Even vs. Odd Heads: Does Symmetry Force Fairness?
You flip a coin 10 times. The coin has an unknown probability $p$ of landing heads on each flip, and the flips are independent.
Suppose you are told that the probability of getting an even number of heads (0, 2, 4, 6, 8, or 10) is exactly equal to the probability of getting an odd number of heads. Does this guarantee the coin is fair -- i.e., must $p = 1/2$?
Hints
- Think about computing P(even) - P(odd) directly. Can you write it as a single sum over all $k$ from 0 to 10, using $(-1)^k$ to flip the sign of odd terms?
- The binomial theorem says $\sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} = (a+b)^n$. Notice that $(-1)^k p^k = (-p)^k$ -- can you put this sum in that form?
- Once you recognize the sum equals $(1-p-p)^{10} = (1-2p)^{10}$, the rest is immediate: a real number raised to an even positive power equals zero only when the base is zero.
Worked Solution
How to Think About It: At first glance this might feel like a symmetry argument -- if even and odd are equally likely, maybe the coin is just balanced. But be careful: symmetry in outcomes does not always imply symmetry in the underlying parameter. The question is whether P(even) = P(odd) has a unique solution in $p$, or whether some biased coin could also satisfy it.
The cleanest way to attack this is to write out P(even) - P(odd) as a closed-form function of $p$ and ask when it equals zero. If that function has only one root in $[0,1]$, the implication holds.
Approach: Express the difference P(even) - P(odd) using the binomial theorem, reduce to a power of $(1 - 2p)$, and check for roots.
Formal Solution:
Let $X \sim \text{Binomial}(10, p)$. Define:
$$D(p) = P(X \text{ even}) - P(X \text{ odd}) = \sum_{k=0}^{10} (-1)^k \binom{10}{k} p^k (1-p)^{10-k}$$
The key trick: the factor $(-1)^k$ lets us collapse this sum using the binomial theorem. Notice that:
$$\sum_{k=0}^{10} (-1)^k \binom{10}{k} p^k (1-p)^{10-k} = \bigl((1-p) + (-1) \cdot p\bigr)^{10} = (1 - 2p)^{10}$$
So:
$$D(p) = (1 - 2p)^{10}$$
Setting $D(p) = 0$:
$$(1 - 2p)^{10} = 0 \implies 1 - 2p = 0 \implies p = \frac{1}{2}$$
Since $(1 - 2p)^{10}$ is a perfect 10th power, zero is its only root, and it occurs uniquely at $p = 1/2$.
Answer: Yes -- P(even heads) = P(odd heads) if and only if $p = 1/2$. The condition uniquely forces the coin to be fair.
Intuition
The core trick here is the binomial theorem used as an alternating-sign evaluator. When you compute $\sum_k (-1)^k \binom{n}{k} p^k (1-p)^{n-k}$, you are essentially evaluating the generating function of the Binomial distribution at $-1$ (in the sense that you are replacing $p$ with $-p$ in the standard expansion). The result $(1-2p)^n$ is sometimes called the "signed moment generating function trick" and appears constantly in combinatorics and probability -- whenever you want to separate even-indexed and odd-indexed terms in a sum.
The broader lesson is about injectivity: just because a condition sounds like it should have many solutions does not mean it does. Here, the symmetry condition P(even) = P(odd) looks like it might hold for a range of biased coins, but the algebraic structure forces exactly one solution. This comes up in quant interviews a lot -- problems that appear to have ambiguous answers are often designed to test whether you can nail down uniqueness rigorously rather than just guessing 'yes it must be fair' by intuition.