Two Losing Games That Win Together

You have a score, called the capital, which starts at 1,000. Each round you play one game and win or lose one unit. Game A is a slightly unfair coin: it wins 49.5% of the time. Game B is stranger. It looks at the capital and divides it by 3. If there is no remainder, B wins only 9.5% of the time. Otherwise it wins 74.5% of the time. Both numbers have the same small edge, 0.005, taken off, which is why neither game is fair.

The picture shows the capital against the number of rounds. The thick line is the exact expectation: the average over every way the rounds could have gone. The thin line is one sample, a single history drawn from a seed. The dashed lines are game A on its own and game B on its own, and both slope downwards.

Choose "Random mix" and the thick line changes direction. Nothing about either game was altered. The only difference is that each round a coin now decides which of them to play.

Look at the numbers for B. At a capital that is a multiple of 3 it wins one time in ten, which is terrible. At every other capital it wins three times in four, which is excellent. If B could be played only at the good capitals it would be a fine game. It loses on average because it cannot choose where to be played: each win or loss moves the capital, and the capital decides which odds come next.

Played alone, B keeps returning the capital to the bad value. From a remainder of 2, which is a good capital, B wins three times in four, and a win lands exactly on a multiple of 3. Over a long run the capital sits at a multiple of 3 more than a third of the time, 38.4% at the standard edge, so the bad odds are used more often than an even spread would give.

Game A has no such preference. It pushes the capital up or down by one with almost equal chance wherever it is, and that spreads the capital out over the three remainders evenly. Mixed in, A shuffles the capital between B's turns, so B finds it on a multiple of 3 less often: 34.5% of the time for the fair mix, and 30.7% for the pattern ABB. B's good odds are used more, and its bad odds less, than when B plays alone.

The expected gain of a round is twice the win chance minus one. For game A that is a fixed number, a little below zero. For game B the win chance changes with the remainder of the capital, so what B is worth depends on how often each remainder comes up. The share of rounds at each remainder is not part of the rules of B. It belongs to the whole sequence of games that were played before.

That is why the paradox is only apparent. "Two losing games cannot make a winning one" is true whenever each game's odds are fixed numbers, because a mixture of two fixed losses is a loss. It fails here because B's odds are not fixed: they depend on the state, and the state depends on what else was played. Changing the other games you play changes the shares, and the shares are half of what a game is worth.

The laboratory makes this testable. Replace B by a plain coin that wins exactly as often as the real B does in the long run, and every mixture loses again, even though the average of the two games is unchanged. What was doing the work was never the average. It was the dependence on the capital.

Let $c$ be the capital and $r=c\mathrm{mod}3\in \{0,1,2\}$ its remainder. Fix an edge $\epsilon \ge 0$ and let game A win with probability $a=\frac{1}{2}-\epsilon$. Let game B win with probability $b_0=\frac{1}{10}-\epsilon$ when $r=0$ and $b_1=\frac{3}{4}-\epsilon$ when $r\ne 0$. A win adds 1 to $c$ and a loss subtracts 1, so a win sends $r$ to $r+1$ and a loss sends it to $r-1$, both taken modulo 3.

For a strategy that plays A with probability $p$ at random, the win probability at remainder $r$ is $w_r=pa+(1-p)b_r$, and the sequence of remainders is a Markov chain on three states. For a periodic pattern of length $L$, the state is the pair (position in the pattern, $r$), a chain on $3L$ states. Every win probability lies strictly between 0 and 1, so the chain is irreducible and has a unique stationary distribution $\pi$, with $\pi P=\pi$ and $\sum \pi =1$.

The long-run expected gain per round is

The kernel solves for $\pi$ by Gaussian elimination in exact rational arithmetic, and for a rational $\epsilon$ the answer $g$ is a rational number. Starting from capital $c_0$, the expected capital after $n$ rounds is $c_0$ plus the sum, over the first $n$ rounds, of the expected gain of that round, and it is also computed exactly, by carrying the distribution over the finite state set forward one round at a time.

