I have taken quite the liking to roulette lately, ( I know... Shame on me...) As I think it has become painfully obvious there seems to be no good way to play anything systematically with a high success rate. I have a casino relatively close to me, (the state border... as gambling is illegal in my specific state) so the last few weekends I have been making a mad dash over there. Admittedly I am relatively new to casinos in general so if I miscalculate anything or something is just... wrong, please feel free to call me out on it, and any information and constructive criticism is appreciated.

On to the theories!!!

I have been a more slow and steady type of player. Not looking to take 100k in a night or anything crazy, however I want to make say a consistent $100 an hour if possible. If my probability math is off please let me know! My requirements for the table are $3 min and $1000 max. it is a speed roulette where ball rolls roughly every minute to keep things going.

Option 1. Black/Red

So I feel this should be shot down quickly, but I just want to verify my math wasn't wrong.

So example being is last ball was red, probability should be in favor of black? .526^2 that it would be red/green or red/red. playing through this theory I would wait till 4 red would load up and I would put $25 on black, which what I came up with is around a 4% chance of it failing or .526^5. Then I chase the $25 if it does happen to fail, betting $50 after that, then $100, etc. till I don't have bank to keep up or I reach table max. I was getting roughly $200 an hour doing this method. I did get pinched once with 9 blacks popping up and finally shifting on the 10th (should have literally been a 1:1000 right?). This could be cured by simply lowering my initial bet of $25 to just $5 and proceeding as before by doubling each fail which should be around $80 an hour.

Option 2. Green chasing.

The particular game of roulette that I favor allows me to see the previous 70 - 75 rolls. I use green in this theory just because I can more easily identify the last time a green was rolled, although it could obviously be done with regular numbers, was considering stretching it to 4 numbers instead of just 1 or 2. Moving on, If hypothetically no greens were rolled in the previous 70 attempts (highly unlikely in itself) what would happen if I could come up with a way to double my reward gained from my input. So for example, to explain this out after not seeing both greens in 70. I would play as follows: 18 rolls at $1 each, 9 rolls at $2 each, 6 rolls at $3 each, 4 rolls at $4 each, 5 rolls at $5 each, etc etc.. just keeping it at half points. Long handing it out, although there were several different ways you could do it, at 81 rolls my total input was $765 maxing at a $30 bet and having a total reward of $1,080. Total rolls on this theoretical scenario is about 150 rolls. If my math plays out on those 150 rolls on 1 green bet, it would have a probability of failing of about 1.2%, doing this on both green, means a fail of .02%. You can just play both and when you hit one, just keep playing the other till you hit it. I haven't actually penciled out what it would look like to actually reach the max bet of $1000, however I know I would input a ridiculous amount I don't have anything close to having... Moving this along, this is one I have yet to actually see fail when I have played it (haven't been counting the attempts) I am highly interested to see what holes you guys could put in this one as this weekend I am wanting to test it out.

Option 3. Combo

Simply put, it would just be combining the two of these. The red/black method (which I am aware would also work with high/low and even/odd) would have a semi consistent "low risk, low reward" system while having a heavy hitting system with the greens over time.

Much appreciated on any tips and suggestions!!!

Roulette is definitely one of those games described as "random bets on random numbers", even if you try to develop some complex strategy. In the very long run you will average losing 1 unit of every 37 that you bet on a single-zero game and 2 units of every 38 that you bet on a double-zero game, no matter what bets you place. This assumes that you always avoid the five-number bet (1-2-3-0-00), which has an even higher expected rate of loss.

However, variance can be very high, and you may temporarily come out ahead or lose all your money right from the beginning. There are strategies that can increase or decrease this variance, but they won't reduce or increase your long-term average rate of loss. Only way to avoid those losses is the simple one: Don't play.

MANY people have fallen into this way of thinking but let us save you a lot of time and a lot of money! It simply does not work. You will find patches of success but every now and then you will get destroyed. The house edge against you is 5.26% on roulette so for every $100 you wager you can expect to hand the casino $5.26 of your hard earned money. You may not realize that loss for many sessions but eventually it will come.

In your example you say you wait for 4 consecutive reds to come in and then start betting on black. What are the odds of a black coming up next? 47.36% the same as it was one each of the previous 4 spins. Now you might say to yourself "But wait! The odds of a coin flip coming up heads 5 times in a row is .50 x .50 x .50 x .50 x .50= 0.03125 which is 3.125%. That may be true if you are betting that wager from the start before spin #1, in other words you said...I bet the next 5 flips will all be tails! It would have a 3.125% of happening. However once the first 4 flip have been completed the % chance of the next flip being heads is 50/50.

There are ways to beat the house but in my opinion roulette is not one of them.

GL,

RD

Quote:DecahOption 1. Black/Red

So I feel this should be shot down quickly, but I just want to verify my math wasn't wrong.

So example being is last ball was red, probability should be in favor of black? .526^2 that it would be red/green or red/red. playing through this theory I would wait till 4 red would load up and I would put $25 on black, which what I came up with is around a 4% chance of it failing or .526^5.

Let me rephrase what everyone else has said.

The probability of five reds coming up in a row (on a double-zero wheel) is (9/19)

^{5}.

However, the probability of four reds coming up in a row followed by a black is also (9/19)

^{5}.

This is what most people don't seem to realize.

ok, $775 would be what you lose in 5 spins using a Marty, not winning your $25Quote:DecahOption 1. Black/Red

So I feel this should be shot down quickly, but I just want to verify my math wasn't wrong.

So example being is last ball was red, probability should be in favor of black? .526^2 that it would be red/green or red/red. playing through this theory I would wait till 4 red would load up and I would put $25 on black, which what I came up with is around a 4% chance of it failing or .526^5.

Then I chase the $25 if it does happen to fail, betting $50 after that, then $100, etc. till I don't have bank to keep up or I reach table max.

(25+50+100+200+400)

that probability is (20/38)^5 = about 1 in 24.8

so you win 23.8 times at total of $25 = $595 winnings on average

and lose $775 one time on average for a net loss -$180

looks like you need to do better than average

to avoid that average loss

sure looks fun making that $400 bet

have fun

thank you for sharing

Sally

Oh, I do not like the green bet option

but adding in the even/odd and high/low

should give more action per hour I would think.

This sums it up the best to me... you think red has some bigger probability of hitting, but it's the exact same as black, and every spin before it.Quote:ThatDonGuyLet me rephrase what everyone else has said.

The probability of five reds coming up in a row (on a double-zero wheel) is (9/19)^{5}.

However, the probability of four reds coming up in a row followed by a black is also (9/19)^{5}.

This is what most people don't seem to realize.