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Guf
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« Reply #3 on: May 26, 2008, 06:33:19 PM » |
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Hello Roddo
I have had the same idea. Tried it. And cried.
Before you go on, I must say that the tactic WORKS, but only if the limits wasn't so limited ;-)
I made a simulation program and ran it in there. I'll give a screenie of it later when I find it.
What you need to understand is, the RNG of i.e. Playtech or .NET's RNG is very unlikely to be predicted, even though you compensate for it by betting on the numbers which least likely will show, there will always be a chance of 2/37 of being pushed 35 coins back.
If so, you would probably say then just bet on the numbers that just appeared as they would have a theoretical lower chance of showing up, yes that is true, though it still will not compensate at all for the 0.
There is a 2/37 chance of loosing with this strategy, that is 5.405% chance of loosing, and IF you loose, which you would do every now and then, you would have to earn the money back, but how? ---
I'd rather show you the simulation program, but at the moment I am not in the hands of it, so I'll show it to you in text.
I'll show you a totally plausible play cycle.
You play the Ross' strategy (2/37 betting system)
This is just an example, you can try it out in the real deal later yourself and experience it, though I wouldn't if I were you.
You place 35 chips on 35 different numbers. No side bets or anything.
Let's say you win 35 times in a row.(Really good cycle) balance: +35
Now you lost and you loose 35 chips. Balance 0
We can only work from the theoretical point of view as that is the RNG is not static seeding.
Remember out of 37 spins you are destined to loose 2/37 spins, even though you bet on the least possible numbers to show.
So right now we have spun 36 times out of 37 spins. We won 35 in the first 35 spins, and we lost 35 in the 36th spin. Now there is one more loose to go.
Status: 36/37 spins run so far. 35 out of 35 wins. 1 out of 2 losses.
There is many ways of getting the money back(Martingale doesn't work). You can triple all the stakes so there is 3 chips on each number, leaving 105 chips on the table. Now you will win 3 chips at a time so you only need to spin 12 times to get the 35 chips back plus 1.
In short, a double up wont work here in this strategy since you would have to spin 18 times to win the 35 coins back +1 and if you run into another loose you would have to
So right now what would you do? If you knew you would loose you wouldn't bet of course, but as we don't know when we will loose we will bet 105 so we can get our 35 lost coins back.
But we looses in spin 37 as told, and now we're behind -105. If you would have doubled only, you would be down with 70 though.
So now a totally new play cycle starts. We are down with -105 and we need to catch it up before we loose again (the 2/37 chance).
So now we could triple our bets, which is the only way this Ross tactic would work.
So now we bet 9 on each number, 315 coins all in all. Now we will get 108 coins after 12 spins, which will leave us with a profit of 3 coins.
Say we win 10 times and looses on the 11th. We get 90 coins but looses 315 coins. Now we have -105 -315 +90 = -330.
Now we want to earn our money back before we get the second and last loss of the play cycle (37 spins is a play cycle).
So we should triple our bets to 27 on each number, but we can't because of the limits of the table (1-25$).
So now you are left with a debt on 330! And the chances for those 2 numbers to come are pretty high (5,4%).
This is just an example of a run. I have seen many of these runs while I simulated it, it can occur very often. The chance of getting 2 of your numbers in a row is 2/37 = 0,054^2 = 0,29%. That sounds small, but that is just in a row! For 2 numbers to occur within 12 numbers the chance is 16,2%! That can you survive though. You can survive 3 times even, but you can't survive 4 times. The chance of those 2 numbers showing up within 12 spins is 4,05%. So the chance of you losing is 4,05%. That chance IS better than flat betting on 35 numbers, which would be 5,4% of losing.
A fast summary of what happened with this tactic (if it was to be applied to betting patterns)
bet 35 coins on 35 numbers
win 35 times. Balance: +35
Loose: Balance: 0
Tripled up to win back the lost money.(105 coins)
Loose: Balance: -105
Tripled up again to win back the lost money.(315 coins)
win 10 times. Balance: -15
Loose: Balance: -330
And then you stop because you can't triple any bets anymore leaving you uncapable of earning any money back.
You might be confused by all this, totally understandable, but after you've made a lot of your own simulations (if you do that before you start betting for real money) then come back to this post and read it again, and I am definite you'll get it totally :-)
Basically this tactic would be good if there weren't that small limits on numbers. But there is, so I've stopped dreaming about it ;-)
I just ran the simulation:
The starting balance is 1 000 000, the reason why the balance already is so crazy is because it had to triple a LOT and that makes it win more with the last spin.
Total bet: is the highest triple up. The amount of chips needed.
Round: Rounds/Spins.
Losses: times lost. That is every loss that it had lost.
"Double"up: is how high, how many times it had to triple up to be able to overcome the losses.
Just so you, or whoever might want to know how it runs;
It starts betting 35 coins on 35 numbers out of 37 numbers. If it looses it triples up and plays exactly 12 spins. If it hasn't lost in the next 12 spins it goes down to 35 coins again. That's pretty much it.
The betting pattern (if lost) goes like this:
35
35*3
35*3*3
35*3*3*3
and so on...
That's it. And if it wins it will of course continue with the pattern of betting 35 coins on 35 numbers.
I just wanted to save you from losing a lot of money ;-) If you got any objections then don't hesitate to write to me.
Regards
Guf
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