Dear Uncle Colin,

I bought a 750-piece jigsaw from a charity shop and found it had only 740 pieces! If I bought the same jigsaw again, and found it was missing ten random pieces, how likely is it I’d be able to complete the puzzle?

Just Imagine Going Shopping As Well

Hi, JIGSAW, and thanks for your message!

This looks to me like a birthday problem. I can solve it exactly, but I’m going to ballpark it first, then solve it exactly.

### Ballpark

The probability of the first missing piece being present in the second puzzle is $\frac{740}{750}$, or $\frac{74}{75}$. If that were constant, the probability of all ten pieces being present would be $\br{\frac{74}{75}}^{10}$.

This looks like Ninja work. Taking logs gives $10\ln\br{1 - \frac{1}{75}}$, which is about $-\frac{2}{15}$. Unlogging gives about $\frac{13}{15}$, or roughly 87%.

This is an overestimate – the presence probability of the later pieces varies depends on the presence of the earlier one – given all of the previous ones are there, the probability drops, and my guess would be closer to 80%.

### Doing it properly

The true probability is $\frac{740}{750}\times \frac{739}{749} \times \dots \frac{731}{741}$.

That’s $\frac{740!}{730!}\times\frac{740!}{750!}$. Gosh! To get a exact answer, I’d stick that directly into Wolfram|Alpha, or the original fraction into a calculator.

It works out to be 0.874 – so my first answer was considerably closer!

Hope that helps. Enjoy the jigsaw!

- Uncle Colin