Today I ran across a year 2000 article in the New York Times titled
"Paradox in Game Theory: Losing Strategy That Wins." Given two games which lose most of the time, a strategy can be devised so that we can win most of the time.
The article has a (fairly) detailed description of two example games:
The paradox is illustrated by two games played with coins weighted on one side so that they will not fall by chance to heads or tails.
In game A, a player tosses a single loaded coin and bets on each throw. The probability of winning is less than half.
In game B, there are two coins and the rules are more complicated. The player tosses either Coin 1, loaded to lose almost all the time or Coin 2 loaded to win more than half the time. He plays Coin 1 if his money is a multiple of a particular whole number, like three.
If his money cannot be divided evenly by that number, he plays Coin 2. In this setup, the second coin will be played more often than the first.
After reading this story,
I wrote a quick Python script to demonstrate this game. The problem description isn't definitive on things like the exact probabilities of the coin flips, so we have to make a few assumptions. I set the probability of the coin flip in Game A to p1=0.33, and in Game B, the probability of coin1 to p=0.1 and coin2 to p=0.66. These probabilities are pulled from the air, but they do meet the requirement set forth in the problem description, so they should work.
So the two games alternating should outperform both Game A and Game B when run indivdually. Let's run the program a few times to see if a combination of Games A and B are better than Game A or Game B.
[jcchurch@mcart python]$ python badgames.py
Game A, 1000 runs:
Before: 1000 MONEY
After: 674 MONEY
Game B, 1000 runs:
Before: 1000 MONEY
After: 978 MONEY
Alternating: 1000 runs:
Before: 1000 MONEY
After: 850 MONEY
[jcchurch@mcart python]$ python badgames.py
Game A, 1000 runs:
Before: 1000 MONEY
After: 734 MONEY
Game B, 1000 runs:
Before: 1000 MONEY
After: 890 MONEY
Alternating: 1000 runs:
Before: 1000 MONEY
After: 870 MONEY
[jcchurch@mcart python]$ python badgames.py
Game A, 1000 runs:
Before: 1000 MONEY
After: 636 MONEY
Game B, 1000 runs:
Before: 1000 MONEY
After: 932 MONEY
Alternating: 1000 runs:
Before: 1000 MONEY
After: 820 MONEY
[jcchurch@mcart python]$ python badgames.py
Game A, 1000 runs:
Before: 1000 MONEY
After: 604 MONEY
Game B, 1000 runs:
Before: 1000 MONEY
After: 918 MONEY
Alternating: 1000 runs:
Before: 1000 MONEY
After: 846 MONEY
I just ran the problem described in the New York Times 4 times and got 4 negative results. Game B always outperforms Game A and the combined Games A and B. This article looked very interesting and very improbable. If it's too good to be true, check for yourself.