Random Thoughts on Roulette

[fergus lee]

I hadn't intended posting again until there was a sensible discussion but the masturbating Dice Man's distortion of my position needs answering.All the more so since Steve A and the usually cynical Danny C have bought it hook,line and sinker!I'll leave Danny C to wallow in his cesspit of cynicism and answer Dice Man.
Your analysis deals with the position of Andy and Bob before,unlike you , they decide to think more deeply about the problem.For example ,your first scenario where they have 14 numbers in common is only bet when any one of the 4 numbers ! OR 4 OR 7 OR 10 is the previous winning number.if you reead my post you will find that the numbers given to bet on the next spin are 20-23-26-21-24-and 27 and not the 20 numbers you claim.Listen to the argument and not your prejudice.
I can only think that you confuse Probability with the Expectation.Calculate theExpectation and you will find that it is,according to the rules of Probability Theory ,iactually a winning system.And for free on this site.The calculation for the first set is 400/1369 and for the second (which includes 1-4-7-10 ) is 20x18 over 37x37.
Gambling is about betting on a future event the outcome of which is UNCERTAIN Probability Theory is about UNCERTAINTY the result of each spin is UNCERTAIN so why do people claim CERTAINTY where there is only uncertainty ?Saying that you are likely to lose betting on roulette is one thing ,claiming that you MUST lose is unjustified .I think mathemaicians realise this which is why they are frightened to tell us how long their short run or long run is.It is only theory.Probabillity THeory only tells us what SHOULD happen and Not what WILL happen.
I think your Trashcat owes me an apology.

[DannyC]

Fergus Lee after continous debate on the topic I can now only result in listening to who else wants to take the time to answer your replies, as not many people seem to be listening to mine.

Roulette is Random, some people (mention no names) are 'deluded' into thinking there are systems, because they have seen a couple of numbers reappear or a certain colour always coming out... and THAT is exactly what the casino's want to hear. So you may continue ever deluding yourself and other people, that roulette is a game which can be beaten, and its no good me even asking you to PROVE or SHOW me these systems, because I have analysed the game to its bare bones, and its unpredictably random.

Now go to the casino tonight, play off your £50 you cashed in at the post office, and let me know whether you win or not!

[Steve A]

...or better yet, send us a postcard from Hawaiii taunting us all after your system pays off big time. Bon chance!

[Dice Man]

OK right, here is my response. Your post (#24) isn't clear enough. Here is how I interpret it. Tell me what's wrong with this interpretation (which I'm forced to make because, as mentioned, it is too unclear).

They thus each have a20/37 chance.The chance that they will each bet the winning number on any one spin is20/37 multiplied by 20/37 which is 400/1369...

Moreorless agreed. I have the number at 100/333, slightly higher than yours. But it is pretty minor and in your favour, then you say (in the same sentence)...

...so will require them to bet no miore than ten numbers between them to tilt the odds in their favour.

What is this? How does this follow? I think there's probably a couple of chapters of your book missing between the two halves of this sentence.

So now whenever some numbers appear, they change their strategy overall? You seem to be saying that they now have a new strategy which goes like this:

If 14/17/20/23 was last then they bet on 2-5-8-26-19-20-21-22-23-24
If 15/18/21/24 was last then they bet on 3-6-9-27-19-20-21-22-23-24
If 25/28/31/34 was last then they bet on 1-4-7-10-19-22-25-26-27-28
If 26/29/32/35 was last then they bet on 2-5-8-20-23-25-26-27-28
If 27/30/33/36 was last then they bet on 3/6/9/21/24/25/26/27/28
If 1/4/7/10 was last then they bet on 20-21-23-24-26-27
If 2/5/8/11 was last then they bet on 19-22-25-28-21-24-27
If 3/6/9/12 was last then they bet on 19-20-22-23-25-26-28
If 13/16/19/22 was last then they bet on 2-3-5-6-8-9-26-27

OK so if I now understand correctly, what has happened is that Andy and Bob came up with the theory I initially investigated, then from that they worked out this new theory, and now one of them goes home while the other one bets on this new strategy. I haven't worked out how you managed to go from one to the other but let's investigate the second betting system:

Surely this one os more obviously ridiculous than the first one?

