How's this roulette strategy?

16 February 2014

By John Grochowski

QUESTION: Can you please give me your expert opinion on outside bets on black or red using the following roulette system?

Place $1 on red and $2 on black. Ball lands on red. Next bet is $1 on red again because red won previously but place $4 on black (doubling your previous bet because you lost).

Ball lands on black this time.

Next bet is $1 on black and $2 on red and so on.

It would seem that you are mitigating your losses as you win. What do you think of this strategy? Does it change the game in any way?

ANSWER: The key problem with betting red and black at the same time is that when one of the zeroes turns up, you lose both bets. That happens an average of once per 19 spins of the wheel, and it's too much to overcome. Too many roulette players do not take those zeroes into consideration when they wager on those outside propositions.

The end result of those zeroes appearing is that over a long period, the house will keep an average of 5.26 percent of your total wagers. That's the same house edge as if you were betting only red, or only black. It doesn't matter what wagering system you use, the house edge is always going to be there. These games cannot be changed in any way by your betting methods. The house edge is always 5.26 percent except in the case of “surrender,” which some casinos allow.

Surrender means that on the outside bets of red/black, high/low or even/odd if those zeroes show the casino will only take half your bet. This rule cuts the house edge in half.

OK, let’s do a breakdown of your starting wagers, $1 on red and $2 on black, and imagine a sequence of 38 spins in which each number turns up once. Since your wagers total $3, in those 38 spins, you are risking $114.

On each of the 18 red numbers, you keep your $1 bet and get $1 in winnings, so that at the end of the trial you have $36 resulting from your red bets. On each of the 18 black numbers, you keep you $2 bet and get $2 in winnings, giving you $72 resulting from your bets on black. You lose both bets on the green numbers, 0 and 00.

That means you have $108 of your original $114. The house has taken $6. Divide the house’s $6 by $114, then multiply by 100 to convert to percent, and you see the house has kept 5.26 percent of your money. I wish there were a way to get around this but there isn’t.

What if you reduce each wager by $1, leaving you with a $1 bet on black and no bet on red? Now in our 38-number sequence you risk only $38. On the 18 black numbers, you keep your $1 bet and get $1 in winnings for a total of $36. You lose on the other 20 numbers.

So at the end of the sequence, you have $36 of your original $38. The house has $2, and as you’ve probably deduced, that comes to 5.26 percent. Again, you have no way to get around the house edge.

Whether you bet $1 on red and $2 on black, or just reduce to $1 on black, the house keeps an average of 5.26 percent of your wagers. However, with the reduced wager, you have less money at risk so that in dollars, your average losses are smaller.

I do recommend that when you sit down to play roulette, you ask the dealer if the casino offers surrender. Then those outside bets become more attractive than the inside bets where there is no surrender.

Luckily, roulette is a leisurely game, so the number of decisions you face is limited. A blackjack player will usually play twice as many decisions as a roulette player.