# The Last Banana

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.

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