Law of Averages/Stats

Hobart

I am working on a betting theory. Can anybody explain to me, in simple english terms, how the law of averages works. For example. A roulette whell has 36 numbers and 2 colours* . Am I correct in saying that, in relation to the law of averages, over a period of 20 spins that each colour should come up an even amount of times? i.e. Red 10 and Black 10?






*I am purposely not including 0 and its colour in this example.

davros

Originally posted by Hobart
Am I correct in saying that, in relation to the law of averages, over a period of 20 spins that each colour should come up an even amount of times? i.e. Red 10 and Black 10?

Not over 20 spins, no. The law of averages tells you that the more spins you make, the closer your split will come to 50-50. We are talking about a lot of spins here, i.e. tending towards infinity. The law says nothing about short runs.

Capt'n Midnight

On average it should balance out on a fair wheel. Also the where the next ball lands is not influenced by where it landed the last time.

ie. it's 32:1 that it lands on the green and the house gets their 3%, the chances of it landing on the green twice in a row is 1024:1. But if it has already landed on the green it's now only 32:1 to land on it again.

It's 1024:1* that a fair coin will end up heads 10 times in a row. But it's still 50:50 on then next throw.

*easier to compare with the smaller 32 number wheel

Some people here could work out the odds of the wheel balancing out. But the easiest way to say it is that if you are on the very last spin and there is one more white than black then, the next ball will still be random. If you work out the odds you will find that having the sequence converge at a specific spin to be very low. During the series of spins the numbers should match on a couple of occsions (guess - not having worked out the odds).

Really like saying if you have a group of people what are the chances of them being 50:50 male/female ?

dod

Roulette contenders hardly nettle
Casinos when they test their mettle,
Foretelling where the ball will settle.

davros

Originally posted by Capt'n Midnight
Some people here could work out the odds of the wheel balancing out.

I fancy a crack at that. Someone correct me if I'm wrong:

20 spins, what is the prob of exactly 10 reds and 10 blacks?

Number of possible outcomes is 2^20 = 1,048,576

Number of ways of arranging 10 black spins and 10 red spins is 20!/(10!.10!) = 184,756

So, prob of exactly 10 reds and 10 blacks in 20 spins is 184,756/1,048,576 = 0.176

Capt'n Midnight

Found this
http://www.stp.dias.ie/APG/coin_tossing.html

If the coin is fair, every outcomes is equally likely, and so the probablility of getting r heads is nCr /2n.
in the above case n=20 r=10

The applet is best viewed with Netscape 4 (Unix or Windows), several other browsers will not display the applet

Hobart

Originally posted by Capt'n Midnight
On average it should balance out on a fair wheel. Also the where the next ball lands is not influenced by where it landed the last time.

ie. it's 32:1 that it lands on the green and the house gets their 3%, the chances of it landing on the green twice in a row is 1024:1. But if it has already landed on the green it's now only 32:1 to land on it again.

It's 1024:1* that a fair coin will end up heads 10 times in a row. But it's still 50:50 on then next throw.

*easier to compare with the smaller 32 number wheel

Some people here could work out the odds of the wheel balancing out. But the easiest way to say it is that if you are on the very last spin and there is one more white than black then, the next ball will still be random. If you work out the odds you will find that having the sequence converge at a specific spin to be very low. During the series of spins the numbers should match on a couple of occsions (guess - not having worked out the odds).

Really like saying if you have a group of people what are the chances of them being 50:50 male/female ?

Ok. I accept, and actually understand that. I mean it makes sense that every spin has an equal chance if the wheel is fair. So the last spin should have no influence on the next, right? Why then , do we have a law of averages? Or am I missing something obvious here?

seamus

Originally posted by Hobart
Ok. I accept, and actually understand that. I mean it makes sense that every spin has an equal chance if the wheel is fair. So the last spin should have no influence on the next, right? Why then , do we have a law of averages? Or am I missing something obvious here?

The law of averages is more of a mathematical concept. It uses things like infinity to get the point across.

It may seem obvious that flipping a coin, the same no. of heads and tails should come up, but in practice, they don't.

It basically tries to assert that although you may flip the coin ten times, and heads comes up each time, in fact there is still exactly the same chance of tails coming up for each throw.

