Roulette Red Odds

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By Ion Saliu, Founder of Roulette Mathematics

While the odds in roulette are very consistent across almost all bets, this particular bet is an exception. This is because payouts in roulette are generally designed to be fair for a wheel with 36 pockets, with the house edge coming from the fact that there are actually 37 (European roulette) or 38 (American roulette) pockets for the ball to. This applies for any game of chance, including roulette. For example, hitting the same color for n times in a row on a single-zero roulette has the probability For n = 3, this is 11.51%, for n = 5 is 2.72%, while for n = 10 it goes down to 0.07%. If we bet $1 on red and green or black occurs, then we lose the dollar that we bet. This results in net winnings of -1. The random variable X defined as the net winnings from betting on red in roulette will take the value of 1 with probability 18/38 and will take the value -1 with probability 20/38.

The probability that the ball eventually ends up in one of the slots depends on how many slots there are: They are 37 (in French/European roulette) or 38 (in American roulette) colored and numbered pockets on the wheel. If you were simply betting on red, and if there were only black or red slots, then you would be winning with a 50% chance. Let's say I'm betting on a 37 number roulette wheel (18 red 18 black and one green 0)?' On the surface, the best probability for the roulette player to be ahead is in one trial (spin): 48.6% to win (versus 51.4% to lose), as far as even-money betting is concerned.

This question was posted in mathematical newsgroups (alt.math.recreational, alt.math.undergrad, alt.sci.math.probability): 'Winning and Quitting on Red/Black in Roulette'.

Red
    'Obviously in roulette betting on in the long run you are going to lose your money but at some point chances are you'll be in profit. To take an extreme example if you had $1000 you could reasonably expect to be up $1 at some point. Is it possible to generalize this? I want to win W dollars at which point I will quit. How much cash C would I need to have probability P of succeeding? Let's say I'm betting on a 37 number roulette wheel (18 red 18 black and one green 0)?'

On the surface, the best probability for the roulette player to be ahead is in one trial (spin): 48.6% to win (versus 51.4% to lose), as far as even-money betting is concerned. I don't agree that it is the best strategy (betting all your bankroll on one spin).

Theoretically, no bankroll will put a player ahead guaranteed, IF flat-betting and playing very long consecutive sessions. There are moments, however, when the roulette player can be ahead by at least one betting unit. Even in even-money bets, the player has a good chance to be ahead by at least one unit after 5, or 10, or even 100 spins. But more than 20 spins are NOT recommended; the probability (odds) to lose go(es) above 50%! Think about it!

The main thing, mathematically, is the number of player's wins in N trials. To be ahead, means the player has won at least one more roulette spin (number of successes) than the number of losses in N trials. The question then becomes:

'What are the probabilities for the player to be ahead in various numbers of trials?'

Everybody can use my probability software SuperFormula, option L: At Least M successes in N trials.
Winning probability: p = 18/37; M must be at least (N/2) + 1.
Here is a number of cases from the player's perspective.

The figures are applicable to all even-money roulette bets: black or red; even or odd; low or high (1-18 or 19-36).

1 trial (spin)
- probability (odds) to win: 48.6%; odds = 1 in 2.05
- probability (odds) to lose: 51.4%; odds = 1 in 1.95
(the probability to lose is 19/37; adding zero to unfavorable cases).

2 trials (spins)
- probability (odds) to win 2 of 2: 23.7% (1 of 2 doesn't mean 'being ahead')
- probability (odds) to lose 1 of 2: 76.3%

3 spins
- probability (odds) to win at least 2 of 3: 48%
- probability (odds) to lose at least 2 of 3: 52%

10 spins
- probability (odds) to win at least 6 of 10: 34.4%; odds = 1 in 2.91
- probability (odds) to lose at least 6 of 10: 41.1%; odds = 1 in 2.43

20 spins
- probability (odds) to win at least 11 of 20: 36.5%
- probability (odds) to lose at least 11 of 20: 46.2%

100 spins
- probability (odds) to win at least 51 of 100: 35.5%; odds = 1 in 2.82
- probability (odds) to lose at least 51 of 100: 56.8%; odds = 1 in 1.76.
It's getting worse for the player...

The roulette strategy (or system) is a totally different ball game! But there are professional gamblers out there, including roulette players! They must have strategies, some roulette systems deduced from some figures like the ones above! The player can be ahead at any point in the game. If so, maybe it's time to move to another (or casino) table: It improves the odds of winning!

