A Contract That Loses Four Times in Five and Is Worth 1.80

A contract settles on one of five equally likely outcomes. Four of them pay 1.00 and the fifth pays 5.00. It is quoted at 1.50. Take it?

Yes, and the reason it feels wrong is worth more than the answer. Four times in five this contract loses money. That premise is true, it is not a trap in the sense of a miscalculation, and the conclusion drawn from it is still wrong.

Count five contracts rather than one

Averages are easier to distrust than tallies, so run one full cycle of outcomes. Four small outcomes return 1.00 against an outlay of 1.50, so each of those is down 0.50 and together they are down 2.00. The large outcome returns 5.00 against 1.50, so it is up 3.50:

Five contracts, up 1.50. Share that over the five and the edge is:

The same 0.30 arrives from the average payoff, which is the route most people take first:

Relative to the 1.50 committed, that is exactly one fifth. Twenty percent on the money, on an instrument that shows a loss on four settlements out of five.

Two different numbers, both called typical

The reflex answer is not bad arithmetic, it is the wrong statistic. Asked whether an instrument is worth 1.50, the mind reaches for the outcome it expects to see, and the outcome it expects to see is the modal one. That works whenever the distribution is roughly symmetric, which is most of ordinary life, and it fails precisely when a rare outcome is large enough to carry the average on its own. Here the single 5.00 supplies 5 of the 9 units of total payoff across a cycle, more than the four small outcomes combined.

How long before the edge shows up

The average is a promise about the long run, and it is fair to ask how long. After $n$ contracts of which $k$ settled large, the profit is:

which is positive exactly when $8k > n$, so you need the large outcome to have arrived more often than one time in eight. It arrives one time in five, and the whole edge lives in the gap between those two fractions. That reframing is more useful than the average, because it turns the question into a count you can watch.

With $k$ binomial on twenty trials at one fifth, being ahead after twenty contracts means $k \ge 3$, and:

So a little under four times in five you are showing a profit after twenty contracts. Not certainty, and nothing here promises certainty. Twenty contracts is also the point at which the picture starts to look like the arithmetic rather than like a losing streak, which is a real reason people abandon positive-expectation trades: the modal experience of a good contract is a small loss.

Push the requirement higher and the picture gets stranger. Asking for a ninety-five percent chance of showing a profit takes 63 contracts, and the sequence of those probabilities is not even increasing in $n$: it reads 0.9563 at seventy contracts and 0.9435 at eighty. The threshold in (4) is $k > n/8$ and $k$ is an integer, so the bar steps up whenever $n$ crosses a multiple of eight while the distribution of $k$ only drifts smoothly. Waiting longer can genuinely make you less likely to be ahead, which is the sort of thing that only shows up when you compute the tail rather than reason about it.

There is a cleaner way to say the same thing, using the spread of a single settlement. The payoff has variance $\tfrac45(1) + \tfrac15(25) - (9/5)^2 = 64/25$, so its standard deviation is exactly 1.60. The edge is 0.30, so a single contract is more than five times as noisy as it is profitable, and the cumulative edge catches up with the spread of the running total only at $0.30\,n = 1.60\sqrt{n}$, that is $n = 256/9 \approx 28$ contracts. That figure and the 0.7939 above are two views of the same fact.

Where taking it stops being obvious

The obvious way is the price. The contract is worth 1.80 and nothing more, so any quote above that turns the verdict around instantly. There is no sense in which the large outcome makes the instrument worth chasing at 2.00.

The less obvious way is size. Expected value is additive and utility is not, so the calculation above is a statement about repeated play at stakes that do not matter. Multiply every figure by a million and a single cycle of four small outcomes puts you two million down before the large one arrives. At those stakes the right question is about the variance of the tally and the depth of the account, and (4) is where you would start: the drawdown before $k$ catches up is what has to be survivable.

Underneath both sits the question of whether the five outcomes really are equally likely and independent across contracts. Both are stated here. Drop equal likelihood and the mean moves directly. Drop independence and (5) stops applying, since the binomial count is exactly the assumption that one settlement says nothing about the next.

Sources and further reading

• The quantity in equation (3) — Expected value
• Why the mean and the most common outcome part company — Central tendency
• The promise the average makes about the tally — Law of large numbers
• The count in equation (5) — Binomial distribution

Nothing above was computed in floating point. The average, the edge, the twenty percent and the per-cycle tally are exact fractions, the five outcomes were enumerated rather than sampled, and the tail in (5) was computed exactly and then cross-checked against a simulation that measured 0.7946. A four-hundred-thousand-contract Monte Carlo returned an edge of 0.30168 against 0.30 and a loss rate of 0.79958 against 0.80.