To put it another way, let's say you're trying to figure out, using quantitative data, the likelihood of a heads coming up. Let's assume that you have no idea what it is beforehand.

You throw it ten times, and get 7 heads and 3 tails. You may then think that the chance of getting a heads is 0.7. Let's say you continue on and do 100 throws, coming up with 45 heads and 55 tails. You may now think that the odds of getting a head are 0.45.
Continue on more and do 1000 throws. Let's say you get 507 heads and 493 tails. Now you may think the odds of getting a heads are 0.507.

But basing mathematical proofs or concepts on quanitative data can be misleading. The law of averages simply lets us state that given an infinite no. of throws, our data would then tell us that the odds of getting a heads is 0.5.
Put simply, its a nice simple and convenient way of telling ourselves, that for x unweighed possibilites, given an infinite no. of throws/rolls/spins/whatevers, then 1/x of our infinite data will be x, therefore the possibility of x coming up is 1/x.

It may seem obvious, but sometimes the obvious things need to be stated

Hobart

Thanks Seamus. That seems quite easy to understand.

DeVore

all the law of averages really says about this sort of situation is that 10/10 Red/Black is the *most likely outcome* of all possible outcomes (this is often called the "Mode outcome")

It doesnt mean its likely, it doesnt mean you should expect it but statistically as your number of attempts approaches infinity, by the law of averages deviation from this outcome becomes statistically less likely.

Ignore what CM is saying, while he's correct he's getting confused and confusing you in the process.

Given 20 trys the likely hood is VERY HIGH that you will not get 10 red and 10 blue.
In fact the chance of significant deviation from the norm is large and that deviation will be considerable when treated as a fraction of the number of attempts.

Given 2000000000000000000000000000000000 tries, you still arent certain of getting 1000000000000000000000000000000000 red and 1000000000000000000000000000000000 blacks but the variation from that as a *proportion* of the attempts will be tiny.


DeV.

Hobart

Yea I accept that. What I am "trying" to do is to come up with a betting model wherby I cannot loose. Probably impossible but it goes something like this.

In the following example I intend to win €1

1) *** Most importantly *** you must decide befoe you bet how much you want to win. When you win this amount you must stop betting for that session.

2) I place a bet of €1 on black on a roulette wheel.

3) It comes up black. I win €1 and I stop betting

or

3) It comes up red and I must now place a bet of €2 on black to supplement the €1 I have lost in the first bet and win €1.... and so on and so on....

Now the "catch" with this method is that, in theory, you must have an infinate amount of money. Also you could face a situation wherby you would be placing a €1000 bet for a €1 gain.

However If I had an infinate amount of money is there any way I could loose?

ecksor

Argh, you can't have an infinite amount of anything. An 'unlimited supply' of money would be a better term.

seamus

Originally posted by Hobart
However If I had an infinate amount of money is there any way I could loose?

No, but there's nothing to say you might necessarily win either.

If you had an unlimited supply of cash, let's say €x, and kept betting ad inifinitum, losing each time. By your theory, you would bet €1 the first time, €2 the second time, €3 the third time, etc etc. So as you say, you could find yourself at a point where you're betting €x/2 for a €1 gain. That is, up until this point, you have lost €(x/2 - 1) and are betting just to regain that money, and €1 more, which gives you €x/2. If you lose that bet, then the most you can possibly bet on the next hand is €x/2, which means that you can only regain your losses, and not increase your cash.

The problem with the word unlimited is that as x tends to infinity, then all values tend to infinity. So x/2 == inifinity, as does (x/2 -1). Which means that if you have an umlimited supply of cash, then you can never have more than you have. Infinity has that effect on things

I think gambler's ruin kind of illustrates the point a bit better. Afaik, it's mostly dependent on the fact that even though the law of averages states that if you keep betting on the same thing over and over, you should end up with the same amount of cash as you started with, the nature of averages, and the fluctuations in the amount of times a possibility occurs before it reaches exactly 1/x means that you can never have enough resources to break even. But I could be wrong on that point.

Capt'n Midnight

Originally posted by Hobart
Yea I accept that. What I am "trying" to do is to come up with a betting model wherby I cannot loose. Probably impossible but it goes something like this.

