Roulette Odds 6 Red In A Row

  1. Odds On A Roulette Table
  2. Best Odds For Roulette
  3. Roulette Odds Red
  4. Roulette Odds 1 12

As before, this depends on the likelihood of losing 6 roulette spins in a row assuming we are betting red/black or even/odd. Many gamblers believe that the chances of losing 6 in a row are remote, and that with a patient adherence to the strategy they will slowly increase their bankroll. As before, this depends on the likelihood of losing 6 roulette spins in a row assuming we are betting red/black or even/odd. Many gamblers believe that the chances of losing 6 in a row are remote, and that with a patient adherence to the strategy they will slowly increase their bankroll. 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). There are 37 numbers on the wheel. There are 18 blacks and 18 reds. Therefore, the winning odds of red are 18/37, which is 0.4865. The odds of black are the same. The same goes for any “even chance” bet that includes highs and lows, and evens and odds. The odds are not changing, no matter how many times red or black have appeared in a row.

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Peter
Hi,
I'm currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
Peter
OnceDear
Administrator
Thanks for this post from:
Hi,
In 1500 spins, there are 1499 possible pairs of spins
For any pair of spins, the chances of both coming up red is (18/38) * (18/38) or 0.2243767313 assuming double zero wheel.
So you can expect to see 2 consecutive reds 1499 * 0.2243767313 times or 336 times.
Of course a sequence of 3 reds is treated as two sequences of 2
If you want to work it out for sequences of 3, then...
In 1500 spins, there are 1498 possible triplets of spins
For any triplet of spins, the chances of all three coming up red is (18/38) * (18/38) * (18/38) or 0.10628371482 assuming double zero wheel.
So you can expect to see 3 consecutive reds 1498 * 0.10628371482 times or 159 times.
Of course a sequence of 5 reds is treated as three sequences of 3.
Meanwhile... Remember, you don't have a winning system. You never will. :o)
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
SM777

Hi,
I'm currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
Peter


Don't worry about it, and move onto your next project. Your idea won't beat roulette.
ThatDonGuy
It depends on whether or not how many times you count a sequence of, say, 5 reds as 'a sequence of 2 reds.'
Zero, since you are counting how many times red comes up exactly twice in a row?
One?
Two, since it is 'two pairs of reds' (1 & 2, 3 & 4)?
Four, since spins 1 & 2 are consecutive, as are 2 & 3, 3 & 4, and 4 & 5?
But as SM777 has pointed out, if you are looking for some mysterious way to beat roulette, don't bother. Remember, after two (or four, or 257, or zero) consecutive reds, the probability that the next spin will be red on a double-zero wheel is still 9/19 (and black is 9/19, and green is 1/19).
mustangsally

I would like to know how I can calculate the expected number of times a sequence of red comes up.

it is easy.
just remember dealing with averages or expected numbers is not the same as the probability.
2 totally different animals
Roulette*****
for 3 or more run (in Excel for example)
parameters:
p=(18/38)
length=3
trials=1500
q=(20/38)Row
THE FORMULA (without proof - that is internet stuff)Odds on rouletteRoulette
=(p^length)*(1+((trials-length)*q))
for exactly length of 3
calculate 4 and 3
subtract 4 from 3
here is my Excel in Google if want to see (easy)
https://goo.gl/98yjKp
Quote: Peter

So for example, in 1500 roulette spins,
how often do I expect a sequence of 2 reds in a row to come up,
how often do I expect a sequence of three reds in a row to come up and so on.

here is a table of data.
I simulated this 1 million times (1 million sets of 1500 spins)
and calculated it also (rounded to 4 decimals)
super close I do say
.simsimsimcalccalc.
lengthfreqexact 33 or moreexpected #exact 3length
1197014850197.0149374.1835374.1856197.05791
29328373593.2837177.1686177.127793.28112
34416857044.168683.884983.846744.15633
42091215620.912239.716339.690320.90224
598986169.898618.804218.78819.89445
646894284.68948.90568.89374.68376
722177192.21774.21614.21002.21717
810508691.05091.99841.99291.04958
94989500.49900.94750.94340.49689
102360310.23600.44860.44660.235210
111119770.11200.21260.21140.111311
12531590.05320.10060.10010.052712
13250120.02500.04740.04740.024913
14119000.01190.02240.02240.011814
1555200.00550.01050.01060.005615
1626020.00260.00500.00500.002616
1712320.00120.00240.00240.001317
186130.00060.00120.00110.000618
192720.00030.00050.00050.000319
201400.00010.00030.00030.000120
21590.00010.00010.00010.000121
22300.00000.00010.00010.000022
23200.00000.00000.00000.000023
24110.00000.00000.00000.000024
2550.00000.00000.00000.000025
2650.00000.00000.00000.000026
2700.00000.00000.00000.000027
2800.00000.00000.00000.000028
2910.00000.00000.00000.000029

