Any probability experts here?
08 January 2004, 02:31 PM
#1
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Hi,
I've recently been playing roulette online using a demo account. The wheel they use is a 'French' wheel meaning it only has one zero on it, which slightly improves the odds.
Anyway, I've been playing the 'Martingale' system and have done pretty well from it, but I realise how it can go wrong. For those that don't know the system, this is it:
You put $1 chip down on red, for example, then spin the wheel. If red comes up you get back $2 meaning a profit of $1. Then you do the same again and so on. If however, black comes up then you lose your chip. So this time you put down $2 on red - if red comes up you get back $4 giving a profit on that spin of $2 and clearing the loss on the black that came up. If black comes up again you put down $4 looking to make $8, and so on. Basically you continue doubling up as the spins go against you until red comes up. The sequence is $1, $2, $4, $8, $16, $32, $64, $128
This leads me to the problem. The table limit per spot is $100. Therefore, as soon as you get the 'wrong' colour 7 times in a row you're screwed and you've lost $127 (1+2+4+8+16+32+64).
So, to you experts out there - how often is this situation likely to occur? Seven of the same colour in a row? (don't forget the zero too!) Is it rare enough to just continue through it?
I made $470 the other day (yes, it took ages
) - too good to be true? Almost certainly!
I've recently been playing roulette online using a demo account. The wheel they use is a 'French' wheel meaning it only has one zero on it, which slightly improves the odds.
Anyway, I've been playing the 'Martingale' system and have done pretty well from it, but I realise how it can go wrong. For those that don't know the system, this is it:
You put $1 chip down on red, for example, then spin the wheel. If red comes up you get back $2 meaning a profit of $1. Then you do the same again and so on. If however, black comes up then you lose your chip. So this time you put down $2 on red - if red comes up you get back $4 giving a profit on that spin of $2 and clearing the loss on the black that came up. If black comes up again you put down $4 looking to make $8, and so on. Basically you continue doubling up as the spins go against you until red comes up. The sequence is $1, $2, $4, $8, $16, $32, $64, $128
This leads me to the problem. The table limit per spot is $100. Therefore, as soon as you get the 'wrong' colour 7 times in a row you're screwed and you've lost $127 (1+2+4+8+16+32+64).
So, to you experts out there - how often is this situation likely to occur? Seven of the same colour in a row? (don't forget the zero too!) Is it rare enough to just continue through it?
I made $470 the other day (yes, it took ages
) - too good to be true? Almost certainly!
08 January 2004, 02:43 PM
#2
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im not an expert and this is off the top of my head but would it be somnething like this... .47x.47x.47x.47x.47x.47x.47 that leaves 0.005ish so i dont know if thats right to be honest.
(roughly .47 as the 0 would be roughly 3% chance of occuring)
(roughly .47 as the 0 would be roughly 3% chance of occuring)
08 January 2004, 03:54 PM
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Open another account so you can keep going on another name.
Just be careful, if you get 20 in a row against you, your next stake needs to be $520,000. It's called averaging! Unlikely to happen, but certainly possible
M
Just be careful, if you get 20 in a row against you, your next stake needs to be $520,000. It's called averaging! Unlikely to happen, but certainly possible
M
08 January 2004, 04:05 PM
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I am assuming the zero is neither black nor red?
if so:
chances of 7 reds in a row = (18/37)^7 = 0.006449046
7 blacks is the same
chances of 7 zero's in a row = (1/37)^7 = 1.05339E-11
if so:
chances of 7 reds in a row = (18/37)^7 = 0.006449046
7 blacks is the same
chances of 7 zero's in a row = (1/37)^7 = 1.05339E-11
08 January 2004, 04:07 PM
#7
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Milo,
Yes, eventually it happens. But the thing is, if you've taken in more than $127 in profits then you can just keep going. Do you see what I mean? If it only happens every so often then overall you're going to profit from it.