You could loose easily - when the other person is ahead they stop taking bets off you. Also the house gets about 3% - ie everything on the table when the ball lands in the green - virtually impossible to gain 3% on long runs.

Safest thing to do is to find some mug who believes in law of averages and bet against the numbers evening out after n throws. The 0.176 posted earlier suggests as long as you offer odds of 5:1 you should do well in the long run.

If you have infinite/unlimited amounts of money then no-limit poker is your game (just make sure that all the other players have finite amonuts of money.)

Hobart

Originally posted by ecksor
Argh, you can't have an infinite amount of anything.

Numbers?

An 'unlimited supply' of money would be a better term.

Ok, point taken and noted.

Originally posted by seamus
No, but there's nothing to say you might necessarily win either.

Why not? Is there ever a situation whereby a wheel turns up the same colour in, say a situation where the wheel is turned 100 times? I know it is possible, but what are the odds?

If you had an unlimited supply of cash, let's say €x, and kept betting ad inifinitum, losing each time. By your theory, you would bet €1 the first time, €2 the second time, €3 the third time, etc etc.

Well actually €4 for the 3rd, €8 for the 4th etc.....

So as you say, you could find yourself at a point where you're betting €x/2 for a €1 gain.

I don't think so. And I'm far from a matamatician (and a speller) but I thought I would have had to be betting €*2 for a €1 gain based on the fact that my gain is to be €1. e.g I get to the stage wherby I have lost 3 bets so I have placed €1+€2+€4 (€7) so my next bet has to be €8.

That is, up until this point, you have lost €(x/2 - 1) and are betting just to regain that money, and €1 more, which gives you €x/2. If you lose that bet, then the most you can possibly bet on the next hand is €x/2, which means that you can only regain your losses, and not increase your cash.

You have lost me here seamus.

Capt'n Midnight

I see you have given up on the wheel (and the 1/36 chance of loosing your bet) and gone back to double or quits, in which case the winner is the person who stops betting while they are winning. Unless you can force the other person to keep betting when you are loosing - you loose.

Hobart

Originally posted by Capt'n Midnight
I see you have given up on the wheel (and the 1/36 chance of loosing your bet) and gone back to double or quits, in which case the winner is the person who stops betting while they are winning. Unless you can force the other person to keep betting when you are loosing - you loose.

What are you talking about? Did you read my initial post? I said that, yes there is a wheel, yes there is 36 segments, but that there is only 2 colours. Did you miss that point? I am talking about an evens bet here.

seamus

Originally posted by Hobart
Why not? Is there ever a situation whereby a wheel turns up the same colour in, say a situation where the wheel is turned 100 times? I know it is possible, but what are the odds?

Of course it's possible, but the law of averages says that over an infinite number of bets, with an unlimited supply of money, you have to break even.

Well actually €4 for the 3rd, €8 for the 4th etc.....

Ah yes, I had though of that originally, but I got distracted. That means that you would be unable to guarantee the ability to increase your winnings after y bets, where 2^y >= x/2. That is, the chance exists that after y bets, you begin to win again, and keep winning until you recoup your losses, but there's nothing certain.

Let's look at it this way. You want to find out what your chances are that at some point you will come out on top. That chance will always be 0.5. If you always bet on heads, and heads has a probability of 0.5 of coming up, then what you're saying is that you want heads to appear x times, and tails to also appear x times, where x >= 0, but that you also want heads to appear just that one extra time at the end.

Now, the law of averages implies that flipping the coin 2x times, the probablility of there being x heads and x tails is 1, i.e it is certain. So, then the odds of that happening AND you then getting a heads is (1 x 0.5), which is....0.5.

Which means that no matter how long you bet for, the law of averages implies that at no point is it certain that one possibility will occur more than another (the others).

The notion of "quit while you're ahead" is common wisdom. Practical applications suggest that at some points you will win, and at some points you will lose (provided you bet on the same thing every time), but since neither your time nor your money is unlimited, you are better to quit and change your game or your tactics while you're winning, i.e. before you begin to lose again.

Hobart

Originally posted by seamus
Of course it's possible, but the law of averages says that over an infinite number of bets, with an unlimited supply of money, you have to break even.

