128 players are put in a random knockout bracket. Every game knocks one player out, and the better player always wins. You pick two of the players at random. What’s the chance they play each other at some point?
An 8-player bracket, to picture it.
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Answer: 1/64 ≈ 1.56%
In 8 players, 7 games.
Step 1 · Games played
128−1=127
Each game knocks out exactly one player, and all but the winner get knocked out.
Step 2 · Pairs of players
(2128)=128×127÷2=8,128
Each game is played by one pair of players.
Step 3 · Chance your pair is one of the games
127÷8,128=641
The bracket is random, so every pair is equally likely to be one of the 127 pairs that meet.