Each of those numbers on the left (the antecedents) have a 1/9 chance of coming up on the previous spin. Each of the numbers they bet on will pay out x36 and has a 1/37 chance of coming up. 3/9 of the time they bet on 10 numbers, 2/9 of the time they bet on 9 numbers, 1/9 on 8 numbers, 2/9 on 7 numbers and 1/9 on 6 numbers.


It seems a little unnecessary to even do this calculation but here it is:

EV = (3/9 x 10/37 x 36/10) + (2/9 x 9/37 x 36/9) + (1/9 x 8/37 x 36/8 ) + (2/9 x 7/37 x 36/7) + (1/9 x 6/37 x 36/6)

EV = 1080/3330 + 648/2997 + 288/2664 + 504/2331 + 216/1998

EV = 0.324324 + 0.216216 + 0.108108 + 0.216216 + 0.108108

EV = 0.973

There's that magic number again. If we were betting £40 a time then we would be turning each bet into 0.973 x 40 =

= £38.92

So this system is exactly as profitable as the last system and any system.

The calculation for the first set is 400/1369 and for the second (which includes 1-4-7-10 ) is 20x18 over 37x37.

I'd like to see how you got that...it looks like utter rubbish.

Saying that you are likely to lose betting on roulette is one thing ,claiming that you MUST lose is unjustified .I think mathemaicians realise this which is why they are frightened to tell us how long their short run or long run is.It is only theory.Probabillity THeory only tells us what SHOULD happen and Not what WILL happen.

Nobody is saying that any one person or any one spin must lose. Only that any system would lose if tested for infinite spins. You might think that that doesn't mean anything really but it definitely does manifest itself in quite a short period of spins. If you bet £1 on 1000 spins you would need to get the right number at least 28 times to break even or make a profit. We can use the binomial distribution to work out the probability that you break even or make any money to be 45.1%. (I used a binomial calculator because I couldn't be bothered to do the working myself for this). Over 10,000 spins you need to win 278 times to break even and will only do so 32.5% of the time. So it's not like anybody is claiming that you can't win at roulette because nearly a third will show a profit over 10,000 spins. (Only 7.2% will over 100,000 spins though) Because it'd be pretty difficult to play so many trials you're bound to across nearly as many people who have won on roulette than people who have lost. Plus the people who win are likely to be more vocal so we'll encounter more people who say they've won that people who say they've lost (add some more who really have lost but don't realise it) .

[Steve A]

Some will win and some will lose, that is a statistical certainty, but wholly unpredictable as to who will be the winners or losers. The whole point is that you cannot devise a betting system that alters the house edge - it is mathematically impossible unless you choose to believe the patently nonsensical approach that allows for one spin of the wheel effecting the outcome of later spins. If you are the one out of several billion people who has managed to do this, within a year you will own the planet. Call me then and crow about it, but otherwise stop deluding yourself.

[fergus lee]

Oh,Dear, I seem to have upset Danny C and Steve A.Their posts reveal themselves to be fundamentalists,people who are so certain that theirs is the only "Revealed Truth " and so utterly convinced that they are right that they are not even prepared to even listen to other views.

I accept ,Dice Man , that I was unclear.I didn't appreciate that others may not be familiar with the roulette layout.Your listing of my numbers is correct.You correctly deduce that if any of the first numbers is the last winning number then you should bet on the next spin only the numbers which follow them on the same line.

Again, I didn't make it entirely clear as to how these were deduced.Andy and Bob bet twenty numbers each and then realised that this was a losing proposition and decided to bet only those numbers which their bets had in common.I think.though that I did make clear that they only bet the numbers which you have now listed and not your forty numbers.

If you separate the first set of five rows from the bottom four you may better understand the change of plan,the reasoning for which follows.For this we must go back to their original bets of twenty numbers each.

