Random Thoughts on Roulette

[fergus lee]

My point,Kasino King,was that bettors of two adjacent dozens would get better value by using one of the even-money outside bets But feel free to think that there is no difference between losing £40 rather than £25 .
Your" simpler " suggestion is not similar to mine but a bastardisation of it.Your idea of betting eighteen reds on the layout is precisely what I was warning against !
I accept that my ideas require some thought but they are intended mainly for those punters who already know something about the roulette layout and the odds on offer.
I have already started on a book of ideas on roulette betting which will give examples of my ideas such as ( wait for it !) the previous winning number.I may need to self-publish as some people regard me as a delusionist !

[papeter]

I am new to this forum and have to say that I make a bloody good second income on the outside bets. You can't beat the system but you can make a good living off of it. I just bet on red and black. I use discipline, have a stop loss in place and above all I cut my losses early if things are not going my way. Someone showed me a progression system with controlled risk some years ago and that's what I have been using.

I never ever bet using online RNG casino roulette wheels because they are cleverly programmed to know when you are on a winning streak. Apart from the British controlled sites the ones based in offshore tax havens I would leave well alone. I only ever use live online european roulette wheel casinos.


papeter

[fergus lee]

GLAD TO hear that you are winning at roulette,Papeter,but your post seems a bit confusing.How can you claim that roulette can't be beaten and also that you have beaten it?
Bear in mind that no one can fairly claim that they have won at gambling unless they have won more than they are prepared tp lose.e.g. a Bettig Bank.
Any system can win sometime and "Progressive" systems win lots of times but they involve horrific losses when they hit a losing streak.Unless your system wins using level stakes it is unlikely to win using progressive stakes.The main problem is that no one can forecast with certainty WHEN a winner will occur.
"Stop LOsses"are for losers f0r the same reason , you might be stopping just before a winning streak.Hindsight is a wonderful thing.if only we had it!

That said, carry on betting and good luck!
We are all entitled to our opinion and to put our money where our mouths are.
Just remember one thing."Systems "are ten a penny but cost pounds when they hit the buffers.The recurring crises in the financial markets are caused by " clever clog" systems like yours. Do bear that in mind .

[fergus lee]

In my last post I mentioned that I was writing a book on roulette ideas.Since the book may never be finished never mind published I will give you a flavour of what it will be like.(Crap?)
Using the last winning number I introduce two fictitious mates Andy and Bob.Andy always bets all of the RED/ODD and BLACK/EVEN numbers while BOb always bsts the same column and dozen of the last winning number.
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 so will require them to bet no miore than ten numbers between them to tilt the odds in their favour.They soon suss that to have this Edge then when any of the four numbers
14/17/20/23 is the last winning number then they should only bet the numbers
2-5-8-26-19-20-21-22-23-24 and similarly if the last winning number is one of the four
15/18/21/24 then they should bet only 3-6-9-27-19-20-21-22-23-24
And if 25/28/31/34 then they should bet only 1-4-7-10-19-22-25-26-27-28
26/29/32/35 then they should bet only 2-5-8-20-23-25-26-27-28
27/30/33/36 then they should bet only 3/6/9/21/24/25/26/27/28

Using a different calculation to bet the other numbers this gives
1/4/7/10 then bet 20-21-23-24-26-27
2/5/8/11 then bet 19-22-25-28-21-24-27
3/6/9/12 then bet 19-20-22-23-25-26-28 and if
13/16/19/22 then bet 2-3-5-6-8-9-26-27

And that's it.Bet level stakes ONLY thst is one chip only on each number!

DISCLAMER
I don't claim that this is a winning system only that it has an edge. It is the mathematicians who claim that if you have an edge then you will win " In the long run ".
Feedback would be appreciated particularly if my calculations are wrong.I am neither a "brainbox" or a mathematician. VALUE BETTORS will no doubt, prefer to bet when only six or seven numbers are indicated.Checking 1274 bets showed a profit of 376 chips with the sixes showing a better profit.A Betting Bank of 1274 chips is indicated to get you over the long losing run.

[fergus lee]

People who talk to themselves end up in the funny farm so this will be my last post in this thread
Anyway, I think I have shown that mathematicians are wrong in claiming that you MUST lose when playing roulette. It depends on your pespective.Being smugly dismissive of roulette players and calling them "delusional " is an unwarranted accusatoin.I think we are owed an apology.
'BYE 'BYE

[Dice Man]

Originally Posted by fergus lee

.I think we are owed an apology.

Sorry. I only look in this forum once every few months because it is usually full of idiotic babble. Good to see some numbers finally being introduced to back up somebody's claims here.

However, I don't see anything like a conclusive argument in your favour.

If Andy and Bob were playing independently of each other than they would each have a 20/37 chance of winning each time. But if they're playing together then the probability of both of them winning definitely isn't 400/1369, because there are only a few numbers that can come up for them both to hit. I'm using this diagram to get my info. You might be able to memorise all this stuff but I've never even played roulette so I don't. I think this is the same as the English version except we have no 00, which I will also ignore.



Here are their strategies. Andy is always betting the following numbers:

Andy- "Blackevens and redodds":
1,2,3,4,5,6,7,8,9,10,12,20,21,22,23,24,25,26,27,28

Bob is betting on the column and dozen of the last winning number. As each number is random he has a 1/9 chance of playing each of the following strategies each turn (although what does he do if the last number was 0? I'm assuming he bets the same dozen and column as the last non-0 winner then).

