Oct 08, 2026 by Offer

Millennium Probability Puzzle 2027: Unfair Coins Odd vs Even

The Millennium unfair coins odd even heads result: P(odd) equals P(even) exactly when the product of (1 - 2pi) over all coins is zero — i.e. when some coin has pi = 1/2. The difference factorises via a generating-function argument. Commonly reported by candidates.

What This Question Assesses

This tests whether you reach for the right mathematical tool. Expanding cases for n = 1, 2, 3 and guessing the pattern is slow and error-prone; the generating-function approach solves all n at once. The interviewer is watching for that leap to the closed form — exactly the structural thinking quant work requires.

Millennium Unfair Coins Odd Even Heads: How to Answer

Step 1 — Define the difference. "Let D = P(odd) − P(even). I want the condition for D = 0."
Step 2 — Find the multiplicative structure. "Consider one coin: it contributes +pᵢ to odd-count parity flips... more cleanly, encode parity with the polynomial: for coin i, the factor (q + pᵢx) tracks heads with x. Setting x = −1 gives (qᵢ − pᵢ) = (1 − 2pᵢ), and the coefficient difference between odd and even powers of the full product is exactly D."
Step 3 — State the result. "D = ∏(1 − 2pᵢ). So P(odd) = P(even) iff this product is zero, iff some coin has pᵢ = 1/2."
Step 4 — Sanity-check. "If all coins are fair, the product is 0 — correct, symmetry gives equality. If one coin always lands heads (p = 1), the factor is −1 and parity just flips deterministically — also consistent."

An example line: "The difference between odd and even probabilities factorises as the product of (1 − 2pᵢ) over the coins, so equality holds exactly when one factor vanishes — when some coin is fair at p = 1/2."

Millennium Unfair Coins Odd Even Heads: Common Mistakes

Trying induction on n without the closed form. It works but is messy; the generating-function argument is cleaner and generalises. If you start inducting, pivot when you see the product structure.
Forgetting the degenerate cases. pᵢ = 0 or 1 still fit the formula — check your condition against them rather than assuming interior probabilities.
Confusing P(odd) = P(even) with P(odd) = 1/2. They coincide here (probabilities sum to 1), but stating the condition as "the product vanishes" is the precise answer the question asks for.

The best probability answers find the structure that makes the problem trivial — here, that structure is the factorised difference.

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FAQ

What is the generating-function trick here? Encoding each coin as (qᵢ + pᵢx) and evaluating at x = −1 turns "odd minus even" into a product of (1 − 2pᵢ). It converts a combinatorial parity question into algebra.

Does the result depend on n? No — the condition is simply that at least one coin has pᵢ = 1/2, regardless of how many coins there are or what the other probabilities are.

What if two coins have p = 1/2? The condition still holds — the product is still zero. Extra fair coins do not change the equality.

Can this be generalised? Yes: for "number of heads ≡ r mod m" questions, use roots-of-unity filters — the same idea with complex evaluations of the generating function instead of x = −1.

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