Parrondo's paradox, a paradox in game theory, has been described as: A combination of losing strategies becomes a winning strategy.[1] It is named after its creator, Juan Parrondo, who discovered the paradox in 1996. A more explanatory description is:

There exist pairs of games, each with a higher probability of losing than winning, for which it is possible to construct a winning strategy by playing the games alternately.

Parrondo devised the paradox in connection with his analysis of the Brownian ratchet, a thought experiment about a machine that can purportedly extract energy from random heat motions popularized by physicist Richard Feynman. However, the paradox disappears when rigorously analyzed.[2] Winning strategies consisting of various combinations of losing strategies were explored in biology before Parrondo's paradox was published.[3]

## Illustrative examples

To clarify the concept of Parrondo's paradox, simple examples are presented initially. It is important to note, however, that according to Harmer and Abbott,[4] other than the last two examples (i.e. the saw-tooth example and the coin-tossing example), the other examples do not truly exhibit Parrondo's paradox in its strictest definition. In particular, since Parrondo's paradox was physically motivated from the flashing Brownian ratchet, a skip free process must be used to properly preserve this ratchet action and so only a +1 or -1 unit payoff structure can be used. See the coin-tossing example for an explicit illustration of this.

### The simple example

Consider two games Game A and Game B, with the following rules:

1. In Game A, you lose $1 every time you play. 2. In Game B, you count how much money you have left ⁠ ⁠— if it is an even number you win$3, otherwise you lose $5. Say you begin with$100 in your pocket. If you start playing Game A exclusively, you will obviously lose all your money in 100 rounds. Similarly, if you decide to play Game B exclusively, you will also lose all your money in 100 rounds.

However, consider playing the games alternatively, starting with Game B, followed by A, then by B, and so on (BABABA...). It should be easy to see that you will steadily earn a total of $2 for every two games. Thus, even though each game is a losing proposition if played alone, because the results of Game B are affected by Game A, the sequence in which the games are played can affect how often Game B earns you money, and subsequently the result is different from the case where either game is played by itself. ### Watering a plant Now consider the case of a potted houseplant. There are two games that can be played with the plant: 1. Game A: pour a continuous stream of water into the pot. 2. Game B: pour no water into the pot. From the point of view of keeping the plant alive, both are losing games; over time, the plant will either take on too much water and rot, or it will dry up. Paradoxically, however, the plant can be kept alive by judiciously switching between the games, alternately watering the plant and turning off the water. ### Pedestrian example Consider the situation of a pedestrian trying to get to a grocery store. The pedestrian can play two games: 1. Game A: cross every street regardless of traffic. 2. Game B: stand still. If the pedestrian plays only A, then they will eventually be struck by a vehicle. However, if they only play B, then they will never move from their position, failing to get to their destination. Parrondo's paradox suggests a solution: counterintuitively, if the pedestrian waits at every red light for traffic to pass, but crosses the street when the light turns green, then they may safely make their way to the grocery store. ### Another pedestrian example This time, the pedestrian is trying to get to the office on time. The two games are: 1. Game A: put the left foot on the ground. 2. Game B: put the right foot on the ground. If the pedestrian plays only A or B, they will effectively hop along on one foot. This is a losing strategy, as it is not an efficient way to travel, and they will be late for work. Rigorous analysis shows that there is a solution, namely to alternate between the two games, which permits the pedestrian to walk, run, or even leap all the way to the office. ### Example of a student writing a test Suppose a student is to write a multiple-choice test; for simplicity, assume the test is a list of statements which are either true or false. To prepare for the test, the student comes up with two games, or strategies: 1. Game A: given a statement, answer "true". 2. Game B: given a statement, answer "false". By sticking to Game A or B, the student will always answer true, or always answer false. If the distribution of the correct responses is close to even, however, then both of the strategies will give a grade of around 50%, which may be unacceptable. A more sophisticated approach would have the student carefully alternate between the two games: for each statement, they might consider whether they believe it to be true or false, and answer accordingly. In other words, paradoxically, if the student knows the subject matter, then they can get a better score this way than by always guessing the same answer regardless of the question. ### 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 profile between the two points. Here also, the marbles will roll towards either end depending on the local slope. 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 alternating from one profile to the other. 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 we again apply the first profile and repeat the steps (points C, D and E now shifted one step to refer to the final valley closest to A). 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 if we define having marbles at point A as a win and having marbles at point B as a loss, we clearly win by alternating (at correctly chosen times) between playing two losing games. ### The coin-tossing example A third example of Parrondo's paradox is drawn from the field of gambling. Consider playing two games, Game A and Game B with the following rules. For convenience, define ${\displaystyle C_{t}}$ to be our capital at time t, immediately before we play a game. 1. Winning a game earns us$1 and losing requires us to surrender \$1. It follows that ${\displaystyle C_{t+1}=C_{t}+1}$ if we win at step t and ${\displaystyle C_{t+1}=C_{t}-1}$ if we lose at step t.
2. In Game A, we toss a biased coin, Coin 1, with probability of winning ${\displaystyle P_{1}=(1/2)-\epsilon }$, where ${\displaystyle \epsilon }$ is some small positive constant. This is clearly a losing game in the long run.
3. In Game B, we first determine if our capital is a multiple of some integer ${\displaystyle M}$. If it is, we toss a biased coin, Coin 2, with probability of winning ${\displaystyle P_{2}=(1/10)-\epsilon }$. If it is not, we toss another biased coin, Coin 3, with probability of winning ${\displaystyle P_{3}=(3/4)-\epsilon }$. The role of modulo ${\displaystyle M}$ provides the periodicity as in the ratchet teeth.