Yes, eventually it happens. But the thing is, if you've taken in more than $127 in profits then you can just keep going. Do you see what I mean? If it only happens every so often then overall you're going to profit from it.
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08 January 2004, 04:11 PM
#8
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ajm,
Yes, the zero is neither black nor red and you lose if the ball lands on it.
How do you express your results in a percentage form? As in, during the course of 1,000 spins how likely is it that you'll get seven of the same colour, or seven of the same colour including the zero, in a row?
Yes, the zero is neither black nor red and you lose if the ball lands on it.
How do you express your results in a percentage form? As in, during the course of 1,000 spins how likely is it that you'll get seven of the same colour, or seven of the same colour including the zero, in a row?
08 January 2004, 04:16 PM
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yes - i know the system well
with your odds tho, you stand to only gain $1 per run of bets. so if u think about it on a smaller scale, u will be making $32 and $64 bets to win a slightly less than 50% chance of getting $1.
on the grand scheme of things, u could turn a profit on this. bear in mind that u need to have 7 in a row NOT coming up a minimum of 127 times (it will be a lot more in reality, because you're not going to win first time every time - statistically you're probably hoping the 7 in a row does NOT come up several hundred times or even thousand times). given that there is a near 1% chance of it coming up, it may well come up in a few hundreds "turns" (where a turn is a run of bets in which u hope to turn the $1 profit).
with your odds tho, you stand to only gain $1 per run of bets. so if u think about it on a smaller scale, u will be making $32 and $64 bets to win a slightly less than 50% chance of getting $1.
on the grand scheme of things, u could turn a profit on this. bear in mind that u need to have 7 in a row NOT coming up a minimum of 127 times (it will be a lot more in reality, because you're not going to win first time every time - statistically you're probably hoping the 7 in a row does NOT come up several hundred times or even thousand times). given that there is a near 1% chance of it coming up, it may well come up in a few hundreds "turns" (where a turn is a run of bets in which u hope to turn the $1 profit).
08 January 2004, 04:16 PM
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multiply by 100 to get percentage
If you include the zero as a loss then the results are slightly different.
The chances of it going against you would be (19/37)^7 = 0.009415928 (oops just noticed someone already worked this out above!)
If you include the zero as a loss then the results are slightly different.
The chances of it going against you would be (19/37)^7 = 0.009415928 (oops just noticed someone already worked this out above!)
08 January 2004, 04:27 PM
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OK, then that answers it. If the chances of it happening are 0.942% it's going to happen more often than taking in the required $127 to cover it through successful bets.
However, you can spin the wheel without placing a bet. Therefore, you could wait until the wheel has recorded, let's say, 3 of the same colour in a row before placing a chip down. By doing that you'd be saying that the wheel has to go against me 10 times in a row for me to lose.
Is that false logic? Every spin is completely independent of the last.
However, you can spin the wheel without placing a bet. Therefore, you could wait until the wheel has recorded, let's say, 3 of the same colour in a row before placing a chip down. By doing that you'd be saying that the wheel has to go against me 10 times in a row for me to lose.
Is that false logic? Every spin is completely independent of the last.
08 January 2004, 04:39 PM
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Are you sure about that in an on-line simulation ?
08 January 2004, 04:39 PM
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Ok, I researched this quite extensively when I tried to modify the system a little. Basically I concluded that you should try for runs in your favour. So if red was your b!tch for the night you'd play it and if you landed you'd keep the £2 on and try to double it up. Do this until you've got a tenner (or however big your ***** are) and then bank all or half of it as profit. If your run brakes before you bank you double up to recover your initial investment and start over. This worked really well and I tried it on an online table that had a play for fun feature.
