads minus > tails equals S_n = X₁ + X₂ + ... + X_n where each X_i takes >value +1 or −1. The sum > of n odd numbers shares parity with n. For S_n = 1 (the first >time heads exceeds > tails), we need n odd. > > No even stopping time exists. The probability equals exactly 0. > > Cogitated for 38s