Parrondo's paradox calculator

The games

Each round you win or lose one unit. With a small handicap ε:

Game A: P(win) = 1/2 − ε Game B: P(win) = 1/10 − ε if capital is a multiple of 3, otherwise 3/4 − ε Expected gain per round = 2 × P(win) − 1

Worked example with ε = 0.005

Game A wins with probability 0.495, so it loses 0.01 per round on average.

Game B is harder. Track the capital modulo 3 as a Markov chain. With ε = 0, the chain spends 5/13 of its time at 0 (the bad coin) and 8/13 elsewhere, giving P(win) = 5/13 × 0.1 + 8/13 × 0.75 = 1/2 exactly: a fair game. With ε = 0.005 the long-run win probability drops to about 0.4957, a loss of about 0.0087 per round.

Now flip a fair coin each round to choose A or B. That is the same as a single game B′ whose win probability is 0.5 × 0.495 + 0.5 × 0.095 = 0.295 at a multiple of 3 and 0.5 × 0.495 + 0.5 × 0.745 = 0.62 elsewhere. Its chain spends less time at 0, and the long-run win probability comes out at about 0.5079, a gain of roughly 0.0157 per round. The fixed pattern AABB also wins, at about 0.0147 per round. Not every pattern does: type ABAB into the pattern box and the long-run win probability is about 0.4966, a loss of roughly 0.0067 per round. Strict alternation happens to land B on the bad branch too often.

With ε = 0 the random mix wins with probability exactly 727/1418 ≈ 0.5127, which the calculator reproduces if you set ε to 0.

Reading the simulation

The chart shows mean capital after each round, averaged over many runs from zero. The A-only and B-only lines drift down; the random and pattern lines drift up. Over 1,000 rounds the expected final capital is about −10 for A, −9 for B, +16 for the random mix and +15 for AABB. Single runs are noisy (the standard deviation after 1,000 rounds is about 30), which is why the calculator averages many runs.

Why it works

Game B is losing only because of how often it visits the bad branch. Alone, B tends to fall back to a multiple of 3 after winning, then lose with the bad coin. Game A is close to a fair coin flip, so it shuffles the capital and puts it on a multiple of 3 about a third of the time, less than B does on its own. B then gets to use its good coin more often, and that more than pays for A's small losses.

Common mistakes

• Expecting it to work with any two losing games. Game B's odds must depend on a state that game A changes.
• Reading one simulated run as the answer. Compare the exact drift column, or average many runs.
• Making ε too large. The paradox holds only for small ε; with ε = 0.05, say, every strategy loses.