However, like you I became worried about bad runs and sure enough they came along all too often and very often much much more than 7 in a row
I also tested this probability with a 10p coin and flipped it over and over and over recording 'runs' This isn't even a proper reflection as there is no 'zero' on a coin. ***Before anyone starts when I flip a coin its spins like **** and I allowed it to bounce on the carpet. Chaos theory dictates randomness from here*** The result??? You guessed it, monster (and I mean MONSTER) runs against me of upwards of 15 in a row. Flip a coin long enough and they come
I gave up after that without putting real money into it. If I was you I'd cash up now that you are ahead because probability dictates that if you play it long enough you'll get done
Of course if you want to play with 1% of luck or so against you do it online where you can get 10-20 times the spin rate of real life and thus if you go bust you'll have not made a massive time committment to it
However, like you I became worried about bad runs and sure enough they came along all too often and very often much much more than 7 in a row
I also tested this probability with a 10p coin and flipped it over and over and over recording 'runs' This isn't even a proper reflection as there is no 'zero' on a coin. ***Before anyone starts when I flip a coin its spins like **** and I allowed it to bounce on the carpet. Chaos theory dictates randomness from here*** The result??? You guessed it, monster (and I mean MONSTER) runs against me of upwards of 15 in a row. Flip a coin long enough and they come I gave up after that without putting real money into it. If I was you I'd cash up now that you are ahead because probability dictates that if you play it long enough you'll get done
Of course if you want to play with 1% of luck or so against you do it online where you can get 10-20 times the spin rate of real life and thus if you go bust you'll have not made a massive time committment to it
08 January 2004, 04:42 PM
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Is that false logic? Every spin is completely independent of the last.
....or should that be
08 January 2004, 05:35 PM
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You will always come out worse in the end.....
18 red
18 black
1 green
Prob of red or black is 18/37 (ie 0.486486) on every go....
so you will on average get back around 97% of your stake over time....
Don't play it to make money, you will not do it. Play for fun or not at all.....
18 red
18 black
1 green
Prob of red or black is 18/37 (ie 0.486486) on every go....
so you will on average get back around 97% of your stake over time....
Don't play it to make money, you will not do it. Play for fun or not at all.....
08 January 2004, 07:05 PM
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Thinking that what has gone before has the slightest influence on the next colour out is known as the gamblers fallacy...it is plain wrong.
IIRC, if you cant card count at blacjack(and very few people can do it properly, and even then you can be spotted, and asked to leave), the best chance of winning money is to take everything you can afford to lose, and plce it on black or red once, then walk out. Still likely to lose, mind!!!
IIRC, if you cant card count at blacjack(and very few people can do it properly, and even then you can be spotted, and asked to leave), the best chance of winning money is to take everything you can afford to lose, and plce it on black or red once, then walk out. Still likely to lose, mind!!!
08 January 2004, 08:38 PM
#21
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It's exactly the same software as the real game you'd be playing with them if you were using money
08 January 2004, 08:42 PM
#22
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This reminds me of the old 2:1 horse trick that me and my friends used to talk about (but never tested, so I'm not offering advice here!). Actually, since I'm not into the whole betting thing, I'm not sure if it's 2:1 or 3:1, whichever is the one that gives you three times your money on a win. I'll call it 2:1 here....
Anyway, the idea is you start by betting £1 on a 2:1 horse.
It loses. You lose your £1.
So, on the next 2:1 horse you see, you bet £2.
It loses. You lose that £2 too.
On the next step you bet £4, then £8, then £16 and as many times as you have to, doubling your money each time, until a horse finally comes in - let's say it doesn't come in until the eighth horse on which you've bet £128. Here's how it adds up:
You bet £128, you get £128 x 3 = £384.
BUT you've only lost (on all the other horses) £1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = £255.
Leaving you with a profit of £129.
Mathematically speaking, so long as you continue to double your outlay each time, you always recoup more than you have bet at this odds ratio. The main problem is that if you get up to, say, 25 horses and none have won yet, you're next bet is about £33 million pounds!
Spend it on beer instead!
Anyway, the idea is you start by betting £1 on a 2:1 horse.
It loses. You lose your £1.
So, on the next 2:1 horse you see, you bet £2.
It loses. You lose that £2 too.
