Saturday, 9 March 2013

When Multiple Wrongs Make a Right: Parrondo's Paradox

If you have a maths degree, I recommend that instead of reading this post, you read this page instead. This is a very interesting article but not very accessible to non-maths grads. I have tried to make my version more simple and easier to understand.

Imagine you are at a pretty dodgy-looking carnival stand and there is a game you can play in the hope of winning some money. The odds of winning are less than half, which means if you play for a little while, you will probably end up broke.

Opposite this stand is another one, just as dodgy-looking, and with the same premise. Again, the odds of winning are less than half, so again, you will end up broke if you keep playing.

Common sense, and your maths teacher, would tell you not to play either of these two games and to go and buy some candy floss instead. But then a mysterious man called Parrondo appears and tells you to play both games. He says that by playing them both in a particular sequence it is likely to earn you a tidy profit.

How can this be possible? Two bad things surely can't make a good thing, can they? These two games are not tied to each other, you can play them independently, so surely one can't affect the other's odds?

Let's take a look.

Let's keep game A nice and simple. Let's say it's a coin toss game, where if it's heads you win a pound, if it's tails you lose a pound. But because it has to be a game you're more likely to lose, let's make it a biased coin. We'll say the probability of winning is, oh, I don't know... 0.495 (I'm pretending to have chosen that randomly, I'm guessing you can tell I didn't!) and hence the probability of losing is 0.505. If you kept playing this game for a while, you will gradually lose your money. You are likely to lose a pound for every 100 games you play. Anyone can see that this is a bad game.

Game B will have to be slightly more complicated, but bear with me. Game B involves tossing a coin as well. But there are two different coins, and the one that you use is determined by the amount of money you have at the time. If your balance is a multiple of three, you flip the bad coin, otherwise you flip the good coin. Both coins are biased, but differently.

The bad coin has a a 0.095 chance of winning £1, a 0.905 chance of losing £1.
The good coin has a 0.745 chance of winning £1, a 0.255 chance of losing £1.

This game is also bad, as the bad coin is soooooo bad. Even though you will only have to use the bad coin a third of the time, the chance of losing is so great that it overrides any wins you get from the good coin. Unfortunately, my knowledge of Markov chains and such like is not good enough to compute the expected win/loss, but running a simulation shows roughly the same results as with game A. Although it's not obvious, take my word for it, this is a bad game.

I can sense that some of you are already putting two and two together here: does playing game A somehow make you less likely to have a balance that is a multiple of three, and hence less likely to use the bad coin on game B? Obviously you can't only do game B when you have a certain balance, as the carnies would see what you are up to and throw you out. The idea is to plan the sequence of games before starting so you don't look suspicious. The best sequence to choose happens to be ABBABBABBABB... Although there are others you could use to win.  But how does this work?

I simulated this game 100 times with 102 goes each time. The average (mean) winnings at the end of 102 goes was £9. The highest was £42, and the lowest was -£12. Out of the 102 simulations, only 10 ended in a loss. I think this is enough evidence to suggest that this combination of games is worth playing.

My simulation spreadsheet can be downloaded here.

Let's think about how this works.

Say your balance is a multiple of 3, and you are about to play game A. If you lose, you're more likely to win next time, because you'll be flipping the good coin, and also the one after.

Now say your balance is 1 more than a multiple of 3, and you play game A. If you lose, you will probably lose the next one, but win the one after.

If your balance was 1 less than a multiple of 3, and you play game A, If you lose, you will probably win the next one, and also the one after.

So, without putting probabilities on it, it just sort of looks right that you'd end up winning.

So there we have it, two wrongs can make a right!

Now I'm off to alternately buy lottery tickets and put 10p coins in one of those slidey things.

Emma x x x


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