You’ve just made it into the finals of a chess tournament. Although your opponent is “stronger” than you, you are trying to think of some strategy to beat him on the board.

There will be 2 (two) matches, and if they both result in a tie, a third match will be played to decide the winner, who will take home a $1,000 prize. If the play-off also ends in a draw, you’ll share the prize with your opponent.

The scoring system is the one used in chess tournaments since the middle of the 19th century: players who scored a win in a game are awarded 1 (one) point, while those scoring draws are given a 0.5 (half-point) each. Losing a game, as you might expect, is worth 0 (zero) point.

You know, from previous games against the same opponent, that you can choose between two different strategies: “play fearlessly”, in which you win 45% and lose 55% of the time, or “play defensively”, in which you draw 90% and lose 10% of the time (but selecting this alternative you’ll have no chance to win). Note that both strategies have the same expected value since 0.45 x 1.0 = 0.9 x 0.5.

If you play optimally, what are your chances of beating your opponent (and winning the $1,000 prize)?

The decision tree below shows all possible sequence of events that result in your winning the tournament. To keep the diagram simple, all remaining, non-winning sequences were omitted. Obviously this decision tree applies equally well to any other zero-sum game besides chess that can also result in a draw (e.g. checkers).

It is really surprising that the probability of you, the “weaker” player, winning the finals is around 0.537 (or 0.536625 to be exact), which is greater than 0.5. Each strategy, analyzed separately, confirms that, on the average, your opponent will win more games than you. Yet the odds of grabbing the $1,000 are in your favor! How come?