Ah yes, I had though of that originally, but I got distracted. That means that you would be unable to guarantee the ability to increase your winnings after y bets, where 2^y >= x/2. That is, the chance exists that after y bets, you begin to win again, and keep winning until you recoup your losses, but there's nothing certain.

Let's look at it this way. You want to find out what your chances are that at some point you will come out on top. That chance will always be 0.5. If you always bet on heads, and heads has a probability of 0.5 of coming up, then what you're saying is that you want heads to appear x times, and tails to also appear x times, where x >= 0, but that you also want heads to appear just that one extra time at the end.

Now, the law of averages implies that flipping the coin 2x times, the probablility of there being x heads and x tails is 1, i.e it is certain. So, then the odds of that happening AND you then getting a heads is (1 x 0.5), which is....0.5.

Which means that no matter how long you bet for, the law of averages implies that at no point is it certain that one possibility will occur more than another (the others).

The notion of "quit while you're ahead" is common wisdom. Practical applications suggest that at some points you will win, and at some points you will lose (provided you bet on the same thing every time), but since neither your time nor your money is unlimited, you are better to quit and change your game or your tactics while you're winning, i.e. before you begin to lose again.

Ah. Ok. That's exactly my point (or the answer I was expecting). So at no point does the previous result have any influence on the next one. So my chances of winnig are always 50/50. No matter what the law of averages says. Correct? BTW the notion of "Quit while your ahead" is central to this "plan". Have a look at point No1 in this post for my thoughts on it. However if I was to fashion a situation wherby my betting encompassed my losses, while keeping my goal of winning €1 a priority and I used the system I outlined above, is there any way I could loose?

DeVore

I think people are confusing the issue.

Hobart starts by betting 1 euro and loses.
Then he bets 2 Euro and loses

Then he bets 4 Euro and loses

Then he best 8 Euro and wins 8 euro, which covers the 7 euro he has lost
already and gives him a euro profit.

Now he returns to betting 1 Euro and the process starts from there having been "guarunteed" a euro profit.

Wash.Rinse.Repeat.

This is meant to be some form of "impossible to lose" system.
Theres no such thing but I am struggling to poke a hole in this one.

Obviously there is no guaruntee that you are ever going to win but from the starting
position you can expect to win within a few attempts at 50/50.

I think the flaw lies in the stake versus the odds but I havent been arsed thinking it through.

Anyone else want to poke holes in this "unbeatable betting sequence"?

DeV.

seamus

Originally posted by DeVore
I think people are confusing the issue.

Hobart starts by betting 1 euro and loses.
Then he bets 2 Euro and loses

Then he bets 4 Euro and loses

Then he best 8 Euro and wins 8 euro, which covers the 7 euro he has lost
already and gives him a euro profit.

Now he returns to betting 1 Euro and the process starts from there having been "guarunteed" a euro profit.

Wash.Rinse.Repeat.

This is meant to be some form of "impossible to lose" system.
Theres no such thing but I am struggling to poke a hole in this one.

Obviously there is no guaruntee that you are ever going to win but from the starting
position you can expect to win within a few attempts at 50/50.

I think the flaw lies in the stake versus the odds but I havent been arsed thinking it through.

Anyone else want to poke holes in this "unbeatable betting sequence"?

DeV.

I see it now yeah. The only hole that can be poked is the unlimited cash one. There is no guarantee that just because the last one came up as possibility x, that this one won't arise as possibility x, and so on and so on.

In theory, it's a risky bet, but using common sense, it would seem logical that you have to win. But after just 10 consecutive lost bets, you would have to place a bet of €1024, just to up yourself by €1. If the stakes get bigger (e.g. to win €100, the stakes get bigger, after 10 bets you could be betting €102,400, just to win €100).

So, yes, with an umlimited supply of cash, it should work. But if you had an unlimited supply of cash, why would you bother betting large sums of money to win measly ones? Unless you have a lot of free time that is.

GuanYin

Just looking at DeVore's post, I think the flaw is in the experimental design.

You are devising a system that combines probability and gambling (which, if I may be very base in describing it, is a subset of statistical analysis). The system makes an assumption, that holds for one aspect (probability) but which is unrealistic from the gambling point of view.

Were such a wheel to exist, the odds you would get would be small, and possably determined by previous spins, thus, in some small way, compensating for the "law of averages".