Using Probability Theory we calculate the chance of them both wnning on the same spin as 20/37 multiplied by 20/37 which is 400/1369. This, however , is only the probability and not the mathematical EXPECTATION which will determine whether they are likely to win or lose.so we calculate further.
We multiply their expected 400 wins by the "payback" of 36 chips which is 14400 and then multiply the number of bets 1369 by (say ) ten and this is 13690
Deducting the bottom line -the amount of chips we bet 13690 from the top line of 14400 -the amount we "won" gives a balance in our favour of 440 chips. So this is ,mathematically speaking, a winning method,in the long run of course !
But Andy and Bob found that this only worked for the firt five rows and was a lossmaker for the other four rows and so had to devise an alternative calculation for these.
It was decided they would still use the Red/ODD and BLack/Even numbers but , excluding the zero, bet only those numbers which were NOT in the same Column or Dozen.This gives 16 numbers, and we do a slightly different calculation than the previous one.This is.
20/37 multiplied by 16/37 which gives 320 / 1369.
If we assume a bet of 8 chips on these
We "receive " 36 chips for each of our EXPECTED 320 wins which totals 11520
We deduct from this our 1369 bets at,say, 8 chips which total 10952 which g ives a profit of 568. The margin in our favour I calculate,in both instances to be 5.18%
Despite the criticism, I maintain that this is a mathematical;y valid method.It does, however,depend on my understanding of Probability THeory and EXPECTATION being correct and I have always asked for someone to show me were I have erred.
I hope this makes things clearer!

[Steve A]

I'm sorry you feel that way Fergus, but it is mathematically impossible to alter the house edge on a roulette table. I feel sure that Dice will be able to point out the erroneous assumption you have made somewhere in your maths (as there is no doubt in my mind you have done) but if you choose to keep chasing this particular rainbow it's entirely your prerogative. Better minds than yours have done so over the past 150 years (some have wasted entire lives on it!) with just about as much success. There is no winning mathematical roulette system. Period. If facts like that don't fit in with your view of the world, you'll still probably be better off looking for fairy gold. I'm not a narrow minded or blinkered person myself, despite what you may think - I just hate to see people wasting their time deluding themselves. I'll not post in this thread again as I've said my piece. Feel free to ignore me as I'm sure you probably will, and good luck - you'll need it!

[Dice Man]

OK then. Let me see if I have this straight. We have two systems: the first, Theory 1, where Andy and Bob play at the same time two different strategies. You say that this system is profitable, but that from this we can deduce a more profitable system, Theory 2.

So first I’ll look at your defence of Theory 1.

You say that the probability of both Andy and Bob winning on any one spin is 20/37 x 20/37. I worked it out to be 100/333 a few posts above which is almost exactly the same number. However, I think in principle your calculation should be correct so I may have got something wrong in my calculation. So let’s work with your number on that.

So they both win 400 out of 1369 games, so we’ll suppose they actually play 1369 games and win 400 of them.

We multiply their expected 400 wins by the "payback" of 36 chips which is 14400 and then multiply the number of bets 1369 by (say ) ten and this is 13690
Deducting the bottom line -the amount of chips we bet 13690 from the top line of 14400 -the amount we "won" gives a balance in our favour of 440 chips. So this is ,mathematically speaking, a winning method,in the long run of course !

This is where we have a problem. There are a few things to say about it.

1. When they both win they make 40 bets and they are awarded 72 bets back. So when they win they make +32.
2. You don’t even account for the times when just one of them wins. In that case they make 40 bets and receive 36 back, making it -4.
3. The bit where you say, “and then multiply the number of bets 1369 by (say) ten and this is 13690”. There is no basis for this at all. You even write “say” as if you realise that this number is completely arbitrary and unrelated to anything. Why not, (say) nine and deduct 9 x 1369 from 14400 to get 2079 to make it an even more profitable system? Where does that 10 come from?

