Bertrand’s box

1. Step 1 · Gold coins you could have drawn

   $\displaystyle 2 \text{ in the two-gold box } + 1 \text{ in the mixed box } = 3$

   Every coin is equally likely to be the one you draw. You drew a gold one, so it’s one of these 3, each equally likely.

2. Step 2 · Of those, from the two-gold box

   $\displaystyle 2 \text{ of } 3 = \frac{2}{3}$

   2 of the 3 gold coins sit in the box whose other coin is gold. It’s not 1/2: the two-gold box gives you gold twice as often.

Answer$\displaystyle \frac{2}{3} \approx 66.7\%$