Hobart

Originally posted by seamus
But if you had an unlimited supply of cash, why would you bother betting large sums of money to win measly ones? Unless you have a lot of free time that is.

Thats a fair point. But say you had a stake of e.g. €10,000. It would be possible to make a profit in a very short time (I believe).

Originally posted by skye
Just looking at DeVore's post, I think the flaw is in the experimental design.

You are devising a system that combines probability and gambling (which, if I may be very base in describing it, is a subset of statistical analysis). The system makes an assumption, that holds for one aspect (probability) but which is unrealistic from the gambling point of view.

Please expand on this flaw. I believe that Devore has hit the nail on the head with his description. Gambling is all about probability. Without that you would not have odds!! Yes it is a subset of stats (hence the title of the thread) and that is why favorites exist in racing.

Originally posted by skye
Were such a wheel to exist, the odds you would get would be small, and possably determined by previous spins, thus, in some small way, compensating for the "law of averages".

You fail to see the wood for the trees. Such a wheel does exist. It's called a roulette wheel. The odds are even, i.e. 1/1 (betting on black or red)

GuanYin

I can see fine thank you. Roulette has green and also numbered segments... the betting works slightly differently from the description here.... when I said "were such a wheel to exist" I meant a game where by you get even odds every time for a 50/50 outcome.

Thinking about it, the flaw is the evens bet. If you were to tackle the problem from a statistical analysis point of view, you wouldn't get evens on the bet, in all likelihood, the longer the game went on one way or another (as in the more loses you got in a row) the worse the odds would get.

If you want to design the experiment so you get evens on a 50/50 bet, you are taking a very biased approach to the statistical side of the gambling, thus the experiment is designed to give the outcome you desire. I don't think you would find anywhere giving you even odds on a win/loss outcome every time.

So what I was trying to refer to is you are focusing heavily on the statistical probability of the outcome, and not at all on how the outcomes effect the odds of the game.

Hobart

Originally posted by syke
I can see fine thank you. Roulette has green and also numbered segments... the betting works slightly differently from the description here.... when I said "were such a wheel to exist" I meant a game where by you get even odds every time for a 50/50 outcome.

Thinking about it, the flaw is the evens bet. If you were to tackle the problem from a statistical analysis point of view, you wouldn't get evens on the bet, in all likelihood, the longer the game went on one way or another (as in the more loses you got in a row) the worse the odds would get.

If you want to design the experiment so you get evens on a 50/50 bet, you are taking a very biased approach to the statistical side of the gambling, thus the experiment is designed to give the outcome you desire. I don't think you would find anywhere giving you even odds on a win/loss outcome every time.

So what I was trying to refer to is you are focusing heavily on the statistical probability of the outcome, and not at all on how the outcomes effect the odds of the game.

I was not enquiring about your ability to see, it was more your lack of understanding. Roulette has 1 green segment, 18 Red segments and 18 black segments. Like ALL segments on the wheel the green segment is numbered. The betting works exactly as outlined here. i.e. You can have a 1/1 bet on either Black or Red coming up. If 0/Green comes up all colour bets are lost and (assuming somebody has bet on 0) a 36/1 payout is won. You get evens odds all the time on a roulette wheel, based on a colour pick.

I am not talking about "structuring" an unrealistic bet here. I am not being biased. The odds of the game are set. The bet is set. Why is it that you seem to be having a difficulty understanding this?

davros

Originally posted by DeVore
Anyone else want to poke holes in this "unbeatable betting sequence"?

All the flaws are in the real world, I think:

1. The green 0
2. The house limit on stakes (which will stop your doubling pretty quickly)
3. The potential need for very deep pockets.

seamus

Originally posted by Hobart
Thats a fair point. But say you had a stake of e.g. €10,000. It would be possible to make a profit in a very short time (I believe).

The idea is sound, yes, and while it may be possible to make a 'profit' in a very short time, making anything close to a significant profit is another story.
The only thing in you favour is :- It's far more likely, (not statistically, just given common sense) given (z = x +y) spins, where x = no. of time red appears, and y = no. of times black appears, that at some point, x = y + 1 (at which point you stop betting and start again) than it is that y > x, as z grows larger.

