## The Three Game Paradox

Posted by on Saturday, November 15, 2014 Under: Mathematics

*In my continuing series on curious mathematical paradoxes, I have reviewed several interesting and counter-intuitive phenomena in both mathematics and statistics. Recently a couple of readers of my blog asked me about the Three Game Paradox. I must admit that I don't see this as a true paradox, but rather simple statistics, but it is still interesting and as such I present it here for your pleasure...*

Suppose that you have just joined a chess club, and you are really a novice player. It seems as though everyone else in the club is far more experienced and can beat you regularly. And because you are so easy to defeat, other players enjoy matches with you to improve their own win-loss record.

And so to convince you to play, two of the better players offer you a bet. You will play three games of chess, and if you can win two in a row, they will pay you a $500 prize. The only constraint is that you must play them alternately, so that the first and third game are played against the same opponent and the second game is against the other one.

These two players are themselves of very different skill levels, Gary the Good Player can beat you 50% of the time. Bob the Best Player can beat you 90% of the time. So which order should you play them?

Clearly you want to have the two matches against the worse player, G. The you only need to get lucky once against Bob to win the prize. That makes perfect sense, and most players would choose this option. And they would be wrong.

Consider the ways that you can win when playing the set of Gary, Bob, Gary. There is a small chance that you could win all three games, with a probability of 2.5%. Or you could win the first two matches and lose the third, with probability of 2.5% or lose the first and win the last two, also with probability of 2.5%. The total chance of winning is 7.5%.

But now consider what happens when you play the set of Bob, Gary, Bob. Now the chance of winning all three drops to 0.5%. That isn't looking good for you! But the probability of winning the first two and losing the last is now 4.5% and the probability of winning the last two is also 4.5%. The total chance of winning is now 9.5%.

It turns out to be better to play the best player twice, and in the extreme cases you can actually double your chances of winning using this strategy.

But why should it be so? The trick is that the bet required two consecutive wins. It doesn't matter if you beat the weaker player twice, if the best player beats you on the second game. Whereas if the best player beats you on the first game, you get a second chance on the third game.

Another way of thinking about this is to make Gary a really bad player who you beat every time you play him. Then the question is reduced a choice of playing Bob once or twice, with the bet won if you win even one game. Clearly you are better off with two chances to win than just one.

It is counter-intuitive, but in my opinion not paradoxical. Still it is a fun piece of recreational statistics and one worth knowing.

In : Mathematics