Will the online casino's ban me for this?
Discussion
Jon1967x said:
blinkythefish said:
Road2Ruin said:
Jon1967x said:
You need to qualify that... Over 100 coin tosses your much more likely to have 50 of each than 100 red do long as the order doesn't matter.
No your not. Each time you flip the coin you have an equal chance of getting heads or tails. . The fact your are doing it 100 times doors not alter the odds each time. . However, the probability of the following coin toss being the same as the one before is where the real maths come in. There is only a 1 in 2 chance of that 
meaning that at some time it won't be. But the chance of you getting 50 heads and 50 tails is just as unlikely as 100 of one or another. You may get 60-40 or even 49-51 or 80-20. The more times you flip the more accurate 50-50 looks but that is only because as a percentage of the total number it looks close. If you flipped 1 milion time you might get 50.01% but how much difference in coin tosses were there? Of course I am not Stephen Hawkin so may be talking out of arse..;)There are 100!÷50!÷50! = 1E29 outcomes containing 50 head and 50 tails. So over 100 spins 50R50B has a probability of 1E29:1.3E30=0.079:1.
The 50H50T outcome is much more likely than 100H. However any specific combination of H and T has the same likelihood as all Headd - overall 50:50 has a higher probability because there are more combinations which give this outcome).
This has no bearing on any individual spin, which has a 50% chance of head or tail.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
TwigtheWonderkid said:
Precisely. It's the same as rolling 2 dice. You might roll 12, and you might roll 7. But your 6 times more likely to roll 7 than you are 12. Because there are 6 ways to roll 7, and only 1 way to roll 12.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
It never fails to amuse me how some people are prepared to make themselves look so foolish on threads like this when it is obvious they have next to no mathematical understanding!
Amusing for the rest of us though!
sidicks said:
TwigtheWonderkid said:
Precisely. It's the same as rolling 2 dice. You might roll 12, and you might roll 7. But your 6 times more likely to roll 7 than you are 12. Because there are 6 ways to roll 7, and only 1 way to roll 12.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
It never fails to amuse me how some people are prepared to make themselves look so foolish on threads like this when it is obvious they have next to no mathematical understanding!
Amusing for the rest of us though!
TwigtheWonderkid said:
sidicks said:
TwigtheWonderkid said:
Precisely. It's the same as rolling 2 dice. You might roll 12, and you might roll 7. But your 6 times more likely to roll 7 than you are 12. Because there are 6 ways to roll 7, and only 1 way to roll 12.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
It never fails to amuse me how some people are prepared to make themselves look so foolish on threads like this when it is obvious they have next to no mathematical understanding!
Amusing for the rest of us though!
Jon1967x said:
TwigtheWonderkid said:
sidicks said:
TwigtheWonderkid said:
Precisely. It's the same as rolling 2 dice. You might roll 12, and you might roll 7. But your 6 times more likely to roll 7 than you are 12. Because there are 6 ways to roll 7, and only 1 way to roll 12.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
If I was spinning the roulette wheel 100 times and I had to bet of the final outcome, I'd go exactly 50:50, because that is the single most likely result. Chances are it would be close to that but not spot on, 51:49, 52:48, 60:40 etc. It would not be 100:0, that's almost certain.
It never fails to amuse me how some people are prepared to make themselves look so foolish on threads like this when it is obvious they have next to no mathematical understanding!
Amusing for the rest of us though!
h said:schmunk said:
Ug_lee said:
No idea was around 2006 ish when I did this. I was quite sceptical when a friend showed me the theory. Had nothing to do all weekend so thought I'd give it a go, worse I could lose was £100.
Basically every spin you put £2 on red and black, also £2 on green. Do 10 spins, you get the money you gambled on red or black back every time. No green within them 10 spins meant you upped the amount on green to £4 and do another 10 spins, upping the amount on green every 10 spins by £2. When green did come in (and it did nearly all the time within 30 spins) I would get my money back plus a small increase in the fighting fund.
I'll see if I still have the tally sheet and where it went wrong, something/someone obviously clocked me and I went around 80+ spins without green coming up. By that time I was losing £20+ a spin, it was pretty horrible feeling knowing the winnings were bleeding away.