Take $\epsilon =0$ and game B alone. The three balance equations are ${\pi }_0=\frac{1}{4}{\pi }_1+\frac{3}{4}{\pi }_2$, ${\pi }_1=\frac{1}{10}{\pi }_0+\frac{1}{4}{\pi }_2$ and ${\pi }_2=\frac{9}{10}{\pi }_0+\frac{3}{4}{\pi }_1$, with ${\pi }_0+{\pi }_1+{\pi }_2=1$. The solution is $\pi =(5,2,6)/13$, and substituting shows each equation holds: for the first, $\frac{1}{4}\cdot \frac{2}{13}+\frac{3}{4}\cdot \frac{6}{13}=\frac{20}{52}=\frac{5}{13}$. The gain is $\frac{5}{13}(2\cdot \frac{1}{10}-1)+\frac{8}{13}(2\cdot \frac{3}{4}-1)=-\frac{4}{13}+\frac{4}{13}=0$. Without the edge, B alone is exactly fair, and A alone is fair too. The fair mixture of the two has stationary distribution $(245,180,284)/709$ and gain $18/709>0$: two fair games combine into a favourable one. That is the paradox at its sharpest, and each of those equalities can be checked with a pencil.

With the edge, the same system has rational coefficients with denominator 200, and the laboratory solves it exactly for every edge it offers. The values it prints are those rationals rounded to four places. The claim that these numbers are proved is the claim that an exact solution, checked against the balance equations, is a proof of what it computes. A second computation by a different route, the spanning-tree formula for a three-state chain applied to the product of the transition matrices over one period, agrees with the kernel exactly on every pattern up to length seven.

Now suppose B ignores the capital, winning with a fixed probability $q$, and play any rule at all for choosing between A and B, even one that looks at the whole history. On each round the expected change in capital is $2a-1$ if A is played and $2q-1$ if B is, and the rule chooses between them or mixes them, so the expected change is never larger than the larger of the two. Summing over rounds, the expected gain of any strategy is at most $n\max (2a-1,2q-1)$. If both are negative, so is the total. That is why no mixture wins when the games do not look at the capital, and it is exactly the step that fails for the real game B, where the expected change on a round is not one of two fixed numbers.

The gain is real but narrow. At a fair mixture it exists for edges up to 0.013 on the grid of thousandths the laboratory offers and not from 0.014 up, where the edge is large enough that both games lose more than mixing recovers. The page checks those fifty-one edges and no others, and it says nothing about the exact crossing between them. Other mixtures move the boundary: of the mixtures in steps of five per cent, playing A in two rounds in five does best at the standard edge, at +0.0166 per round, and playing A in nine rounds in ten loses.

Not every way of alternating wins. ABB gains more than the random mixture, +0.0574 per round, and AABB less, +0.0147, while ABBB loses, at −0.0027. A longer pattern, ABABB, does better than all of them at +0.0652. The page shows these exact values but does not claim to know which pattern is best of all, nor the full rule for which patterns win. Whether they win depends on the order of the letters as well as their counts, and the laboratory leaves that question to be explored rather than answered.

The sample run is scenery. It wanders about the exact line by roughly the square root of the number of rounds, which is about 100 units after 10,000 rounds, and another seed gives another path. After 10,000 rounds the fair mixture is expected to gain about 157 units, so a single run can finish well away from that figure, and now and then below its starting capital. The expectation is a statement about the average of many runs, not about yours. The page keeps the exact line and the sample visibly apart for that reason.

Finally, this is a statement about a fixed set of games with unit stakes and unbounded capital. Nothing here says that any real betting system wins: the games have an edge against the player, the capital is allowed to go negative, and there is no house limit. What the page establishes is the narrower mathematical fact that, for this one pair of games, shuffling their order is worth more than either of them alone.