On the next step you bet £4, then £8, then £16 and as many times as you have to, doubling your money each time, until a horse finally comes in - let's say it doesn't come in until the eighth horse on which you've bet £128. Here's how it adds up:
You bet £128, you get £128 x 3 = £384.
BUT you've only lost (on all the other horses) £1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = £255.
Leaving you with a profit of £129.
Mathematically speaking, so long as you continue to double your outlay each time, you always recoup more than you have bet at this odds ratio. The main problem is that if you get up to, say, 25 horses and none have won yet, you're next bet is about £33 million pounds!
Spend it on beer instead!
08 January 2004, 08:42 PM
#23
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Surely if it was a real game there would be no software -- just a bloke spinning a roulette wheel?
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08 January 2004, 09:43 PM
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Casino-on-Net uses the trusted MD5 RNG, which provides absolutely random results.
08 January 2004, 10:09 PM
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You can beat blackjack by counting at the number of picture cards left in the deck compared to those that have been already dealt. You bet differently depending on where the majority of picture cards lie (high bets one way, low the other). This gives you the mathematical advantage BUT it is VERY difficult to do (bearing in mind there are multiple decks in the 'shoe'). Lots of people try it, but as everything moves so fast, if you get slightly out, you are betting the wrong way round and giving the house a massive advantage.
If a casino spots anyone doing this (and winning), they will throw you out on your ear. Dealers say it is very easy to spot.
Various machines have been built to try and give gamblers the edge in this counting business, so they know whether to bet high or low...these I think are actually illegal (fraud, I guess)
If you do a google on "card counting" you'll see lots and lots of books advertising the "easyiest" way to do it...if it was so easy they wouldnt have written books 'cus they'd be millionaires by now!
If a casino spots anyone doing this (and winning), they will throw you out on your ear. Dealers say it is very easy to spot.
Various machines have been built to try and give gamblers the edge in this counting business, so they know whether to bet high or low...these I think are actually illegal (fraud, I guess)
If you do a google on "card counting" you'll see lots and lots of books advertising the "easyiest" way to do it...if it was so easy they wouldnt have written books 'cus they'd be millionaires by now!
08 January 2004, 10:13 PM
#27
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i thought the practice of putting several shuffled packs of card into the big holder and then drawing off a block from the top of this, to deal with during the game, prevented the practice of card counting as you cannot know how many picture cards are in the block of cards being used. and then after the game they draw a differnt block, so you can't use information gained from counting the cards in the previous game?
[Edited by ProperCharlie - 1/8/2004 10:14:58 PM]
[Edited by ProperCharlie - 1/8/2004 10:14:58 PM]
09 January 2004, 11:54 AM
#28
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It can be a bit more simple than that: Lets say you are at a table with 5 other players, that means (in BJ) that one hand will use at least 12 cards, all of which every player gets to see. If a "10" hasn't come out for all those 12 cards you can be sure one is about to come out, as the odds of a 10 coming out is 4/13. Its not a "sure thing", but you can make an educated guess...
09 January 2004, 12:46 PM
#29
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Cleterridge, the horse racing scheme doesn't work because the 2for1 odds aren't going to be the correct odds in all cases..ie in 'fair' roulette (no greens) you are exactly 50/50 to win betting on a colour. If you play one millions times, always betting on one colour using the doubling technique when you lose you will almost certainly break even..assuming the payout is 2 for 1. On the horses you can't cant on that mathmatical certainty..many horses rated 2 for 1 should in reality be longer odds, and ultimately bets made on them will reflect that with their payoff.
Roulette is a suckers game...statistically the best you can hope for is just shy of breaking even(ie being a loser long term)...unless the wheel isn't exactly fair and you know what number/s is/are favoured.
Rgds
Cman
Roulette is a suckers game...statistically the best you can hope for is just shy of breaking even(ie being a loser long term)...unless the wheel isn't exactly fair and you know what number/s is/are favoured.
Rgds
Cman
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