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person A: considering the odds of losing 6 even money bets in a row are 1.53% then in 74 plays of doubling your stake starting at £5 you would once lose a total of £315 (5+10+20+40+80+160). In the other 73 sets of plays, a play being up to 6 bets till you win, you would win £5 (e.g. winning £40 to cover losses of 5+10+20). 73 x 5 is 365 meaning a net profit on 50 (365-315) from this system.
Person B: except you've not counted the zero here which makes 37 outcomes 19/37 losing which is 0.5135... Over six attempts that's 1.83% of a loss which is 1/54.55... I'll give you the benefit of the doubt and round it up and also that a £5 minimum table will have a maximum that high so we'll roll with the figures... That's £315 loss and 54 x £5 = house always wins I'm afraid... Sorry to piss on your bonfire :P
Person A: Thats ridiculous mate, are you seriously suggesting that a 1/37 shot would take the payout from 365 to 270??? The zero pays half your stake back anyway and considering 50% of the time the zero falls when you have £5 down, a further 25% when you have £10, a further 12.5% when you have £20 then it is only going to be one in 8 times that it falls (i.e. once every 296 spins (8 x 37) that it will cost you more than £10!!!!! By which time you have made 4 lots of 50 (296/73), i.e. 200 profit!!!!! And the table plays £5 minimim to £250 maximum, I'm suggesting you don't need to go above £160, in fact after 2 or 3 losses you can start doubling and adding 5 to compensate your time and still place your 6th bet within the £250 table limit!!!
Person B: Ok so I hadn't included the half stake back, It's not 73 wins though It's 53 or 54, I suggest you revisit your probability lessons and check my maths It's all correct... The 50% on zero will only return you a fraction of your loss back but either way house still wins unless It's a true 50/50
Person A: the whole point is every spin is a new spin so it doesnt need to factor in the ongoing probability of 48.5% x 48.5% etc etc that you are proposing. The zero is only a factor when it comes up, when it does you work around it by doubling your stake then adding half the loss on zero to get back on track for a 50/50 bet the next time.
Person B: You have to multiply them to calculate the chance of no loss in 6 chances and find how many times you would win £5 vs lose £315, I'm not multiplying the odds of winning. You can't discount the zero and calculate it on 50/50 chance! That's just simply wrong!... Plus any time you start varying the bet you're no longer following the pre-calculated system so it just becomes a punt
Person A: the chances of a wipeout loss, i.e. hitting one of the 18 numbers that takes all your money is what the calculation is based on, the half loss on zero can be turned around with only adding 50% to your next stake anyway, so if the zero comes on 20 your next bet is 50 not 40, of which the extra 10 has been given back anyway, so you only adding the 10 return to claim back the 10 loss from zero, the system plays on regardless. And thats part of the pre alculated system too so its not a punt. also your mathe is based on playing 6 plays each time, which is why your payout drops from 365 to 270, you havent factored in that in 48.5% of plays you play once and win, in a further 24.25% of plays you play 2 spins, losing first and winning second. So the actual figure is in between 270 and 365, but a lot closer to 365 due to the fact the majority of plays finish in 3 spins or less
Person B: You can work them out seperately, but the chance of hitting either the 18 winning numbers or18 losing numbers still remains at 48.5%. anything that results in less than 6 spins means that you've won £5. if you reach 6 spins with a loss then you lose £315, it's that simple and that's the only thing that my probability calculates, 1/54 times you will lose all 6 times, granted one of those times it might be a zero and you might return half of your stake but that calculation is negligible since it only affects how much you lose
Person A: where the hell are you getting 1 in 53 from. With no zero the maths reads 50%, 25%, 12.5%, 6.25%, 3.125%, 1.5635% with the 1.5635% chance of losing taking 1 out of 74 plays. With zero the same maths reads 48.64%, 23.66%, 11.51%, 5.60%, 2.72%, 1.32%. So the 1.32 still gives you 66 wins for every one loss, which is 330 playing against 315, so you have an edge, plus profit of every time the zero returns half!!!!!
