What you can learn from two finger Morra

You're probably familiar with some form of two-finger morra. In one version of the game, you and an opponent each show one or two fingers at the same time. When the total number of fingers is two or four, the person designated as Odd pays the player acting as Even that number of units. Conversely for totals of one or three.

If you both pick ones and twos at random, in equal proportions, the game is balanced. In the long run, 1-1, 1-2, 2-1, and 2-2 occur equally often. So, on the average, for every four tries, Odd loses $2 on 1-1, wins $3 on 1-2, wins $3 on 2-1, and loses $4 on 2-2 a wash. Analogously, Even wins $2 on 1-1, loses $3 on 1-2, loses $3 on 2-1, and wins $4 on 2-2 also a wash.

Participants must randomize their choices or their opponents will take counteractions and gain an advantage. As an extreme example, say that Odd always shows one finger. Even would soon learn to also show one and win $2 on every coup.

Odd, however, can get a surprising 8.33 percent edge in this game without knowing how Even will play. This can be done by selecting one or two fingers randomly on each round, but maintaining them in an optimum ratio over an extended period.

Of course, edge is statistical and won't guarantee a profit in a session of finite duration. But the more coups played, the closer the outcome relative to the theoretical edge times gross wager.

The optimum strategy for Odd is the solution to a classic "minimax" problem. It can be solved with high school algebra (but you can just skip the next paragraph take the results on faith).

For anyone still here, call F1 and F2 the fractions of rounds in which Odd shows one and two fingers, respectively. Rounds are either F1 or F2, so F1 + F2 = 1. Odd's average net win or loss when Even shows one finger is N1 = -2xF1 + 3xF2; Odd's average net when Even shows two fingers is N2 = +3xF1 - 4xF2. There's no way to predict what Even will choose, so set N1 and N2 both equal to N. This gives what are known as "three equations in three unknowns" (F1, F2, and N), which can be solved "simultaneously."

The solution to the equations is F1 = 7/12, F2 = 5/12, and N = 1/12. That is, Odd should show one finger 7/12 of the time and two fingers the remaining 5/12. Average profit will then be 1/12 of a unit per round; this is an edge of 1/12 or 8.33 percent.

Similar math gives an optimum strategy for Even. The fractions are also one finger 7/12 of the time and two fingers the other 5/12. The best Even can do is negative, though; that is, hold losses down to an average of 8.33 percent.

Whether you're Odd or Even, it doesn't matter whether your opponent knows you're playing 7/12 and 5/12. Provided choices on individual rounds are random within the bounds of that 7-to-5 ratio, Odd will average gains and Even losses of 8.33 percent.

A casino version of two-finger morra might use a deck of 70 aces and 50 deuces. The casino gets the edge by being Odd. You bet on "one" or "two;" the dealer exposes a card. No sensible solid citizen would play the game because house advantage is too big at the start of a shoe. And no casino would run it because, as the shoe is depleted, percentages of remaining cards could shift such that Even could get an edge; for instance, were all aces gone, players would win four units on every round by betting on "two."

If this game would be neither played nor offered, why worry about it? First, to highlight that what you think you see in a casino isn't always what you get. Second, to illustrate that in games where probabilities vary when samples are drawn from a population and not replaced, players may get an edge by tracking or counting the action and acting accordingly. Third, to show that luck may override skill on any coup, but optimum strategies when they exist yield the best possible long run results. And fourth, to give you pause before luring someone into a contest where you have a hidden edge by recalling this rhyme by Sumner A Ingmark:

A lopsided bet is truly a very,
Unkindly technique to dupe the unwary.