# A Triplets Puzzle

A simplified version of my New Scientist triplet puzzle:

— Rob Eastaway (@robeastaway) June 10, 2020

Amy had triplets on her birthday. What's the chance that on a future birthday, the triplets' ages will add up exactly to Amy's? (Don't post answers underneath, I'll post answer later).

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You know you’re in for a treat when @robeastaway asks a birthday question. I don’t know if the poll shows up embedded, but you’ll see that the answers are split pretty evenly between $\frac{1}{3}$ and $\frac{1}{2}$, each of which is about twice as popular as “100%”. You might wish to have a go at it first. Spoilers below the line.

(Tacitly implied in the question: Amy and all of the children live for at least as long as is necessary to make the puzzle work.)

Although almost every voice in my head was insisting “of course it’s a third! of course it’s a third!”, one very patient and persistent naysayer pointed out “if it was a third, he wouldn’t be asking.”

### Getting an answer

Suppose Amy’s age when she gives birth is $a$. After $x$ years, on their mutual birthday, Amy will be $a+x$ years old and the triplets $3x$; those will be equal when $a + x = 3x$, or $x = \frac{a}{2}$.

That is to say, when the triplets’ age is half Amy’s when she gave birth, the sums will be the same - and that can only be true if Amy gives birth when her age is an even number of years. The probability is 50%.

### Checking that

Despite the persistent voice, the but-obviously chorus wanted me to check. The algebra says, if Amy gave birth on, let’s say, her 30th birthday, the triplets would catch her up 15 years later.

On Amy’s 45th birthday, the triplets would all be 15 - it checks out.

If she was 31, the twins would turn 15 (totalling 45) on her 46th birthday and 16 (total 48) on the 47th, and the two would not meet.

32 would work. 33 wouldn’t. 50% seems to hold up.

### Intuition

So the algebra and the testing check out - but neither answers ‘why’ to my satisfaction. So I started thinking about other numbers of children. If Amy had only one child, the total of its age would never catch up with hers - it would always be $a$ years younger.

Had she had twins, the total would catch up when both twins are $a$ years old and Amy celebrates her $2a$th birthday - which is certain to happen.

Every extra child means the total catches up an extra year-of-age for every year that passes - so triplets catch up after $\frac{a}{2}$ years!

(Interestingly, if Amy had *no* children on her zeroth birthday, their ages would equal hers the day she was born. Which makes sense: she was zero, and their total age is always zero…)

A lovely puzzle - was your intuition better than mine? Do you have a better explanation? I’d love to hear it.

## A selection of other posts

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