A staple, because it connects random walks, martingales and a real intuition about bankroll.
The setup
You start with k units. Each round you win 1 with probability p or lose 1 with probability 1 - p. You stop at 0 (ruin) or at N (target).
The fair case, p = 1/2
Probability of reaching N before 0 is k/N.
The cleanest derivation is the martingale one: your wealth is a martingale, so its expectation is unchanged at the stopping time. If P is the probability of hitting N first:
k = P times N + (1 - P) times 0, so P = k/N.
Expected number of rounds is k(N - k).
This is worth internalising because it is much larger than intuition suggests. Starting at 50 with a target of 100, you expect 2,500 rounds. The game drifts nowhere, so it takes a long time to get anywhere.
The biased case
With p not equal to 1/2, let r = (1 - p)/p. Then the probability of reaching N before 0 is:
(1 - r^k) / (1 - r^N)
The consequence that matters: a small edge compounds into near-certainty over enough rounds, and a small disadvantage into near-certain ruin. The dependence is exponential in the number of units, not linear.
The casino result
Let N go to infinity - the opponent has unlimited funds.
- If p is at most 1/2, ruin has probability 1. Even a perfectly fair game ruins you eventually against an infinitely rich opponent, because your wealth is bounded below at zero and the walk is recurrent.
- If p is greater than 1/2, you survive with probability 1 - r^k.
This is the honest mathematical statement of why bankroll management matters more than edge, and it connects directly to the Kelly criterion.
Interview variants
The same problem appears as a random walk between two barriers, as a tennis or table-tennis scoring question, as a market-maker inventory question, and as a "how long until this drunk falls off the cliff" puzzle. Recognise the shape and the formula is immediate.
Practise in stochastic processes.