Person B: I'm getting 1/53 by using the correct maths
Person A: 48.64 % to the power of 6 mate!!!!!! your maths is unsubstantiated, you haven't shown your working, this would get a zero in the international baccalaureaute 2002 maths exam and you know it, miss ebbs would be angrier than when that huge cardboard calculator model misteriously dissapeared from her room!!!!!
Person B: 48.64 is the chance of a win!!! 51.36 is the loss, 51.36 to the power of 6 gives you 1/54....You're over complicating a maths problem to try and represent it in a positive way, you lose 1/54 and you lose more than the 53 wins It's that simple
Person A: there is no way a 1.36% change over 6 runs moves the payout by 30%, where is your working????
Person B: My working is all in my first post..
Person A: yeah so your working suggests 1/54.5, which means a return of 322.50, against a loss of 315, so it is profitable, then my argument is also that the zero returns half the stake therefore it is higher than that. In the 54.5 spins you will average a zero hit 1 and a half times, therefore you need to half your average stake per spin, and multiply it by 1.5 and add that to the 322.50 returns, which gives you a clear edge nearer to 665. As I say the other way of looking at it is an edge of 19% with no zero is never going to be fully wiped out by a house edge of 3% on each spin!!!! and I know your going to say you will only recoup 270 from 54 spins against 315 so I'll work my maths into a zero avoidance system and get it to you tomorrow, also obviously you could double your stake and add 5 each time whihc would mean your 54 spins would recoup far more money, against lesser losses due to hightened stakes for each of your 53 wins agaisnt the one loss!!!
Person B: I'd love to argue this all night but you've clearly got lost in your own sales patter... 322.5/£5 is 64.5 so you've already fallen at the first hurdle... And 1 in 54.5 losses is 53.5 wins so 53.5 * £5 = 267.50... And It's not 54.5 spins It's 54.5 sets of up to 6 one of which you will lose all of them... For your zero argument if you take an average of 3 spins for the 53.5 remaining sets you can expect 4 zeros overall which sounds great until you put it in context ie. It could fall first so results of spins 0, lose, lose, win profit for that result? An extra £2.50... Now I'm no expert but I'm sure if i wanted to spend more time researching and calculating I could work out the difference the zeros are likely to make to returns but I'm not going to flow a dead horse
Person A: I maintain that the zero return money just gets added on to win back on the next spin making it redundant, unless you hit 6 zeros in a row it doesn't feature!!! There is no way I'm that l;ucky with recent run of form to be taking the profit I have without an edge!!!! I find it hard to doubt my figures and recent run, however I also find it hard to doubt a man who worked out that he could defraud pacific pokers friend referal money by opening accounts in fake peoples names, debiting them all off one bank card, and rigging games against himself in heads-up, you can see my dillema here!!! Maybe I'll come down 3rd weekend in October and can show you in person mate, I'm busy saving Africa till then!!!
right I'll have another go, the zero returns half your stake anyway, so you only have to put 1/72th of your even money stake down to make the zero spin break even. In my system of 74 plays you play 160 twice (£2.10 x 2), £80 four times (£1.05 x 4), 40 eight times (0.55 x 8), 20 sixteen times (0.27 x 16), 10 thirty two times (0.135 x 32) and £5 sixty four times (£0.05825 x 64). Total all those up and deduct from my 360 and you only lose £25 meanign you are still clear of your 315 hurdle rate.
Person B: except you've not counted the zero here which makes 37 outcomes 19/37 losing which is 0.5135... Over six attempts that's 1.83% of a loss which is 1/54.55... I'll give you the benefit of the doubt and round it up and also that a £5 minimum table will have a maximum that high so we'll roll with the figures... That's £315 loss and 54 x £5 = house always wins I'm afraid... Sorry to piss on your bonfire :P
Person A: Thats ridiculous mate, are you seriously suggesting that a 1/37 shot would take the payout from 365 to 270??? The zero pays half your stake back anyway and considering 50% of the time the zero falls when you have £5 down, a further 25% when you have £10, a further 12.5% when you have £20 then it is only going to be one in 8 times that it falls (i.e. once every 296 spins (8 x 37) that it will cost you more than £10!!!!! By which time you have made 4 lots of 50 (296/73), i.e. 200 profit!!!!! And the table plays £5 minimim to £250 maximum, I'm suggesting you don't need to go above £160, in fact after 2 or 3 losses you can start doubling and adding 5 to compensate your time and still place your 6th bet within the £250 table limit!!!
