Fair Price of a Coin-Flipping Game

Expectation · Easy · Free problem

You are offered a game: flip $n = 10$ fair coins. The payoff equals $H^2$, where $H$ is the number of heads.

  1. What is the fair price of this game (i.e., the expected payoff)?
  2. Generalize: for a payoff function $f(H)$, write down the fair price formula.
  3. Explain intuitively why the fair price of the $H^2$ game is not simply $(E[H])^2$.

Hints

  1. The fair price is always the expected payoff under the true probabilities -- no discounting needed for a one-period game with no time value.
  2. For the $H^2$ payoff, use the identity $E[H^2] = \text{Var}(H) + (E[H])^2$ to avoid summing 11 terms.
  3. The gap between $E[H^2]$ and $(E[H])^2$ is exactly $\text{Var}(H)$ -- this is not a coincidence, it is the definition of variance, and it connects directly to Jensen's inequality for convex functions.

Worked Solution

How to Think About It: The fair price of any game is its expected value under the true probability measure -- the amount at which a risk-neutral player is indifferent. For a binomial payoff, you weight each possible outcome by its probability and sum. The interesting subtlety here is Part 3: $E[H^2] \neq (E[H])^2$ because of variance. This is Jensen's inequality -- squaring is a convex function, so the expectation of the square exceeds the square of the expectation by exactly $\text{Var}(H)$.

Quick Estimate: With $n = 10$ fair coins, $E[H] = 5$ and $\text{Var}(H) = np(1-p) = 10 \times 0.5 \times 0.5 = 2.5$. So $E[H^2] = \text{Var}(H) + (E[H])^2 = 2.5 + 25 = 27.5$. Done. No need to enumerate all 11 outcomes.

Approach: Use the variance decomposition $E[H^2] = \text{Var}(H) + (E[H])^2$ for Part 1; state the general formula for Part 2; explain via Jensen for Part 3.

Formal Solution:

Part 1 -- Fair price of the $H^2$ game.

$H \sim \text{Binomial}(n=10, p=0.5)$. We use: $E[H^2] = \text{Var}(H) + (E[H])^2$

For a $\text{Binomial}(n, p)$: $E[H] = np = 5$ and $\text{Var}(H) = np(1-p) = 2.5$. Therefore: $\boxed{E[H^2] = 2.5 + 25 = 27.5}$

Part 2 -- General fair price formula.

For any payoff function $f$, the fair price is: $P = E[f(H)] = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{2^n} f(k)$

This weights each value of $f$ by its binomial probability. For continuous or smooth $f$, you can often compute moments directly (as in Part 1) rather than summing all terms.

Part 3 -- Why $E[H^2] \neq (E[H])^2$.

The gap is the variance: $E[H^2] - (E[H])^2 = \text{Var}(H) = 2.5 > 0$. More generally, for any convex function $g$, Jensen's inequality says $E[g(H)] \geq g(E[H])$. Since $g(x) = x^2$ is convex, $E[H^2] \geq (E[H])^2$ always, with equality only when $H$ is constant (zero variance). The gap reflects the fact that the game pays off extra on the upside (a run of many heads contributes disproportionately because of the square) and is not offset symmetrically on the downside.

Answer: - Fair price of the $H^2$ game with $n=10$:

7.5$ - General formula: $P = \sum_{k=0}^{n} \binom{n}{k} 2^{-n} f(k)$ - The excess $E[H^2] - (E[H])^2 = \text{Var}(H) = 2.5$ comes from Jensen's inequality applied to the convex function $x^2$.

Intuition

The variance decomposition $E[X^2] = \text{Var}(X) + (E[X])^2$ is one of the most useful identities in probability. It says the second moment has two contributions: one from the mean (where the distribution is centered) and one from the spread (how much it disperses around that center). When pricing nonlinear payoffs, this spread component always matters -- it is why options have time value even when the stock is at-the-money.

This is Jensen's inequality in action. Convex payoffs -- like $H^2$, or a call option's payoff $\max(S-K, 0)$ -- are worth more than their 'plug-in' value at the expected outcome, because upside realizations contribute more than the downside realizations take away. Concave payoffs (like $\sqrt{H}$) are worth less. This asymmetry is the core of option pricing: volatility creates value for long convexity positions, which is why vega is always positive for vanilla options.

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