Good Gamblers Know their Beans about the Casinos

7 June 2006

By Alan Krigman

Picture two carnival booths. One has a jar with 11 jellybeans, 10 red and one green. The other has a jar with 110 jellybeans, 100 red and 10 green. Both tout chances for $1 to reach in without looking and draw a jellybean. Reds lose, greens get you a $10 bill. These alternatives can teach a lot about casino gambling.

Many solid citizens would choose the booth with the larger jar, believing their chance would be better because it contained more winning jellybeans. This is wrong. The probabilities don't depend on the actual numbers of reds and greens in the jars, but on their proportions. The chances, one out of 11 and 10 out of 110 are equivalent. As fractions they're 1/11 and 10/110, the latter reducing to 1/11 using that "cancelling" trick you learned in the third grade. Higher math (fourth grade) would tell you they both equal 9.091 percent, rounding off at the third decimal place.

The carnival operator has an advantage in these games, and it's the same in either instance. Intuitively, to see there's an edge, compare the odds of winning and the payoff. With the small jar, 10 ways to lose and one to win put the odds against you at 10-to-1; with the large jar, the odds are 100-to-10 and division by 10--cancellation again--shows this is the same as 10-to-1. If you win, you get $10. But this is only a $9 profit because the operator locked away your $1 when you bought the ticket. The payoff is accordingly 9-to-1. The carnival's edge is hidden in the offset between the 10-to-1 odds and the 9-to-1 payoff ratio.

You can find the value of the edge, as a percent, using the standard formula: multiply the probability of winning times the payoff, subtract the probability of losing times the bet, then divide the difference by the amount bet. Here, it's [(1/11)x$9 - (10/11)x$1]/$1. This works out to -(1/11) or 9.091 percent. Remembering about cancelling, you realize you'd get the same answer with probabilities of 10/110 and 100/110.

You can also find the value of the edge using the artifice of a hypothetical "statistically-correct cycle." Assume that 11 people played the small jar, of whom one drew the green and the other 10 reds. The operator grossed $11 in ticket sales and gave back $10, netting $1. The profit was $1 for $11 bet, 1/11 or 9.091 percent of the handle. For 110 players, it would be $110 in sales minus $100 in payoffs, $10 profit for $110 bet again 9.091 percent.

Slot aficionados may prefer to cast the edge in terms of "payback percentage." The formula in this case requires multiplying chance of success times the amount returned to winners, then dividing by the wager. It's [(1/11)x$10]/$1, which equals 10/11 or 90.909 percent. Note that edge plus payback percentage is 100 percent.

The jellybean jars also have implications for games involving withdrawal with and without replacement. The previous gambles assume that what's drawn by one player is put back for the next.

If it isn't, and you know what's missing, the game changes. You wouldn't play the small jar knowing the green was gone. Or, you might gladly play knowing it had seven reds and one green, since bucking 7-to-1 odds for a 9-to-1 payoff would give you the edge. The shifts would be similar but more subtle in the large jar.

With no replacement, when you don't know what's in or out, the game doesn't change. Say you go second on the small jar. The first player draws from 11 jellybeans, you from 10. The chance of the first person winning then you winning is (1/11)x(0/10) = 0. The probability of the first person losing and you winning is (10/11)x(1/10) = 1/11. Add these together to get 1/11, as before.

The arithmetic is messier for the big jar but results are the same. Pretend you're third. You could have win-win-win, win-lose-win, lose-win-win, or lose-lose-win. The respective probabilities are (10/110)x(9/109)x(8/108), (10/110)x(100/109)x(9/108), (100/110)x(10/109)x(9/108), (100/110)x(99/109)x(10/108). Trust me or multiply the terms then add them up, but it's 1/11 or 9.091 percent. Proving the poet, Sumner A Ingmark, piquant in penning: