Roulette: Three Frequently Asked Questions

If a color has been winning a lot, do your odds improve by betting with it or against it on the next round?

Neither. Roulette is a game of independent trials. Apparent "patterns" have no meaning and probabilities associated with the wagers are invariable. The chance of hitting red or black, for instance, is always 47.37 percent in a double-zero game.

It's easy to believe that colors clump because the displays at the tables typically exhibit runs in the red or black columns. But this can be expected in random series of fixed-odds events.

As an example, would you think you were onto something if you saw eight or more sets of black-black or red-red, including cases of over two in a row, in 24 spins? There's 70 percent chance it will happen at random. How about four or more instances of at least triplets? The chance is slightly over 50 percent. And the probability of five or more successive reds or blacks is nearly 25 percent. Try it at home with 20 series of 24 flips of a coin, jotting down heads or tails. You may be surprised at how many pairs and longer runs you get, knowing every flip is 50-50.

Is there a difference between betting on a combination of numbers -- say, covering four spots at a "four corners" intersection -- and distributing the same total over each outcome separately?

With two exceptions, there's no difference. Say you put $4 on the corner of 17, 18, 20, and 21. A hit gets you 8-to-1 on the $4 so you'd collect $32 and recover your $4 -- you'd have $36. Instead, suppose you bet $1 on each of any four numbers. If one hit, you get 35-to-1 on the winner and lose on the other three -- you collect $35 and get back $1 so you'd again have $36.

One of the exceptions occurs in casinos that take only half the bet on "even money" wagers like red or black when the spin comes up 0 or 00. Here, a single $18 bet pays $18 and returns your $18 on a hit, so you have $36; you lose $18 on the complementary numbers, and $9 on 0 or 00. With $1 on each of the 18 individual numbers, a hit pays $35 and returns $1 so you'd have $36; a complementary number costs $18, but 0 and 00 cost $18 and not $9.

The other exception involves the combined bet on 0, 00, 1, 2, and 3. This 33-to-5 shot pays 6-to-1 so winning with $5 at risk gets $30 plus your original $5, leaving you with $35. With $1 each on the separate numbers, a hit nets $35 and $1 back so you have $36.

Can you beat the house at roulette starting with the table minimum on an even-money bet, then doubling-up after every loss and returning to the lowest level when you win?

This "Martingale" betting system offers a high likelihood of a small profit, balanced against a low but finite chances of a major rout. The weakness is that while the probability of an extended string of losses keeps decreasing, it never gets to zero. And the amount of money at risk escalates rapidly.

For double-zero roulette, neglecting the half-back on 0 or 00, the chance of either color losing on any spin is 52.63 percent. The chance of five losses in a row is about 4 percent. Had you started with $5, after five losses you'd have just bet $80 and would be down $155. The chance of 10 successive losses is 0.163 percent; you'd have just bet $2,560 and would be $5,115 behind. Admittedly, 0.163 percent is small. But, if 100,000 players try this approach, it means 163 can expect to lose 10 times running.

Such questions show you can size and distribute roulette bets to tailor sessions to your personal gambling preferences. And in limited cases, you can find the best ways to make similar wagers. So perceptive players can bend the rules of arithmetic. But not break them. As the poet, Sumner A Ingmark, warned:

Mathematics leaves no grounds,
For going out of nature's bounds.