Oct 08, 2026 by Offer

Millennium Probability Interview 2027: Fair Coin Game Puzzle

The Millennium fair coin game probability p solution: write p in binary and flip the coin to generate bits — stop at the first flip differing from p's bit; you win if your bit is 1 there. This wins with probability exactly p, averaging two flips. Commonly reported by candidates.

What This Question Assesses

This is a creativity-under-constraints test. The interviewer wants to see whether you can construct a mechanism, not just compute a probability. The follow-up discussion — expected number of flips, edge cases — matters as much as the construction.

Millennium Fair Coin Game Probability P: How to Answer

Step 1 — State the binary expansion idea. "Every p in (0, 1) has a binary expansion 0.b₁b₂b₃.... I will generate a uniform random number U in [0, 1] bit by bit using fair coin flips, and declare a win if U < p."
Step 2 — Give the stopping rule. "Flip the coin to get bit u₁, u₂, ... Compare with b₁, b₂, ... at the first index k where uₖ ≠ bₖ: if uₖ = 0 and bₖ = 1, then U < p and I win; if uₖ = 1 and bₖ = 0, then U > p and I lose."
Step 3 — Argue correctness. "The generated bits are uniform, so P(U < p) = p exactly — the comparison resolves with probability 1 since the bits differ eventually almost surely."
Step 4 — Discuss efficiency. "The expected number of flips is 2 — at each bit there is a 1/2 chance of resolving — so the game is practical, not just theoretical."

An example line: "I would expand p in binary and generate a uniform random number with the coin, stopping at the first bit where they differ — I win exactly when my number lands below p, which happens with probability p, and it takes two flips on average."

Millennium Fair Coin Game Probability P: Common Mistakes

Restricting to a fixed number of flips. With n flips you can only hit multiples of 1/2ⁿ. The question demands arbitrary p, so the construction must handle non-dyadic values — the sequential comparison does.
Hand-waving the correctness argument. "It obviously gives probability p" is not a proof. The uniformity of the generated bits plus the comparison logic is the argument; state it.
Ignoring the p = 0 or p = 1 edge cases. The question specifies 0 < p < 1, but noting the exclusion shows you read carefully.

This puzzle rewards the habit of translating a probability target into a generative process — exactly the skill quant interviews are designed to surface.

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FAQ

Why does the expected number of flips equal 2? At each bit position, the generated bit matches p's bit with probability 1/2 (continue) and differs with probability 1/2 (stop). The stopping time is geometric with mean 1/(1/2) = 2.

What if p has two binary expansions? Dyadic rationals have terminating and non-terminating expansions; either works because the generated uniform bits hit the ambiguous boundary with probability zero. Mention it only if asked.

Can I use a different construction? Yes — any method generating the right distribution works, such as rejection sampling on a fine grid. But the binary comparison is the cleanest and most standard answer.

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