How to cheat a bad mathematician
That was the title of Juan Parrondo's talk in Turin where he first showed the games. It was never published, but it is still cited as the start of the story.
Two simple coin games, each of which takes your money in the long run. Before every round you toss one more, fair coin and use it to pick which of the two to play. What happens to your money over thousands of rounds? Lock in your answer, then play yourself, run thousands of players and see where the result comes from.
Game A loses $1 per 100 rounds on average, game B $0.87. But if you pick A or B at random before each round, you win $1.57 per 100 rounds on average. The secret is that game B depends on your capital: on its own it lands you in a bad spot for itself too often, and game A knocks you out of it. Here you play yourself, work out the exact Markov chain and meet the same trick in the physics of molecular motors.
The casino has two tables. At each one you bet $1 on a coin toss: win and your capital goes up by 1, lose and it goes down by 1. The only difference is which coin they toss.
We count from zero: +5 means you are $5 ahead, −4 means $4 behind. You can borrow as much as you like, so the game never stops.
One coin, slightly unfair: it wins with chance 49.5%. Each round takes 1 cent on average, $1 per 100 rounds.
Two coins, and which one is tossed depends on your capital. If it is a multiple of 3 (…, −6, −3, 0, 3, 6, …), they toss a bad coin that wins with chance 9.5%. Otherwise a good one, with chance 74.5%. Played on its own, B also loses on average, a little less than A: $0.87 per 100 rounds.
Before each round you toss a fair coin: heads you play A, tails you play B. What happens to your capital over thousands of rounds?
"Two losers make a loser"
In fact the capital grows: +$1.57 per 100 rounds on average. Game B is bad only because on its own it often puts you on a multiple of 3, where its coin is bad. Game A is almost fair, and it shuffles the capital so that B more often plays with its good coin.
These odds come from Gregory Harmer and Derek Abbott, 1999: 1/2 − ε, 1/10 − ε and 3/4 − ε with ε = 0.005.
Pick an option before reading on. Before each round a fair coin decides: A or B. You play thousands of rounds.
Reminder: only A loses $1 per 100 rounds, only B $0.87.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Don't think of game B as one coin. The capital decides which of its two coins is tossed. So what matters is how often the capital is a multiple of 3.
Play only B in your head with a capital of 5. The good coin most often lifts you to 6. What most often happens at 6, and where do you go back to?
Game A is almost fair and ignores the capital. How does it change the share of rounds in which B sits on a multiple of 3? And what happens to B's wins if that share gets close to a third?
Switch on one of the ways and play 100 or 1,000 rounds: only A, only B, in turn ABAB, in pairs AABB or at random. One run is a lottery, so compare it with the dashed line, the exact average. And in "By hand" mode pick the table before each round: 50 rounds, how high can you get?
Every player starts from zero and plays 1,000 rounds. Solid lines are the players' average capital in the simulation, dashed lines the exact average from the Markov chain. On the right, the exact gain per 100 rounds next to the simulation, and the share of players who finished ahead.
| Way | Exact, /100 | Simulation | Ahead |
|---|
Game B's odds depend on where the capital stands. So what matters is not only which coins are in the game, but how often each of them is tossed.
“In plain words” is hidden on all puzzles.
Game B nearly always plays its good coin, 74.5%. There is only one bad coin, on multiples of 3. If the capital landed on them in one round out of three, B would win. But game B itself keeps bringing you back to those numbers. Game A is almost fair and gives nothing on its own, but it nudges the capital randomly up or down and knocks it out of the trap. Why B drives you into the trap and what it costs is taken apart below.
Here is the trap: from 5 the good coin most often lifts you to 6, and from 6 the bad coin drops you back to 5. The capital runs back and forth between 5 and 6, and the bad coin is tossed again and again. To break out upwards you have to pass 6, where the chance is only 9.5%. So in game B the capital sits on a multiple of 3 not in a third of the rounds but in 38.4%.
| B's rounds played… | on a multiple of 3 | B wins |
|---|
A few percentage points decide everything, because the coins are so different: each extra percent of time on the trap costs about 0.65 percentage points of wins. When a coin picks the game, half the rounds are A, and the capital steps onto the trap less often. B's rounds now win 52.1%, A's 49.5%, together 50.8%. That is what gives +$1.57 per 100 rounds.
“Rigorous” is hidden on all puzzles.
| Remainder | Win chance | Share of time |
|---|
The coin is chosen only by the capital's remainder mod 3: 0, 1 or 2. A win moves the remainder one step clockwise, a loss one step back. This is a Markov chain with three states. Over thousands of rounds the share of time in each state stops changing: that is the stationary distribution π. It comes from the balance condition: as much probability flows into a state each round as flows out. Then the average gain per round is π₀ · (2p₀ − 1) + π₁ · (2p₁ − 1) + π₂ · (2p₂ − 1), where pᵢ is the chance of winning in state i.
There is a shorter test. To climb by 3, the capital has to pass each remainder once: 0 once and "not 0" twice. For game B, growth or decline comes down to comparing p₀ · p₁² with (1 − p₀) · (1 − p₁)²: the product of chances of three steps up against three down. In the randomly mixed game each coin is the average of A's and B's: 29.5% on a multiple of 3 and 62% elsewhere. The trap gets softer, and the up product wins.
| Order | B on a multiple of 3 | Per 100 rounds |
|---|
A periodic order is worked out the same way: take the chain over one full period (four rounds for AABB) and its stationary distribution. In ABAB, game A comes right before every B and shuffles poorly: before a B round the capital is on the trap 38% of the time, almost as without A. The pair AA moves the capital by −2, 0 or +2 and knocks it off the trap better.
