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.

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.^[4] 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.