Person B: Ok so I hadn't included the half stake back, It's not 73 wins though It's 53 or 54, I suggest you revisit your probability lessons and check my maths It's all correct... The 50% on zero will only return you a fraction of your loss back but either way house still wins unless It's a true 50/50
Person A: the whole point is every spin is a new spin so it doesnt need to factor in the ongoing probability of 48.5% x 48.5% etc etc that you are proposing. The zero is only a factor when it comes up, when it does you work around it by doubling your stake then adding half the loss on zero to get back on track for a 50/50 bet the next time.
Person B: You have to multiply them to calculate the chance of no loss in 6 chances and find how many times you would win £5 vs lose £315, I'm not multiplying the odds of winning. You can't discount the zero and calculate it on 50/50 chance! That's just simply wrong!... Plus any time you start varying the bet you're no longer following the pre-calculated system so it just becomes a punt
Person A: the chances of a wipeout loss, i.e. hitting one of the 18 numbers that takes all your money is what the calculation is based on, the half loss on zero can be turned around with only adding 50% to your next stake anyway, so if the zero comes on 20 your next bet is 50 not 40, of which the extra 10 has been given back anyway, so you only adding the 10 return to claim back the 10 loss from zero, the system plays on regardless. And thats part of the pre alculated system too so its not a punt. also your mathe is based on playing 6 plays each time, which is why your payout drops from 365 to 270, you havent factored in that in 48.5% of plays you play once and win, in a further 24.25% of plays you play 2 spins, losing first and winning second. So the actual figure is in between 270 and 365, but a lot closer to 365 due to the fact the majority of plays finish in 3 spins or less
Person B: You can work them out seperately, but the chance of hitting either the 18 winning numbers or18 losing numbers still remains at 48.5%. anything that results in less than 6 spins means that you've won £5. if you reach 6 spins with a loss then you lose £315, it's that simple and that's the only thing that my probability calculates, 1/54 times you will lose all 6 times, granted one of those times it might be a zero and you might return half of your stake but that calculation is negligible since it only affects how much you lose
Person A: where the hell are you getting 1 in 53 from. With no zero the maths reads 50%, 25%, 12.5%, 6.25%, 3.125%, 1.5635% with the 1.5635% chance of losing taking 1 out of 74 plays. With zero the same maths reads 48.64%, 23.66%, 11.51%, 5.60%, 2.72%, 1.32%. So the 1.32 still gives you 66 wins for every one loss, which is 330 playing against 315, so you have an edge, plus profit of every time the zero returns half!!!!!
Person B: I'm getting 1/53 by using the correct maths
Person A: 48.64 % to the power of 6 mate!!!!!! your maths is unsubstantiated, you haven't shown your working, this would get a zero in the international baccalaureaute 2002 maths exam and you know it, miss ebbs would be angrier than when that huge cardboard calculator model misteriously dissapeared from her room!!!!!
Person B: 48.64 is the chance of a win!!! 51.36 is the loss, 51.36 to the power of 6 gives you 1/54....You're over complicating a maths problem to try and represent it in a positive way, you lose 1/54 and you lose more than the 53 wins It's that simple
Person A: there is no way a 1.36% change over 6 runs moves the payout by 30%, where is your working????
Person B: My working is all in my first post..
Person A: yeah so your working suggests 1/54.5, which means a return of 322.50, against a loss of 315, so it is profitable, then my argument is also that the zero returns half the stake therefore it is higher than that. In the 54.5 spins you will average a zero hit 1 and a half times, therefore you need to half your average stake per spin, and multiply it by 1.5 and add that to the 322.50 returns, which gives you a clear edge nearer to 665. As I say the other way of looking at it is an edge of 19% with no zero is never going to be fully wiped out by a house edge of 3% on each spin!!!! and I know your going to say you will only recoup 270 from 54 spins against 315 so I'll work my maths into a zero avoidance system and get it to you tomorrow, also obviously you could double your stake and add 5 each time whihc would mean your 54 spins would recoup far more money, against lesser losses due to hightened stakes for each of your 53 wins agaisnt the one loss!!!
