Coin Flip Probabilities with Four Fair Coins

Probability · Easy · Free problem

You flip four fair coins.

  1. What is the probability of getting exactly 2 heads?
  1. What is the probability of getting an even number of heads (0, 2, or 4)?

Hints

  1. Each of the 16 equally likely outcomes has probability
    /16$. How many of them contain exactly 2 heads?
  2. For part 2, try computing the count for each case (0, 2, 4 heads) using $\binom{4}{k}$ and summing. Then ask whether the answer surprises you.
  3. For the general result on even heads, try evaluating $(1+1)^n + (1-1)^n$ and think about what each term contributes to the sum of even-indexed binomial coefficients.

Worked Solution

How to Think About It: Both parts are straightforward binomial counting problems. For part 1, you need to count the ways to choose 2 coins out of 4 to land heads. For part 2, you can grind through the cases -- or notice a slick symmetry argument that makes the answer obvious without any arithmetic. In an interview, seeing that shortcut is what separates a good answer from a great one.

Quick Estimate: There are

^4 = 16$ equally likely outcomes for 4 fair coins. Intuitively, exactly 2 heads should be the most common single outcome -- there are more ways to get a 50/50 split than any extreme. And an even number of heads should be roughly half the outcomes by symmetry. So we expect something around $6/16$ for part 1 and close to
/2$ for part 2 -- let's verify.

Approach: Count favorable outcomes using the binomial coefficient $\binom{n}{k}$, then divide by total outcomes

^n$.

Formal Solution:

Part 1 -- Exactly 2 heads:

There are $\binom{4}{2} = 6$ ways to choose which 2 of the 4 coins land heads. The total number of outcomes is ^4 = 16$.

$P(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^4 = \frac{6}{16} = \frac{3}{8}$

Part 2 -- Even number of heads:

The favorable outcomes are 0, 2, or 4 heads. Count each:

$\binom{4}{0} + \binom{4}{2} + \binom{4}{4} = 1 + 6 + 1 = 8$

$P(X \in \{0, 2, 4\}) = \frac{8}{16} = \frac{1}{2}$

Alternatively, the slick way: evaluate $(1+x)^4$ at $x = 1$ and $x = -1$. We get $(1+1)^4 = 16$ and $(1-1)^4 = 0$. Note that $(1+x)^4 + (1-x)^4 = 2 \sum_{k \text{ even}} \binom{4}{k} x^k$, so at $x = 1$:

$\sum_{k \text{ even}} \binom{4}{k} = \frac{(1+1)^4 + (1-1)^4}{2} = \frac{16 + 0}{2} = 8$

This gives $8/16 = 1/2$. And crucially, this works for ANY $n \geq 1$: for $n$ fair coins, the probability of an even number of heads is always exactly

/2$, because $\sum_{k \text{ even}} \binom{n}{k} = 2^{n-1}$ for all $n \geq 1$.

Answer: - Exactly 2 heads: $P(X = 2) = 3/8$ - Even number of heads: $P(X \in \{0, 2, 4\}) = 1/2$

Intuition

The result for part 2 -- that the probability of an even number of heads is always exactly

/2$ for any number of fair coins -- is a beautiful example of a symmetry argument beating a direct calculation. You can see it without any algebra: flipping the last coin toggles the parity of the head count. Since that last flip is fair and independent, exactly half of all outcomes have even parity. This parity-flipping argument works for any $n \geq 1$ and requires zero computation.

In broader quant work, this kind of parity and symmetry reasoning shows up constantly -- in options pricing (put-call parity), in random walk analysis (reflection principle), and in combinatorics problems on trading floors. The deeper lesson is to always look for a symmetry that collapses the problem before reaching for the calculation. If you can split the sample space into two halves that are mirror images of each other, you often get the answer instantly.

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