It is clear that by playing Game A, we will almost surely lose in the long run. Harmer and Abbott[1] show via simulation that if ${\displaystyle M=3}$ and ${\displaystyle \epsilon =0.005,}$ Game B is an almost surely losing game as well. In fact, Game B is a Markov chain, and an analysis of its state transition matrix (again with M=3) shows that the steady state probability of using coin 2 is 0.3836, and that of using coin 3 is 0.6164.[5] As coin 2 is selected nearly 40% of the time, it has a disproportionate influence on the payoff from Game B, and results in it being a losing game.

However, when these two losing games are played in some alternating sequence - e.g. two games of A followed by two games of B (AABBAABB...), the combination of the two games is, paradoxically, a winning game. Not all alternating sequences of A and B result in winning games. For example, one game of A followed by one game of B (ABABAB...) is a losing game, while one game of A followed by two games of B (ABBABB...) is a winning game. This coin-tossing example has become the canonical illustration of Parrondo's paradox – two games, both losing when played individually, become a winning game when played in a particular alternating sequence.

The apparent paradox has been explained using a number of sophisticated approaches, including Markov chains,[6] flashing ratchets,[7] simulated annealing,[8] and information theory.[9] One way to explain the apparent paradox is as follows:

• While Game B is a losing game under the probability distribution that results for ${\displaystyle C_{t}}$ modulo ${\displaystyle M}$ when it is played individually (${\displaystyle C_{t}}$ modulo ${\displaystyle M}$ is the remainder when ${\displaystyle C_{t}}$ is divided by ${\displaystyle M}$), it can be a winning game under other distributions, as there is at least one state in which its expectation is positive.
• As the distribution of outcomes of Game B depend on the player's capital, the two games cannot be independent. If they were, playing them in any sequence would lose as well.

The role of ${\displaystyle M}$ now comes into sharp focus. It serves solely to induce a dependence between Games A and B, so that a player is more likely to enter states in which Game B has a positive expectation, allowing it to overcome the losses from Game A. With this understanding, the paradox resolves itself: The individual games are losing only under a distribution that differs from that which is actually encountered when playing the compound game. In summary, Parrondo's paradox is an example of how dependence can wreak havoc with probabilistic computations made under a naive assumption of independence. A more detailed exposition of this point, along with several related examples, can be found in Philips and Feldman.[10]

## Applications

Parrondo's paradox is used extensively in game theory, and its application to engineering, population dynamics,[3] financial risk, etc., are areas of active research. Parrondo's games are of little practical use such as for investing in stock markets[11] as the original games require the payoff from at least one of the interacting games to depend on the player's capital. However, the games need not be restricted to their original form and work continues in generalizing the phenomenon. Similarities to volatility pumping and the two envelopes problem[12] have been pointed out. Simple finance textbook models of security returns have been used to prove that individual investments with negative median long-term returns may be easily combined into diversified portfolios with positive median long-term returns.[13] Similarly, a model that is often used to illustrate optimal betting rules has been used to prove that splitting bets between multiple games can turn a negative median long-term return into a positive one.[14] In evolutionary biology, both bacterial random phase variation[15] and the evolution of less accurate sensors[16] have been modelled and explained in terms of the paradox. In ecology, the periodic alternation of certain organisms between nomadic and colonial behaviors has been suggested as a manifestation of the paradox.[17] There has been an interesting application in modelling multicellular survival as a consequence of the paradox[18] and some interesting discussion on the feasibility of it.[19][20] Applications of Parrondo's paradox can also be found in reliability theory.[21]

## Name

In the early literature on Parrondo's paradox, it was debated whether the word 'paradox' is an appropriate description given that the Parrondo effect can be understood in mathematical terms. The 'paradoxical' effect can be mathematically explained in terms of a convex linear combination.