Person B: I'd love to argue this all night but you've clearly got lost in your own sales patter... 322.5/£5 is 64.5 so you've already fallen at the first hurdle... And 1 in 54.5 losses is 53.5 wins so 53.5 * £5 = 267.50... And It's not 54.5 spins It's 54.5 sets of up to 6 one of which you will lose all of them... For your zero argument if you take an average of 3 spins for the 53.5 remaining sets you can expect 4 zeros overall which sounds great until you put it in context ie. It could fall first so results of spins 0, lose, lose, win profit for that result? An extra £2.50... Now I'm no expert but I'm sure if i wanted to spend more time researching and calculating I could work out the difference the zeros are likely to make to returns but I'm not going to flow a dead horse
Person A: I maintain that the zero return money just gets added on to win back on the next spin making it redundant, unless you hit 6 zeros in a row it doesn't feature!!! There is no way I'm that l;ucky with recent run of form to be taking the profit I have without an edge!!!! I find it hard to doubt my figures and recent run, however I also find it hard to doubt a man who worked out that he could defraud pacific pokers friend referal money by opening accounts in fake peoples names, debiting them all off one bank card, and rigging games against himself in heads-up, you can see my dillema here!!! Maybe I'll come down 3rd weekend in October and can show you in person mate, I'm busy saving Africa till then!!!
right I'll have another go, the zero returns half your stake anyway, so you only have to put 1/72th of your even money stake down to make the zero spin break even. In my system of 74 plays you play 160 twice (£2.10 x 2), £80 four times (£1.05 x 4), 40 eight times (0.55 x 8), 20 sixteen times (0.27 x 16), 10 thirty two times (0.135 x 32) and £5 sixty four times (£0.05825 x 64). Total all those up and deduct from my 360 and you only lose £25 meanign you are still clear of your 315 hurdle rate.
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What do you want to discuss? The house has a 2.7% edge on European roulette, Martingaling the stakes can't defeat that. Is the "half your stake back on zero" American roulette? If so I thought they had 2 zero's?
There's one based on the work of the renowned Irish mathematician, Fib O'Nacci.
Try googling his name.
Try googling his name.
| Bet No. | Stake | [b****. stakes[/b] |
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 3 | 2 | 4 |
| 4 | 3 | 7 |
| 5 | 5 | 12 |
| 6 | 8 | 20 |
| 7 | 13 | 33 |
| 8 | 21 | 54 |
| 9 | 34 | 88 |
| 10 | 55 | 143 |
| 11 | 89 | 232 |
| 12 | 144 | 376 |
| 13 | 233 | 609 |
| 14 | 377 | 986 |
| 15 | 610 | 1,596 |
| 16 | 987 | 2,583 |
| 17 | 1597 | 4,180 |
| 18 | 2584 | 6,764 |
| 19 | 4181 | 10,945 |
| 20 | 6765 | 17,710 |
I could extend the table further but have drawn the line at 20 losing bets.
Third column was meant to read '****. stakes'.
In the interest of clarity, perhaps I'd better mention that '****.' is an abbreviation of the word 'Cumulative'.
In the interest of clarity, perhaps I'd better mention that '****.' is an abbreviation of the word 'Cumulative'.
Hmmmm, given two assumptions that I think we can all agree on:
1. at a roulette table the house has an edge on each individual bet,
2. the outcome of each bet is independent, i.e. does not depend in any way on what happened previously,
do you think that it is possible to construct a staking plan that is profitable on avergage in the long run?
1. at a roulette table the house has an edge on each individual bet,
2. the outcome of each bet is independent, i.e. does not depend in any way on what happened previously,
do you think that it is possible to construct a staking plan that is profitable on avergage in the long run?
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