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

The numbers on the left are the ones he bets on, each 1/9 of the time, and the ones on the right after the dashes are the total numbers shared by both his and Andy's choices. I'm not entirely sure I have counted them all correctly, but even one error won't make much difference in the end run I don't think. It does look a bit weird with there being 2 x 10 and 1 x 11. It would be more mathematically harmonious with 3 x 10, but then roulette is a strange thing indeed, as if designed by God to appear beatable but not really be beatable.

So we can work out the probability of them both winning as follows:

P(both win) = (3/9 x 14/37) + (3/9 x 9/37) + (2/9 x 10/37) + (1/9 x 11/37)

= 42/333 + 27/333 + 20/333 + 11/333

=100/333.

OK now I've done that it appears that that is actually a slightly higher probability than you thought it was, which goes in your favour. But that seems to have been just a waste of my time because my main issue was with your inference from this probability to there being an edge.

Ok so I propose to do the following: work out the probabilities that just one of them wins and the probabilities that neither of them win, and look at the actual expected value of this strategy (which I propose is exactly the same as any roulette system, i.e. negative.

So here's my method. Take the first strategy that Bob will be using (the first one in my list)

1,2,3,4,5,6,7,8,9,10,11,12,13,16,19,22,25,28,31,34----14

They have 14 numbers matching. Therefore there are 6 numbers they each have uniquely, and therefore 12 more numbers where only one of them wins. So now like I did before I'll calculate how likely it is only one of them wins:

P(1 wins) = (3/9 x 12/37) + (3/9 x 22/37) + (2/9 x 20/37) + (1/9 x 18/37)

= 36/333 + 66/333 + 40/333 + 18/333

= 160/333

This means that they should both lose 73 times out of 333.

So, now let's consider it in terms of expected value. They're both making 20 bets a turn, so let's say they each bet £1 on each number. If any of their numbers win they receive £36, and they're betting £20, between them £40 a turn, so we need the strategy to be worth more on average than £40 for it to be profitable.

Expected Value = (100/333 x 72) + (160/333 x 36).....+[73/333 x 0]

I put the last bit in brackets because it doesn't add anything to their expected value, it's just there for completeness.

Expected Value = 7200/333 + 5760/333

= 12960/333

= £38.92

So if my calculations are correct it is a losing strategy. I'm now going to masturbate my own ego and show that it's exactly as profitable (i.e. not profitable) as any roulette strategy.

Suppose we have one person betting the whole £40 on any random number (or column, dozen etc, always getting the same odds), they have a 1/37 chance of winning £1440.

Expected Value = 1/37 x 1440

= £38.92

It's actually exactly the same number as 37 x 9 = 333 and 1440 x 9 = 12960.

Therefore your strategy does not work. It may appear to work because over 3/4 of the time Andy and Bob will be raking some prize and only losing outright 1/4 of the time. However nearly half the time (and this represents 2/3 of the time they are "winning") they rake in £36 having bet £40, so they lose money even when one of them wins, even if it seems like they're doing well because they're being paid. The person who bets £40 on one number is going to appear to lose more because they very rarely win anything. If Andy and Bob sat at a roulette table with £4000, after 100 turns we would expect them to have £3892 left, the same as we would expect the £40-bettor. The difference is that the £40-bettor has very high variance. In the 100 turns he'll probably cash in 0,1,2 or 3 times, and it isn't that unlikely for him to lose the whole £4000 (6.5% of the time), whereas it'd be practically impossible for Andy and Bob to lose the whole lot (0.00000000000000000000000000000000000000000000000 000000000000000001224% of the time).

The thing with roulette is, regular, bet-on-one-number-strategy roulette has a high variance and that makes it better for the player if anything. The guy who bets £40 on one number a turn for 100 goes, I suspect (total estimate here), will win some money about 30% of the time I imagine, which is when he gets lucky. But with your system I reckon Andy and Bob will win less than that because they're basically giving away a tiny amount every turn. You would have to be insanely lucky to win with this system over about 10,000 goes, whereas the £40-1-bettor wouldn't have to be as lucky I don't think (still very lucky). I'm not sure about anything I said in this last paragraph though. But the rest of the post is basically a big QED against your system.

Therefore in conclusion:

[Steve A]

Thanks to the Dice Man for his in-depth calculations and explanation, but there is a simpler, common sense way of saying what he says. I'm sure you'd all agree that every individual bet on a roulette table gives the house an edge, yes? In other words, any individual bet is -EV. That being accepted, how can any combination of bets be +EV? Multiple negative values never add up to a positive value. I suspect the best you can hope to do is reduce your variance as much as possible and get lucky now and then!

[Dice Man]

Yeah I agree that every bet on a roulette table is necessarily -EV. I did all these calculations here because I basically couldn't see where fergus lee's conclusion was coming from, so I went through his whole system and explained where the EV is in it as a whole.

I agree with his workings up to this bit,

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 so will require them to bet no miore than ten numbers between them to tilt the odds in their favour.They soon suss that to have this Edge then when any of the four numbers
14/17/20/23 is the last winning number then they should only bet the numbers

But WTF is all that about? Just makes no sense at all.

[DannyC]

Thank god someone else has wrote a bible of a reply instead of myself, that brings a smile to my face. Dicey, you are getting more and more points in my book everyday sweet thing!

Fergus Lee brings a fine example of 'delusion' to the forum.

[geoff365]

The game might be a negative expectancy but it is not by any means equal.