Parrondo's paradox
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Parrondo's paradox is a paradox in game theory and is often described as: A losing strategy that wins. It is named after its creator Juan Parrondo, a Spanish physicist. Mathematically a more involved statement is given as:
- Given two games, each with a higher probability of losing than winning, it is possible to construct a winning strategy by playing the games alternately.
The paradox is inspired by the mechanical properties of ratchets, the familiar saw-tooth tools used in automobile jacks and in self-winding watches.
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[edit] Illustrative examples
[edit] The saw-tooth example
Consider an example in which there are two points A and B having the same altitude, as shown in Figure 1. In the first case, we have a flat profile connecting them. Here if we leave some round marbles in the middle that move back and forth in a random fashion, they will roll around randomly but towards both ends with an equal probability. Now consider the second case where we have a saw-tooth like region between them. Here also, the marbles will roll towards either ends with equal probability. Now if we tilt the whole profile towards the right, as shown in Figure 2, it is quite clear that both these cases will become biased towards B.
Now consider the game in which we alternate the two profiles while judiciously choosing the time between altering from one profile to the other in the following way.
When we leave a few marbles on the first profile at point E, they distribute themselves on the plane showing preferential movements towards point B. However, if we apply the second profile when some of the marbles have crossed the point C, but none have crossed point D, we will end up having most marbles back at point E (where we started from initially) but some also in the valley towards point A given sufficient time for the marbles to roll to the valley. Then again we apply the first profile and repeat the steps. If no marbles cross point C before the first marble crosses point D, we must apply the second profile shortly before the first marble crosses point D, to start over.
It easily follows that eventually we will have marbles at point A, but none at point B. Hence for a problem defined with having marbles at point A being a win and having marbles at point B a loss, we clearly win by playing two losing games.
[edit] The coin-tossing example
Consider playing two games, Game A and Game B with following rules:
- Winning a game will get us $1 and losing requires us to surrender $1.
- In Game A, we toss a biased coin, Coin 1, with probability of winning P1 = (1 / 2) − ε.
- In Game B, we first test whether our earnings are a multiple of 3 or not. If not, we toss another biased coin, Coin 2, with probability of winning P2 = (3 / 4) − ε. If our earnings are a multiple of 3, we toss a third biased coin, Coin 3, with probability of winning P3 = (1 / 10) − ε.
It is clear that by playing Game A, we will surely lose in the long run. For Game B, it is also possible to prove that in the long run it is also a losing game. However, when played alternately, it results in a winning combination as the outcome of one decides the nature of the other.
[edit] Application of Parrondo's paradox
Parrondo's paradox is used extensively in game theory, and its application in engineering, population dynamics, financial risk, etc. are also being closely looked into.
Most researchers discount its use in stock markets as this theory specifically required that Games A and B must be set up to copy a ratchet, which means they must have some direct interaction.
[edit] References
- Parrondo's Paradox Game
- Alternate game play ratchets up winnings: It's the law
- Official Parrondo's paradox page