This is the method you should use to really be working out the edge:

1. Work out how often they both win, how often they both lose and how often one of them wins. I did this a few posts back and worked it out as 100/333, 160/333 and 73/333 respectively.
2. What happens when they both win, one wins and both lose? I just did this above and worked it out as +32, -4 and -40.
3. What is their total edge? We can work this out as (100 x 32) – (4 x 160) – (40 x 73) all over 333. Which is (3200 – 640 – 2920)/333, which is -360/333. Every time they bet £40 between them they expect to lose £1.08. That as a % of their £40 gives their strategy an edge of 0.973, and means that their £40 will become an expected £38.92, the same number that keeps on appearing in this thread.

But Andy and Bob found that this only worked for the firt five rows and was a lossmaker for the other four rows and so had to devise an alternative calculation for these.

I’m not sure how they discovered this. Their strategy fails everywhere, not just because of a few numbers.

It was decided they would still use the Red/ODD and BLack/Even numbers but , excluding the zero, bet only those numbers which were NOT in the same Column or Dozen.This gives 16 numbers, and we do a slightly different calculation than the previous one.This is.
20/37 multiplied by 16/37 which gives 320 / 1369.
If we assume a bet of 8 chips on these
We "receive " 36 chips for each of our EXPECTED 320 wins which totals 11520
We deduct from this our 1369 bets at,say, 8 chips which total 10952 which g ives a profit of 568. The margin in our favour I calculate,in both instances to be 5.18%

Ah OK. Now I think I understand Theory 2. Andy carries on doing what he was doing before, and Bob does the opposite of what he was previously doing. So now they are making 36 bets a turn.

OK, now I realise what your “10” referred to above. 10 was the number of bets they made on each number right? And now in Theory 2 they are betting 8 times on each number. Does it not occur to you that if they bet 8 chips on each number and they both win then they had 16 bets on the winning number and they get 576 chips back? And also that they paid 288 chips in the first place?

Am I right to think that, according to you, if they were to bet 10 chips on each number in Theory 2, then the system would be unprofitable because we would deduct from 11520 the figure of 10 x 1369 which would be a total of -2170? If that’s the case then your position appears to be that they expect to make more money by betting as few chips as possible on each number. You’re basically only using the figure of 8 or 10 (the number of chips bet) to work out their losses, not their profits, which ought to be directly proportional to how much you bet on the number.

Despite the criticism, I maintain that this is a mathematical;y valid method.It does, however,depend on my understanding of Probability THeory and EXPECTATION being correct and I have always asked for someone to show me were I have erred.

Seriously, your understanding of probability theory seems to be badly non-existant. I mean, it just seems like you’ve got all the right numbers, and you’ve just mashed them together in a way that results in a positive number at the end. If you’re betting 8 chips on each number then why would you only receive 36 back whenever two players both win on the same number?

I mean, it’s just terrible. You say at the end that Theory 2 has an edge of 5.18%. I wondered where this came from and I managed to produce that number by just randomly dividing some of your numbers by others. You got that number by dividing 568 by 10952, but the answer is actually 0.05186, so you didn’t even round it up correctly to 5.19%.

The margin in our favour I calculate,in both instances to be 5.18%

Do you mean, in both theories? If we use the exact same method you used in Theory 2 to calculate the edge in Theory 1 we get 440/13690 which is an “edge” of 3.21%. But according to your method, if you bet 9 chips per number in Theory 1 instead of 10 chips, then your edge would be (14400 – [9 x 1369]) / (9 x 1369) which would give it an edge of 16.9%. Doesn’t it sound just absurd that your system becomes more and less profitable depending on how many chips you bet on each number? Particularly that it becomes less profitable the more you bet. Your whole theory is just a total misapplication of mathematics.

[DannyC]

I dont need to write posts as big as dicey's just to tell you that your devised 'system' derails itself recursively.

[Steve A]

Despite all I've said in this thread, I have now found an instance where the odds in roulette are in your favour, rather than the house.
Roulette Edge in YOUR favour
I still think it'd be difficult to make any money, but it just goes to show that's anything's possible!