The Probability of Having Four Babies of the Same Sex

A binomial-distribution proof that a family of 4 children has exactly a 1-in-8 (12.5%) chance of all four being the same sex.
statistics
probability
Author

Abdullah Al Mahmud

Published

September 16, 2026

Let’s say a family has 4 children. What’s the probability all four are the same sex — either 4 girls or 4 boys?

We can find this using the binomial distribution, treating each birth as an independent event (the successive births are unrelated) with two possible outcomes: a boy or a girl.

Direct Solution

There are 4 babies. We need to count how many of the possible birth sequences have all four babies the same sex.

Since there are 4 babies, the size of the sample space is \(2^4 = 16\).

To mention a few, the babies can be like

BBGB, GBGG, GGGB, and so on.

And 4 children out of 4 can be selected in \(\binom{4}{4}=1\) way.

Therefore the probability of choosing 4 girls or boys = \(P(4G \cup 4B) = P(4G) + P(4B) = \frac{\binom{4}{4}}{2^4}+\frac{\binom{4}{4}}{2^4} = \frac{1}{16} + \frac{1}{16} = \frac{1}{8}=0.125\)

Solution with Binomial Distribution

1. The Mathematical Assumptions

  • Let the probability of having a girl (\(G\)) be \(p = 0.5\).
  • Let the probability of having a boy (\(B\)) be \(q = 1 - p = 0.5\).
  • Let the total number of trials (births) be \(n = 4\).

The probability of getting exactly \(k\) successes in \(n\) trials is given by the binomial formula: \[P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k}\]

2. Proving 4 Girls in a Row (\(k = 4\))

To find the probability of a family having exactly 4 girls out of 4 children: \[P(\text{4 Girls}) = \binom{4}{4} \cdot (0.5)^4 \cdot (0.5)^{4-4}\]

  • \(\binom{4}{4} = 1\) (there is only 1 unique way to have 4 girls: \(GGGG\))
  • \((0.5)^4 = 0.0625\)
  • \((0.5)^0 = 1\)

\[P(\text{4 Girls}) = 1 \cdot 0.0625 \cdot 1 = \mathbf{0.0625 \text{ (or } 6.25\%)}\]

3. Proving 4 Boys in a Row (\(k = 0\))

Using the same formula, the probability of having 0 girls (which means exactly 4 boys, \(BBBB\)) is: \[P(\text{4 Boys}) = \binom{4}{0} \cdot (0.5)^0 \cdot (0.5)^{4-0}\]

  • \(\binom{4}{0} = 1\)
  • \((0.5)^0 = 1\)
  • \((0.5)^4 = 0.0625\)

\[P(\text{4 Boys}) = 1 \cdot 1 \cdot 0.0625 = \mathbf{0.0625 \text{ (or } 6.25\%)}\]

4. The Final Proof: “Same Sex”

Because a family can have either all girls OR all boys, and these two outcomes are mutually exclusive, we add their individual binomial probabilities together: \[P(\text{Same Sex}) = P(\text{4 Girls}) + P(\text{4 Boys})\] \[P(\text{Same Sex}) = 0.0625 + 0.0625 = \mathbf{0.125}\]

Converting \(0.125\) to a percentage gives 12.5%. As a fraction, \(0.125 = \frac{1}{8}\) — so roughly 1 family in 8 with four children will have them all the same sex.

A Note on the Assumption

This proof relies on \(p = 0.5\) for a clean, exact result. In reality, the sex ratio at birth is closer to 105 boys per 100 girls (\(p \approx 0.512\) for a boy), which shifts the “4 boys” probability slightly higher than “4 girls.” The gap is small enough that \(\frac{1}{8}\) remains an excellent approximation for four children.