That's when I bailed and the emails started. Never done online casinos since.
This is just a variation on the above-mentioned Martingale system.Basically every spin you put £2 on red and black, also £2 on green. Do 10 spins, you get the money you gambled on red or black back every time. No green within them 10 spins meant you upped the amount on green to £4 and do another 10 spins, upping the amount on green every 10 spins by £2. When green did come in (and it did nearly all the time within 30 spins) I would get my money back plus a small increase in the fighting fund.
I'll see if I still have the tally sheet and where it went wrong, something/someone obviously clocked me and I went around 80+ spins without green coming up. By that time I was losing £20+ a spin, it was pretty horrible feeling knowing the winnings were bleeding away.
That's when I bailed and the emails started. Never done online casinos since.
On a proper roulette table, the red & black bet is pointless as described above - the result would be the same as betting green only. On a real table, you could have some system of ever increasing bets on a single number(e.g. 0), but it would be doomed to failure as the payout is less than the probability of it hitting.
As I understand it the "forcing the green" method relies on the fact that the online random roulette generators aren't as random as they should be. The claimed lack of randomness seems to be that if too much money is being lost to red or black(i.e. someone is using the Martingale System), then the RNG algorithm will increase the probability of a 0, which tips the odds against anyone betting Red or Black. The claim is that if you always bet Red and Black, the algorithm continually detects a Red/Black loss, gets a bit screwed and increases the probability of green to above the truly random level, at which point the green bet in the system above comes in and wins.
I have no idea if this is true or not, as with all gambling systems, there are people out there claiming great success, but that doesn't mean much. If this was genuinely true, it doesn't strike me as a tricky bug to remove from the casino software: if the punter is betting on Red, Black and Green, revert to truly random, or decrease the probability of green.
blinkythefish said:
Not quite, the method is supposed to exploit a flaw in the Casino's software.
On a proper roulette table, the red & black bet is pointless as described above - the result would be the same as betting green only. On a real table, you could have some system of ever increasing bets on a single number(e.g. 0), but it would be doomed to failure as the payout is less than the probability of it hitting.
As I understand it the "forcing the green" method relies on the fact that the online random roulette generators aren't as random as they should be. The claimed lack of randomness seems to be that if too much money is being lost to red or black(i.e. someone is using the Martingale System), then the RNG algorithm will increase the probability of a 0, which tips the odds against anyone betting Red or Black. The claim is that if you always bet Red and Black, the algorithm continually detects a Red/Black loss, gets a bit screwed and increases the probability of green to above the truly random level, at which point the green bet in the system above comes in and wins.
I have no idea if this is true or not, as with all gambling systems, there are people out there claiming great success, but that doesn't mean much. If this was genuinely true, it doesn't strike me as a tricky bug to remove from the casino software: if the punter is betting on Red, Black and Green, revert to truly random, or decrease the probability of green.
If I was raking it in the last thing I'd do is go in a forum and tell the world.On a proper roulette table, the red & black bet is pointless as described above - the result would be the same as betting green only. On a real table, you could have some system of ever increasing bets on a single number(e.g. 0), but it would be doomed to failure as the payout is less than the probability of it hitting.
As I understand it the "forcing the green" method relies on the fact that the online random roulette generators aren't as random as they should be. The claimed lack of randomness seems to be that if too much money is being lost to red or black(i.e. someone is using the Martingale System), then the RNG algorithm will increase the probability of a 0, which tips the odds against anyone betting Red or Black. The claim is that if you always bet Red and Black, the algorithm continually detects a Red/Black loss, gets a bit screwed and increases the probability of green to above the truly random level, at which point the green bet in the system above comes in and wins.
I have no idea if this is true or not, as with all gambling systems, there are people out there claiming great success, but that doesn't mean much. If this was genuinely true, it doesn't strike me as a tricky bug to remove from the casino software: if the punter is betting on Red, Black and Green, revert to truly random, or decrease the probability of green.
There will always be winners, just not as many as there are losers, (and even the winners tend to end up losers) and it suits the casinos that there are, as without some hope of winning or a belief there's a way of beating the system, why would people bet?
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