However, Derek Abbott, a leading researcher on the topic, provides the following answer regarding the use of the word 'paradox' in this context:

## References

1. ^ a b Harmer, G. P.; Abbott, D. (1999). "Losing strategies can win by Parrondo's paradox". Nature. 402 (6764): 864. doi:10.1038/47220. S2CID 41319393.
2. ^ Shu, Jian-Jun; Wang, Q.-W. (2014). "Beyond Parrondo's paradox". Scientific Reports. 4 (4244): 4244. arXiv:1403.5468. Bibcode:2014NatSR...4E4244S. doi:10.1038/srep04244. PMC 5379438. PMID 24577586.
3. ^ a b Jansen, V. A. A.; Yoshimura, J. (1998). "Populations can persist in an environment consisting of sink habitats only". Proceedings of the National Academy of Sciences USA. 95 (7): 3696–3698. Bibcode:1998PNAS...95.3696J. doi:10.1073/pnas.95.7.3696. PMC 19898. PMID 9520428..
4. ^ Harmer, Gregory P; Abbott, Derek (June 2002). "A Review of Parrondo's Paradox". Fluctuation and Noise Letters. 2 (2): 71–107. doi:10.1142/S0219477502000701. Retrieved 30 October 2023.
5. ^ D. Minor, "Parrondo's Paradox - Hope for Losers!", The College Mathematics Journal 34(1) (2003) 15-20
6. ^ Harmer, G. P.; Abbott, D. (1999). "Parrondo's paradox". Statistical Science. 14 (2): 206–213. doi:10.1214/ss/1009212247.
7. ^ G. P. Harmer, D. Abbott, P. G. Taylor, and J. M. R. Parrondo, in Proc. 2nd Int. Conf. Unsolved Problems of Noise and Fluctuations, D. Abbott, and L. B. Kish, eds., American Institute of Physics, 2000
8. ^ Harmer, G. P.; Abbott, D.; Taylor, P. G. (2000). "The Paradox of Parrondo's games". Proceedings of the Royal Society of London A. 456 (1994): 1–13. Bibcode:2000RSPSA.456..247H. doi:10.1098/rspa.2000.0516. S2CID 54202597.
9. ^ G. P. Harmer, D. Abbott, P. G. Taylor, C. E. M. Pearce and J. M. R. Parrondo, Information entropy and Parrondo's discrete-time ratchet, in Proc. Stochastic and Chaotic Dynamics in the Lakes, Ambleside, U.K., P. V. E. McClintock, ed., American Institute of Physics, 2000
10. ^ Thomas K. Philips and Andrew B. Feldman, Parrondo's Paradox is not Paradoxical, Social Science Research Network (SSRN) Working Papers, August 2004
11. ^ Iyengar, R.; Kohli, R. (2004). "Why Parrondo's paradox is irrelevant for utility theory, stock buying, and the emergence of life". Complexity. 9 (1): 23–27. doi:10.1002/cplx.10112.
12. ^
13. ^ Stutzer, Michael. "The Paradox of Diversification" (PDF). Retrieved 28 August 2019.
14. ^ Stutzer, Michael. "A Simple Parrondo Paradox" (PDF). Retrieved 28 August 2019.
15. ^ Wolf, Denise M.; Vazirani, Vijay V.; Arkin, Adam P. (2005-05-21). "Diversity in times of adversity: probabilistic strategies in microbial survival games". Journal of Theoretical Biology. 234 (2): 227–253. Bibcode:2005JThBi.234..227W. doi:10.1016/j.jtbi.2004.11.020. PMID 15757681.
16. ^ Cheong, Kang Hao; Tan, Zong Xuan; Xie, Neng-gang; Jones, Michael C. (2016-10-14). "A Paradoxical Evolutionary Mechanism in Stochastically Switching Environments". Scientific Reports. 6: 34889. Bibcode:2016NatSR...634889C. doi:10.1038/srep34889. ISSN 2045-2322. PMC 5064378. PMID 27739447.
17. ^ Tan, Zong Xuan; Cheong, Kang Hao (2017-01-13). "Nomadic-colonial life strategies enable paradoxical survival and growth despite habitat destruction". eLife. 6: e21673. doi:10.7554/eLife.21673. ISSN 2050-084X. PMC 5319843. PMID 28084993.
18. ^ Jones, Michael C.; Koh, Jin Ming; Cheong, Kang Hao (2018-06-05). "Multicellular survival as a consequence of Parrondo's paradox". Proceedings of the National Academy of Sciences. 115 (23): E5258–E5259. Bibcode:2018PNAS..115E5258C. doi:10.1073/pnas.1806485115. ISSN 0027-8424. PMC 6003326. PMID 29752380.
19. ^ Nelson, Paul; Masel, Joanna (2018-05-11). "Reply to Cheong et al.: Unicellular survival precludes Parrondo's paradox". Proceedings of the National Academy of Sciences. 115 (23): E5260. Bibcode:2018PNAS..115E5260N. doi:10.1073/pnas.1806709115. ISSN 0027-8424. PMC 6003321. PMID 29752383.
20. ^ Cheong, Kang Hao; Koh, Jin Ming; Jones, Michael C. (2019-02-21). "Do Arctic Hares Play Parrondo's Games?". Fluctuation and Noise Letters. 18 (3): 1971001. Bibcode:2019FNL....1871001C. doi:10.1142/S0219477519710019. ISSN 0219-4775. S2CID 127161619.
21. ^ Di Crescenzo, Antonio (2007). "A Parrondo paradox in reliability theory" (PDF). The Mathematical Scientist. 32 (1): 17–22.
22. ^ Abbott, Derek. "The Official Parrondo's Paradox Page". The University of Adelaide. Archived from the original on 21 June 2018.