Your only problem is relying on your €10,000. As I said, after 10 failed bets, your wager is at €1,024. After 14 failed bets, you're at €16,384, and you go home a few thousand euro lighter.
If you came back the next night, breaking even would involve winning, 10,000 times, which, by the laws of averages means a minimum of 10,000 spins of the roulette wheel. And that's a minimum. The chances of you getting 10,000 consecutive blacks is, as my calculator reliably informs me,
5.012372749206452009297555933743e-3011
Good luck with that one.

Seriously, The amount of money you would require to guarantee your success would (probably) be so large that the amount of money you could win in a reasonable amount of time would still be a factor of times smaller than it.

As davros says, in theory, it looks great. In reality, it's not feasible. Bit like living on the moon in that respect

Hobart

Originally posted by seamus
The idea is sound, yes, and while it may be possible to make a 'profit' in a very short time, making anything close to a significant profit is another story.
The only thing in you favour is :- It's far more likely, (not statistically, just given common sense) given (z = x +y) spins, where x = no. of time red appears, and y = no. of times black appears, that at some point, x = y + 1 (at which point you stop betting and start again) than it is that y > x, as z grows larger.

Your only problem is relying on your €10,000. As I said, after 10 failed bets, your wager is at €1,024. After 14 failed bets, you're at €16,384, and you go home a few thousand euro lighter.
If you came back the next night, breaking even would involve winning, 10,000 times, which, by the laws of averages means a minimum of 10,000 spins of the roulette wheel. And that's a minimum. The chances of you getting 10,000 consecutive blacks is, as my calculator reliably informs me,
5.012372749206452009297555933743e-3011
Good luck with that one.

Seriously, The amount of money you would require to guarantee your success would (probably) be so large that the amount of money you could win in a reasonable amount of time would still be a factor of times smaller than it.

As davros says, in theory, it looks great. In reality, it's not feasible. Bit like living on the moon in that respect

Davros makes some good points. However my intention is not to swan into the next casino I come across and start betting huge sums for little gain. The green zero is negligable, in terms of odds, so is not really a factor. The "bottomless" pocket is a theroretical one and the house limit is a misnomer as, if one was serious about taking this on, one could find an establishment willing to suit the pocket.

However I have probably relied too much on the roulette angle. In theory one could apply this method to any sport. For example what about soccer? Manchester United are not going to loose every game in a season. Yes thier odds will reflect that but, based on past performances and current form, they will win more than loose.

Dont forget that the return is in proportion to the wager. Just because you loose €8 in 4 bets, it does not mean that the next time you must have 8 winning spins in a row to get your losses back (if you accept a loss at the initial stage that is). All you need is 1 win of €9 to get your losses and your winning back on track And thats a 50/50 bet (in Roulette red/black).

GuanYin

Originally posted by Hobart
I was not enquiring about your ability to see, it was more your lack of understanding. Roulette has 1 green segment, 18 Red segments and 18 black segments. Like ALL segments on the wheel the green segment is numbered. The betting works exactly as outlined here. i.e. You can have a 1/1 bet on either Black or Red coming up. If 0/Green comes up all colour bets are lost and (assuming somebody has bet on 0) a 36/1 payout is won. You get evens odds all the time on a roulette wheel, based on a colour pick.

I am not talking about "structuring" an unrealistic bet here. I am not being biased. The odds of the game are set. The bet is set. Why is it that you seem to be having a difficulty understanding this?

My bad, I only ever played roulette once and I seem to remember alot of complicated rules on number combinations at weird odds. Didn't see the odd/even or red/black option.

Even still, I'd appreciate you losing the attitude in your post. I wasn't ever being directly insulting to you yet your tone is unappreciated.

Have a look at the Forum rules on posting..

Hobart

Originally posted by syke
My bad, I only ever played roulette once and I seem to remember alot of complicated rules on number combinations at weird odds. Didn't see the odd/even or red/black option.

Even still, I'd appreciate you losing the attitude in your post. I wasn't ever being directly insulting to you yet your tone is unappreciated.

Have a look at the Forum rules on posting..

well..eh.. my apologies if I infringed on the forum rules, it was not my intention, as for the attitude/tone, I don't see it but again apologies if you picked one up, most replies have been informative and it would be remiss of me to pee anybody off here, especially the mod, so again apologies if my "tone" is unacceptable.