have fun
hope this helps some(sum)
Sally
Peter
Thanks all for the answers. Don't worry, I'm not looking to find a way to beat the wheel, I'm actually trying to show a sample size I have is random and I like to take a look at all kinds of different statistics from it. In this case the distribution of number of red numbers in a row.
I guess I wasn't totally clear: I'm looking for 'exactly two in a row', 'exactly three in a row', etc. So a series of four reds in a row will be counted as just that, four in a row. Looks like the best way is to start with 1500 in a row (in a sample size of 1500) and calculate 1499 in a row as suggested with substracting 1500 in a row and work my way down to one.
Or apparently using the below formula. Could you point me in the right direction for the proof? I haven't been able to find anything about this subject on the web, hence I got here.


THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))


Thanks,
Peter
russ451
I have to ask.... Why 1500 instead of an 'even' number like 1,000, or 10,000.
At least it would be easier to turn into percentages.
Russ

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Roulette payouts are an important part of roulette strategy. They work like this. The odds are stated in the form of x to 1, which means you’ll win x dollars for every dollar you’ve bet. For example, the single number bet offers a payout of 35 to 1. If you win, you’ll get your dollar back plus the $35 for the win.

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Payouts on the Outside Bets

On the edge of the table are a series of bets which are “outside” the 38 numbers on the table. Each of these bets refers to a specific set of numbers or colors. If the ball lands on 0 or 00, you’ll lose on any of the outside bets.

The outside bets include:

Red or Black – This bet pays out even odds (1 to 1) if the ball lands on the color you chose.

Odd or Even – This bet pays out even odds (1 to 1) if the ball lands on odd or even, depending on which you chose.

Low or High – This bet pays out even money (1 to 1) if the ball lands on 1-18 if you bet low, or if the ball lands on 19-36 if you bet high.

Columns – The numbers on the layout are organized into three columns of twelve numbers each. A “columns” bet wins if the ball lands on one of the numbers in the column you chose. This bet pays out 2 to 1 when you win.

Dozens – There are 36 numbers on the table, so you can bet on the first dozen (1-12), the second dozen (13-24), or the third dozen (25-36). This bet also pays out 2 to 1.

Payouts on the Inside Bets

Roulette Odds 6 Red In A Row

You can also bet on specific numbers and sets of numbers on the inside of the layout. These bets win less often, but they pay out more when you do win. The house edge on the inside bets is the same as the house edge on the outside bets.

The inside bets for roulette include:

Straight-up – This is a bet on a single number. It pays off at 35 to 1.

Split bet – This is a bet on any two adjacent numbers. You place the chip on the line between the two numbers in order to make this wager. This bet pays out at 17 to 1.

Street bet – This bet covers three numbers. You place your bet on the line outside of the three numbers in the row where you want to win. This bet pays out at 11 to 1.

Corner bet – Some people call this a square bet or a quarter bet. It’s a bet on a corner that makes a square, and it’s a bet on four numbers. A win on this type of bet pays out at 8 to 1.

Five-number bet – You can only make one five-number bet, and it’s the only inside bet that offers different odds from all the others. The problem is that it has a higher house edge, making it the worst bet on the table. This bet is on the numbers 0, 00, 1, 2, and 3, and you place the chip on the outside corner line between the 1 and the 0. This bet pays out 6 to 1, but only masochists place this bet.

Odds On A Roulette Table

Six-number bet – Some people call this a line bet. It covers two adjoining rows of numbers. It pays out at 5 to 1.

How Roulette Payouts Give the Casino an Edge

These payouts all have one thing in common—they pay out less than the true odds of hitting a win. That’s why the casino enjoys a house edge of 5.26% on roulette. Your odds of winning are always less than the payout amounts.

For example, the odds of winning a straight-up bet are 37 to 1. There are 37 numbers on the wheel that lose, and 1 bet on the wheel that will win. But the bet only pays out 35 to 1, not 37 to 1, so the house wins more often than it loses.

A split bet offers you odds of winning of 18 to 1, but it pays off at 17 to 1.

Best Odds For Roulette

I could list all of them, but you get the idea by now. The casino has an unassailable mathematical advantage on every bet. No betting system or strategy can overcome this advantage.

Roulette Odds Red

Of course, in the short run, anything can (and often will) happen. This is called “standard deviation”, and it explains why some people walk away from the roulette table as winners. The mathematically true results only come around the closer you get to an infinite number of spins.

Roulette Odds 1 12

So the best way to approach roulette is as a lark. It’s a fun game. You can relax and socialize while you play. But don’t expect to win, because the odds are against you. And if you do win, walk away and smile, because you beat the odds.

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