What Parrondo's paradox is
Parrondo's paradox is the counterintuitive fact that two games of chance, each of which loses money on average when played by itself, can produce a positive average return when you alternate between them or switch at random. It was formulated by Spanish physicist Juan M. R. Parrondo in the late 1990s and first published in detail by Harmer and Abbott (Nature, 1999). The paradox is a discrete-time analogue of a "flashing Brownian ratchet": individually futile dynamics, combined the right way, extract net drift from noise.
This calculator computes the exact long-run expected change in capital per play for three strategies — Game A, Game B, and a 50/50 random mix — so you can see the paradox emerge from the numbers themselves. Every play the gambler wins +1 or loses −1 unit.
The two games
Game A — a simple biased coin
Game A is a single biased coin. You win one unit with probability p and lose one unit with probability 1 − p. Its expected value per play is simply:
E[Δ]A = p·(+1) + (1 − p)·(−1) = 2p − 1
With the classic value p = 0.495 the game loses on average: 2(0.495) − 1 = −0.010 units per play. Any p below 0.5 is a losing game.
Game B — a capital-dependent pair of coins
Game B chooses which coin to flip based on your current capital. If your capital is an exact multiple of a modulus M (typically M = 3), it uses a heavily biased "bad" coin that wins only with probability p₁. Otherwise it uses a "good" coin that wins with probability p₂. The classic parameters are p₁ = 0.095 and p₂ = 0.745.
The trap is that losing pushes your capital down by one, which moves it toward the punishing multiple-of-M state. Solving the Markov chain on the residue of capital modulo M gives a stationary distribution π over states {0, 1, …, M−1}. The long-run win probability is w = Σ πᵢ·(win prob in state i), and the expected value per play is E[Δ]B = 2w − 1. With the classic parameters the "bad" state 0 is visited about 38% of the time — far more than the naive 1/3 — so Game B loses roughly −0.0087 units per play despite the very favorable good coin.
Why mixing wins
When you randomly play Game A half the time, the extra ±1 steps from A reshuffle the capital distribution so Game B lands on the multiple-of-M state less often. That shifts play toward the good coin. Analytically, the mixed game is another Markov chain whose per-state win probability is ½·p + ½·(Game B coin for that state). Solving its stationary distribution with the classic parameters gives a positive expected value of about +0.0157 units per play — a winning game built from two losing ones. Over 1,000 plays that is roughly +16 units of expected profit, versus about −10 for A alone and −9 for B alone.
Common reference values
- Classic Harmer–Abbott set (ε = 0.005): p = 0.495, p₁ = 0.095, p₂ = 0.745, M = 3 → E[Δ]A ≈ −0.010, E[Δ]B ≈ −0.0087, E[Δ]mix ≈ +0.0157.
- Fair edge case (ε = 0): p = 0.5, p₁ = 0.1, p₂ = 0.75, M = 3 → both A and B become exactly break-even (E[Δ] = 0), while the mix is strictly positive.
- Sanity anchor: for Game A alone, E[Δ] is always exactly 2p − 1, independent of M, p₁, and p₂.