Puzzle 8 of 26 · Medium

The two best in the final

Your answer

128 players are put in a random knockout bracket, and the better player always wins. What’s the chance the two best players meet in the final?

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Answer: 64/127 ≈ 50.4%

The two halves of the bracket.

Step 1 · They meet in the final when they’re in different halves
$\displaystyle \text{each then beats everyone in their half}$
The best two lose to no one else. If they’re in different halves they both reach the final; in the same half they meet earlier.
Step 2 · Spots left for the second best
$\displaystyle 127,\, \text{ with } 64 \text{ in the other half}$
Put the best player anywhere. The second best is equally likely to be in any of the other 127 spots.
Step 3 · Chance
$\displaystyle 64 \div 127 = \frac{64}{127}$
64 of those 127 spots are in the half the best player isn’t in.

Answer $\displaystyle \frac{64}{127} \approx 50.4\%$

Step 3 of 3