A game with coins p₀ on a multiple of 3 and p₁ elsewhere. The balance condition for the three states (q = 1 − p):
Without ε game B (p₀ = 1/10, p₁ = 3/4) is exactly fair: the capital is on a multiple of 3 in 5/13 of the rounds, and the chance of winning is exactly 1/2.
The one-lap test: the game wins when three steps up are likelier than three down.
The exact average gain per round with ε = 1/200, as plain fractions:
Note where exactly the paradox hides. The mixed game has coins that are the averages of A's and B's. But its gain is not the average of the gains, because the share of time in each state changes along with the coins. The gain depends on the coins non-linearly, so the average of two losing games can win.
“Deeper” is hidden on all puzzles.
Juan Parrondo, a physicist in Madrid, was studying a Brownian ratchet in 1996: a particle kicked about at random by heat, with an asymmetric sawtooth potential switched on and off. Neither mode alone carries the particle anywhere, but alternating them does. Games A and B are a discrete version of this picture: A is like free diffusion, B like a saw with a steep and a gentle slope.
He presented the idea in an unpublished talk in Turin the same year. It became widely known after a short paper by Gregory Harmer and Derek Abbott in Nature in 1999, "Losing strategies can win by Parrondo's paradox", which gave the effect its name.
There is no contradiction here. If the games did not depend on the capital, any mix of two losing games would lose: the average of two chances below 1/2 is below 1/2 too. The paradox appears only because B "remembers" the capital, that is, its own past. Abbott himself called such paradoxes apparent: not a logical contradiction, but a result that defies intuition.
There is no free lunch either. The player does not pick the game by the capital: a coin does. But game B itself is a rule that looks at the capital. The information about the capital is built into B, and A only keeps B from getting stuck where the rule works against you.
From one 50-50 mix to the whole family: which share of A is best, and whether it has to be the capital. Then the logic: what makes the paradox possible. And the physics the games grew out of: a ratchet that carries particles although none of its modes does.
Mathematics
Any share of A will do. With ε = 0.005 the mix wins for roughly 7% to 85% of A rounds, and the best share is about 41%: +1.66 per 100 rounds. But the unfairness has to be small: for a 50-50 mix the paradox is gone by ε ≈ 0.013.
In 2000 Parrondo, Harmer and Abbott took the capital out of game B. In the new game B the coin depends on the two previous results: after two losses 9/10 − ε, after a win and a loss in either order 1/4 − ε, after two wins 7/10 − ε. The state is now the game's memory, not money, and the effect is the same: this B and A lose apart and win together when ε is below 1/168. The effect is weaker here, but the capital need not be known.
Logic
| Game | Exact, /100 | Simulation |
|---|
Replace B's two coins with one that wins with the same overall chance, 49.57%, but ignores the capital. On its own the new B loses as much as the old one. But the mix with A now loses: every round's chance is the average of two numbers below 1/2.
Hence a general rule. Two losing actions can win together only if one depends on a state that the other changes. Then "B loses" is true only for the state that B itself creates. In other surroundings the same B wins. Judging such an action apart from everything around it means judging the wrong thing.
Practice
Particles in water are jostled by heat. When the sawtooth potential is on, each one slides to the bottom of its well and stays. When it is off, the particles spread out equally both ways. But switch it on and off in turn, and the particles travel: the nearest crest of the saw is on one side, and while diffusing a particle crosses that one more often.
This was tested by experiment: in 1994 Rousselet, Salome, Ajdari and Prost moved plastic beads in a liquid one way by switching an asymmetric electric field on and off. Small beads diffuse faster and travel faster, which you can see in the simulation too. The energy comes not from heat but from whoever switches the field, so the laws of thermodynamics are not broken.
Physicists and biologists use similar models for molecular motors in cells, such as proteins that walk along microtubules: Dean Astumian in Science in 1997, Frank Jülicher, Ajdari and Prost in a review the same year. The flashing ratchet is one of the basic models in the field.
That was the title of Juan Parrondo's talk in Turin where he first showed the games. It was never published, but it is still cited as the start of the story.
Harmer and Abbott's paper "Losing strategies can win by Parrondo's paradox" in Nature takes one page, page 864 of volume 402. It is what made these games known as Parrondo's paradox.
That is how many of its rounds game B plays on a multiple of 3 instead of a fair third, 33.3%. Those five percentage points are enough to turn a game that could win into one that loses.
Strict alternation ABAB… loses 0.67 per 100 rounds. AABB… wins 1.47, and ABB… as much as 5.74. What matters is not that the games are mixed, but how exactly A knocks the capital before B.
In chapter 46 of the first volume of The Feynman Lectures on Physics, Richard Feynman takes apart a ratchet and pawl that should turn from the thermal kicks of molecules. He shows why at a single temperature such an engine does not work. Brownian motors get around this because outside switching powers them.
In the 2000 version, game B looks not at the capital but at the last two results. There the paradox works when the unfairness ε is between zero and 1/168, that is, up to 0.006. The authors stressed that such games work for a wider range of parameters.
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