As for the game of roulette, you are correct, in that there are many bets one can make. Singles i.e. pick a number and take 36/1. Quarters (bet on any four numbers) at 8/1. And Evens, (as described above).

Capt'n Midnight

Judging by your previous post you should be aware there are 37 slots on the wheel - let me try to rephrase my previous posts - on average everyone looses 1/37th of your stake - even if you win on green the payback is only 36:1 not 37:1.

So the house makes 2.7% of the bets (not winnings) in the long run. So after ten bets the house has made
€56.86 regardless of whether you are ahead or not.

Spin	36/36	37/36	Diff.
0	1	1.028	0.03
1	2	2.056	0.06
2	4	4.111	0.11
3	8	8.222	0.22
4	16	16.444	0.44
5	32	32.889	0.89
6	64	65.778	1.78
7	128	131.556	3.56
8	256	263.111	7.11
9	512	526.22	14.22
10	1024	1052.4	28.44
			56.86

- The 37/36 is the amount you have to bet to get an average return of twice your stake.

========================

Better off betting on the grand national being cancelled three times in a row - or a hole in one during a major golf tourlament. - Can't bet on charts / american wrestling 'cos they are probably rigged.

You need to find an area of betting where there is a lot of "mug money" and thus better odds.

You could try working on odds - ie backing every combination such that you will balance out at the end - but you'd need a lot of maths and lots of hunting for the right event to be able to do this.

Imposter

Originally posted by Hobart
However I have probably relied too much on the roulette angle. In theory one could apply this method to any sport. For example what about soccer? Manchester United are not going to loose every game in a season. Yes thier odds will reflect that but, based on past performances and current form, they will win more than loose.

But here you have to factor in the bookies profit which is far more (per stake) than the casinos. Time is a bigger problem with this too.

However if you ever devise the perfect bet let me know, please.

Hobart

Originally posted by Capt'n Midnight
Judging by your previous post you should be aware there are 37 slots on the wheel - let me try to rephrase my previous posts - on average everyone looses 1/37th of your stake - even if you win on green the payback is only 36:1 not 37:1.

So the house makes 2.7% of the bets (not winnings) in the long run. So after ten bets the house has made
€56.86 regardless of whether you are ahead or not.
Better off betting on the grand national being cancelled three times in a row - or a hole in one during a major golf tourlament. - Can't bet on charts / american wrestling 'cos they are probably rigged.

You need to find an area of betting where there is a lot of "mug money" and thus better odds.

You could try working on odds - ie backing every combination such that you will balance out at the end - but you'd need a lot of maths and lots of hunting for the right event to be able to do this.

Assuming you bet on a single number that is. But if you had a single bet on 18 numbers i.e. black or red, your chances increase.

Originally posted by Imposter
But here you have to factor in the bookies profit which is far more (per stake) than the casinos. Time is a bigger problem with this too.

However if you ever devise the perfect bet let me know, please.

Time is not an issue if you don't make it one. And factoring in the bookies odds is just a calculation.

Ok. I will put my money where my mouth is. Manu play Leeds next in the Premiership. Their next game in the League after that will be Fulham. The odds are not published yet, but, I guarantee that I will win €1 off one of these bets on a Manunited win.

Hobart

Capt'n Midnight

Originally posted by Hobart
Assuming you bet on a single number that is. But if you had a single bet on 18 numbers i.e. black or red, your chances increase.

You put €18 on Black (either €1 on each number or €18 on the colour), You do the same for Red
You put €1 on green (0)

You will win €36 - but the bets cost €37

======================

Re: the Man U bet - what happens if the match is cancelled ?

Hobart

No. Your missing the point again. I put €1 on black at odds of 1/1, there are 18 black and 18 red numbers on a wheel. there is 1 green number (which I am disgarding as the odds of green coming up are 37/1 irrespective of what the house says). This bit is important: I do not bet €18 I bet €1 on a collection 18 numbers at 1/1. Not 18 bets. 1 bet. If any of the 18 black numbers come up I win.


I am not talking about having a single bet on every number. Yes in theory I could bet against a number coming up i.e. put 35 single bets on 35 numbers, hoping that 2 don't come up, but that is not what I am talking about here.



If the match is cancelled the bets are refunded, and I move on to the next match.