Always keep track of the losing and winning streaks. Be strong and put an end to a winning streak. You are ahead, you quit the roulette table. Go to another table and wait until you are ahead. The bankroll is of the essence: It must assure going through long losing streaks. Divide the streaks in 10 spins or 20 spins. Never fight aggressively short or mid-term losing streaks. This is the best approach for those who do not know Ion Saliu's casino gambling systems. A good approach to gambling is the next best thing to a good gambling system! Applicable to blackjack and baccarat, too!

Axiomatic one, everybody knows that the casinos have an edge or house advantage (HA) in all the games they offer, roulette including. The house advantage is created by the payouts in rapport to total possibilities for the respective bet. We can apply this simple formula based on units paid UP over total possibilities TP:

HA = 1 – (UP / TP)
(always expressed as a percentage.)

For example, in single-zero roulette, the one-number (straight-up) bet has payout of 35 to 1. The to qualifier is very important: the casino pays you 35 units and they give you back the unit you bet; thus, you get 36 units. There are 37 possibilities in single-zero roulette: 36 numbers from 1 to 36 plus the 0 number. Therefore, HA = 1 – (UP / TP) = 1 – (36 / 37) = 1 – 0.973 = 0.027 = 2.7%.

Let's calculate HA for the 1 to 1 bets: black/red, even/odd, low/high. HA = 1 – (UP / TP) = 1 – (2 / 2.055) = 1 – 0.973 = 0.027 = 2.7%. There are little differences among bets depending on how many decimal points we work with in our calculations.

The point is, the casinos have an advantage, or the players have a disadvantage. Nonetheless, the players' disadvantage is far better than what they face in state-run lotteries. Yet, most casino gamblers lose big, including at roulette tables. They do not have sufficient bankrolls to withstand long losing streaks.

However, around 45% of the roulette numbers lead the gamblers to profits in a few thousand spins. That is, with a sufficient bankroll, a player has a pretty good chance to make a profit, even if playing a random roulette number, or a favorite number. I analyzed about 8000 roulette spins from Hamburg Spielbank (casino). Quite a few numbers ended up making a profit: roulette systems, magic numbers.

By contrast, the more lottery drawings a player plays, the higher the degree of certainty of a loss. Let's make a comparative analysis to the roulette long series above (spins: total roulette numbers, 37, multiplied by 200). If playing the pick-3 lottery for some 100,000 drawings, it is guaranteed that all pick-3 straight sets will be losers. Some numbers will hit up to 3% to 5% above the norm — but that is not nearly enough to assure a profit. A frequency of 3% to 5% above the norm leads to profits in roulette, however.

Ion Saliu's Paradox and Roulette

Ion Saliu's Paradox of N Trials is presented in detail at saliu.com, especially the probability theory page and the mathematics of gambling formula. If p = 1 / N, we can discover an interesting relation between the degree of certainty DC and the number of trials N. The degree of certainty has a limit, when N tends to infinity. That limit is 1 — 1/e, or approximately 0.632....

If you play 1 roulette number for the next 38 spins, common belief was that you expected to win once. Not! Non! Only if you play 38 numbers in 1 spin, your chance to hit the winning number is 100%. Here is an interesting table, which includes also The Free Roulette System #1 presented at the main roulette site.

Odds Of Roulette Red Or Black

The maximum gain comes when playing 38 numbers in one spin: 36.3%. Obviously, it makes no sense to play that way because of the house advantage. On the other hand, a so-called wise gambler is more than happy to play one number at a time. What he does is simply losing slowly! Not only that, but losing slowly is accompanied by losing more. That cautious type of gambling is like a placebo. A roulette system such as Free System #1 scares most gamblers. 'Play 34 or 33 numbers in one shot? I'll have a heart attack!' In reality, the Free Roulette System #1 offers a 28.8% advantage over playing singular numbers in long sessions. That's mathematics, and there is no heart to worry about, axiomatic one.

You can also use SuperFormula to calculate all kinds of probabilities and advantage percentages. The option L — At least M successes in N trials is a very useful gambling instrument. If you play 19 numbers in one spin, the probability to win is 50%. If you play 19 numbers in 2 consecutive spins, the probability to win at least once is 75%.

Roulette Black Red Odds

Editor's note
• In an apparent change of heart, the Hamburg casino (Spielbank) offers online roulette results for free. The new link is (for the time being!):

www.spielbank-hamburg.de/spielsaal/permanenzen.php4

Roulette Red Or Black Odds

• • Real-life roulette spins are also available from the Wiesbaden, Germany, Casino (Spielbank)

www.spielbank-wiesbaden.de/DE/621/Permanenzen2.php: Wiesbaden Spielbank Permanenzen

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