discuss Casino Question

[Want2learn]

If you are playing American Roulette you have a 1/38 shot of winning . . . 2.6% chance to win

If you have ever played, there are times when you observe certain numbers not appearing. If you could eliminate 2 numbers off the wheel, you are break even. 1-36 covered and 0 00 left off. 1 hits, 35 plus your initial 1 to cover all 36 numbers.

What if you thought to eliminate 5 numbers. 0,00,1,2,3 each of those numbers have a 2.6% chance of hitting, but collectively they are 5/38 or roughly 13.2% chance of hitting. That in turn means you have an almost 87% chance of winning.

In this scenario off a $1 bet . . . you risk $33 and if you hit one of your 33 numbers you get $35 plus your initial $1 for $36 back. That is a 3 unit increase from what you laid out.

But if 0,00,1,2,3 hit you lose $33.


Has anyone ever played roulette this way? Trying to eliminate say 5 numbers?

 

[maya-zir]

I used this for lottery, but it was more advanced way than yours.
It worked but not for me.
When I bought lottery I never win, but when I did it just for fun I always could get at least 3 from 5.
I eliminated 15 numbers from 50, but not randomly, I just suggested that they must use the same balls during some time.
So I suggested that some of balls could weight less than others and also to jump higher and also be above the heavy ones.
But the hole from wich they fall from the sphere is down the sphere so the light balls had less chance to be rejected because they almost always at the top of the bunch of balls.
So I was checking results of hundreds of games making a long research, and I think it worked because I really could improve my ability, but just in theory because each time I filled the form, something went wrong.. I am very unlucky
But trying for fun without buying the real lottery I really began to get at least 3 from 5 correct numbers.
It was 10 years ago.. Have no idea if they change thier tactic and now change balls each game.. Who knows?

I forgot to tell, the rest of numbers (1-35) I divided into 3 groups, for example:
1st groupe: 18 numbers
2nd groupe:17 numbers
3rd groupe: half of numbers from the first group and half of numbers from the second group
And then I made 10 combination with each groupe
Totally it could cost around $60-70 (30 combinations)

 

[Asfas1000]

Won't work, simple math. Don't even try. Every single bet your are placing is a losing bet. Doesn't matter how you arrange your bets, they are still a collection of (long-term) losing bets.

 

[Ategy]

It is actually very easy to beat the house at roulette and a good number of other casino games.

The problem is .. statistically speaking (pure probabilities) that even if you have a system where you win 9 times out of 10 .. then the 10th attempt (the 1 out of 10 that you lose) you end up losing more than all your previous 9 wins combined.

The math behind roulette is probably the simplest of most casino games .. there is no way to beat the house long term .. not even if you have the world's biggest horse shoe up your behind .. lol

PS .. if given the choice never play American Roulette .. always stick to English Roulette with a single zero. American Roulette was about he same when it first came out many years ago and players would only lose half their chips if either 0 came out .. which would lower the long term rake of the house to virtually the same as single zero roulette. But it's not like that anymore!