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Chance of Blue Eyes — Punnett Square

Result

6.25% blue

Result: 6.25 % blue

This page answers one question: how likely is a blue-eyed child. Blue needs a child who inherits no brown allele and no green one, so the two chances multiply. Two brown-eyed carriers (Bb and Gg on both sides) give 6.25 % blue, 18.75 % green and 75 % brown — the full split is in the table below.

Worked examples

Case 1
Parent 1 — brown gene pair
Bb — brown, carries a non-brown allele
Parent 1 — green gene pair
Gg — carries one green allele
Parent 2 — brown gene pair
Bb — brown, carries a non-brown allele
Parent 2 — green gene pair
Gg — carries one green allele

6.25%

Open with these values
Case 2
Parent 1 — brown gene pair
bb — no brown allele at all
Parent 1 — green gene pair
Gg — carries one green allele
Parent 2 — brown gene pair
bb — no brown allele at all
Parent 2 — green gene pair
Gg — carries one green allele

25.00%

Open with these values
Case 3
Parent 1 — brown gene pair
BB — brown, no other allele
Parent 1 — green gene pair
Gg — carries one green allele
Parent 2 — brown gene pair
bb — no brown allele at all
Parent 2 — green gene pair
gg — no green allele

0.00%

Open with these values

How it's calculated

blue = chance of no B × chance of no G

  1. StepPick each parent's pair at the brown gene: blue and green eyes are always bb.
  2. StepPick each parent's pair at the green gene: blue eyes are always gg.
  3. ResultBoth genes are inherited independently, so the two chances are multiplied.

What this number means

Blue is the one colour in this model defined by two absences: a child arrives at it only by inheriting no brown allele and no green one. That is what makes it worth computing, because both gene pairs of both parents have to cooperate for it, while brown is settled at the brown locus alone. Each field contributes one of three numbers — the chance that this pair passes on its recessive allele: 1 for bb or gg, ½ for Bb or Gg, 0 for BB or GG. The two loci count as independent, so the four chances simply multiply. The defaults are the classic cross, both parents Bb and Gg: ½ × ½ = ¼ for no brown, ¼ again for no green, and ¼ × ¼ = 6.25 % blue. The same structure explains the edges of the table. A single BB or GG sets its own factor to 0 and rules blue out completely, while two parents who are bb and gg have nothing else to pass on and the model returns 100 %. What the page cannot supply is its own input. Current genetics describes eye colour as polygenic — OCA2 and HERC2 carry most of it, with a dozen further genes contributing — so the tidy gene pair you are asked to choose is a classroom convention, not something readable off an iris or a family tree. Everything downstream of that choice is exact; the choice itself is the simplification.

A teaching model, not a genetic test

Current genetics describes eye colour as polygenic, with OCA2 and HERC2 doing most of the work and a dozen further genes contributing. The arithmetic on this page is exact; the two-gene assumption underneath it is a classroom convention.

Blue is the figure that uses all four inputs

The brown share follows from the two brown gene pairs alone, so the green fields would never move it. The table above lists brown, green and blue for the same crosses, and the three always add up to 100 %.

Enter gene pairs, not eye colours

A brown-eyed person is BB or Bb, and the two give very different answers. Blue eyes are the unambiguous case, always bb and gg.

Commonly misread

Two brown-eyed parents cannot have a blue-eyed child.

In this model they can, at 6.25 % when both carry a hidden non-brown allele and neither carries green. If either parent is BB the chance is zero.

The three colours in the model cover every real iris.

Two genes with a strict ranking only produce brown, green or blue. Real irises form a continuum of pigment and light scattering, with no option here for hazel, amber or grey.

A result of 6.25 % means the child will not have blue eyes.

It is the share expected across such crosses, not a prediction for one individual child.

Reference table

Cross (brown gene · green gene)BrownGreenBlue
Bb × Bb · Gg × Gg75 %18.75 %6.25 %
bb × bb · gg × gg0 %0 %100 %
BB × bb · Gg × gg100 %0 %0 %
bb × bb · Gg × Gg0 %75 %25 %
Bb × bb · gg × Gg50 %25 %25 %

Questions

Can two brown-eyed parents have a blue-eyed child?

Yes, with a 6.25 % chance if both carry a hidden non-brown allele and neither carries green. If either parent is BB, the chance drops to zero, because two brown alleles mean every child gets one.

Why does this page give only the blue figure?

Because a calculator here answers with one number, and blue is the one that depends on all four inputs — the brown share is decided by the two brown gene pairs alone. The green and brown shares for the same crosses are listed in the table above, and the three always add up to 100 %.

Why enter gene pairs instead of eye colours?

Because eye colour does not fix the gene pair: a brown-eyed person is BB or Bb, and the two give very different answers. Blue eyes are the unambiguous case, always bb and gg. A blue- or green-eyed parent, sibling or child proves that a brown-eyed person carries the recessive allele.

Why is there no option for hazel, amber or grey?

Because two genes with a strict ranking can only produce brown, green or blue. Real irises form a continuum of pigment and light scattering, which is one of the clearest signs that the model is a simplification.

How accurate is this model?

It is a teaching model, not a genetic test. Current genetics describes eye colour as polygenic — OCA2 and HERC2 do most of the work, with a dozen further genes contributing — and the simple dominant-brown pattern is explicitly regarded as too simplistic. The arithmetic on this page is exact; the assumption it rests on is a classroom convention.

Sources and last check

  1. en.wikipedia.org

Information, not professional advice.