This puzzle came to me via Futility CLoset

Two people are stranded on an island with only one banana to eat. To decide who gets it, they agree to play a game. Each of them will roll a fair 6-sided die. If the largest number rolled is a 1, 2, 3, or 4, then Player 1 gets the banana. If the largest number rolled is a 5 or 6, then Player 2 gets it. Which player has the better chance?

I don’t think this is an especially difficult puzzle, but I think the key insight is nice. Spoilers below the line.

The naive instinct, of course, is that player 1 has more options available, so their chance must be larger.

But no! The probability of both rolling a 4 or lower is $\left( \frac23 \right)^2$, or $\frac{4}{9}$, which is less than a half! Player 2 is slightly more likely to win (a ratio of 5:4).

As long as player 1’s win probability is smaller than $\frac{1}{\sqrt{2}}$, player 2 will (more likely) win.