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It was my pleasure chating with you escher, now it’s time to sleep (+1 UTC here LOL)

Every spin is a new event. The odds do not change period. Someone once said you would never get this sequence. 1.2.3.4.5.6.7.8.9 in roulette. Why not? It has the same odds of coming up as any 9 number sequence. Playing roulette blindly….well yes, it is not beatable.

Dude!
If we put 100$ on the table each,
You say that number 5 will come10 times in a row!
I say that number 5 will come at least once.

Hwo has better oods in this bet? You or me?
You’re obviously going to lose in a long run!!
Why?

Now plese don’t tell me that we have the same chances in a long run ( not a chance even in a short run)

I didn’t meant so (tough it looks so), what I meant is playing a serie of let’s say 10 spins on same color or whatever, then this formula is used calculate the probability of wining n-th spin (or loosing n spins in a row).
Yes Red in this example have the same probability of coming out every spin, but in probability teory to calculate n-th same event occuring in x-th trials this formula is used.

Consider the following very different situations. To be consistent let us assume we are betting on black each spin.

1. The probability of losing N spins in a row is (19 / 37) ^ N.

2. The probability of losing N-1 spins and then winning the N-th spin is ((19 / 37) ^ (N - 1)) * (18 / 37).

3. Because the probability of winning on any one spin is 18 / 37, the probability of winning on the N-th spin is also 18 / 37.

Probability always refers to the future, the next N spins. So, when we calculate the probability of losing, for example, 10 spins in a row (situation 1 above) the answer we get is for spins 1-10. If we win spin 1 we obviously cannot lose all 10 spins in this series, so that probability now applies to spins 2-11.

Exactly,
Altrought for the sake of future readers two kinds of probability teories should be distinguished to avoid misunderstandings:
Seqential probability and Absolute probability,
calculating probability that number comes 10 times in a row is what is calculated using a sequential probability.
looking at the spin as independent event is what is called absolute probability.

A misunderstanding is obvious if we start mixing those two!!

There are other kind of probability teories as well which are very interesting to study in case of roulette.(however none of these teories will grant you wining at roulette ), it’s just a way of interpretation and calculation.

I am coming to the conclusion that we have not really been disagreeing; rather I did not understand what you were saying originally. Nevertheless, I do have a couple of final comments.

The first one is trivial. What you call “sequential probability” I have always known as “joint probability”, the probability that two or more events will all occur. If these events are independent (e.g. spins at roulette, where the result of one spin has no effect on the result of the next spin) then the joint probability of all events occurring is simply the product of their individual probabilities (what you have called absolute probabilities) multiplied together.

My second comment is a clarification of my closing statements in my previous post, “... when we calculate the probability of losing, for example, 10 spins in a row ... the answer we get is for spins 1-10. If we win spin 1 we obviously cannot lose all 10 spins in this series, so that probability now applies to spins 2-11.” Of course after the result of spin 1 is known the probability we calculated for 10 losses in a row now applies to spins 2-11 regardless of whether we won or lost spin 1.

Hello again,
Your last statement does not make to much sense to me:

The Midnight Skulker - 05 April 2012 01:26 AM

when we calculate the probability of losing, for example, 10 spins in a row ... the answer we get is for spins 1-10. If we win spin 1 we obviously cannot lose all 10 spins in this series, so that probability now applies to spins 2-11.” Of course after the result of spin 1 is known the probability we calculated for 10 losses in a row now applies to spins 2-11 regardless of whether we won or lost spin 1.

If that would be so straightforward as you stated then sequential teory would be somehow useless,
meaning one could calculate probability only for future spins regardles of past outcomes.
This just does not make sense to me.
Why?
If we calculate the probability for 10 spins and we lose 5 of them, then it is quit logical that there is still 5 remaining spins left together with our calculation which still shows probability of last 5 spins, because each spin according to such calculation has it’s own percentage or chances which increases each spin

Here is a quote from one book that deals with roulette and sequntial probability: (John Solitude is the autor)

The good news, the sequntial probability you would not get your bet right diminishes the longer your play the same chances (without ever becoming certainty)

The bad news, we know in front even remote probabilities will show occasionally and you have absolutley no way of knowing for certain, when this will be the case….

And, we might add, no matter how high your probability percentage of a success, there will ALWAYS remain those annoying remote probability digits behind the comma which would represent ‘the streak from hell’, unless you would lure yourself into some kind of false sense of security by rounding up the figures.

In fact, the more spins you play, the higher the probability becomes you will observe a streak of wich the sequential probability figures for it to emerge are very low.

John also stated:

To make distinction between ‘gamblers fallacy’ and ‘sequential probability teory’ matematics are involved…. etc… etc….etc

However this has nothing to to with absolute probability which does not changes, regardles of past outcomes!

Hello again,
Your last statement does not make to much sense to me:

The Midnight Skulker - 05 April 2012 01:26 AM

when we calculate the probability of losing, for example, 10 spins in a row ... the answer we get is for spins 1-10. If we win spin 1 we obviously cannot lose all 10 spins in this series, so that probability now applies to spins 2-11.” Of course after the result of spin 1 is known the probability we calculated for 10 losses in a row now applies to spins 2-11 regardless of whether we won or lost spin 1.

If that would be so straightforward as you stated then sequential teory would be somehow useless,
meaning one could calculate probability only for future spins regardles of past outcomes.
This just does not make sense to me.

First let me clarify the statement which confused you. The probability of losing 10 bets in a row on black is (19 / 37) ^ 10 = 0.001275, a little more than an eighth on a percent. Consequently, when we first approach a table we know that the probability of losing spins 1-10 is .001275. We bet on black, the ball is spun, and it lands on—something. What I was saying is that the probability we calculated before spin 1 (.001275) is now the probability of losing spins 2-11, the next 10 spins, regardless of what the ball landed on. It is the probability that we will lose spins 1-10 that has changed.

codekiddy - 05 April 2012 01:46 PM

If we calculate the probability for 10 spins and we lose 5 of them, then it is quit logical that there is still 5 remaining spins left together with our calculation which still shows probability of last 5 spins, because each spin according to such calculation has it’s own percentage or chances which increases each spin

But the probability of losing the last 5 spins has changed! It is no longer 19 / 37, it is 1; we are certain we lost because the result of the spin is known. (Had we won the probability that we lost would be 0.)

To demonstrate the difference another way let us break our 10 spins into two groups of 5.
Before spin 1 we calculate the probability of losing spins 1-10 as
((19 / 37) ^ 5) * ((19 / 37) ^ 5).
After losing spins 1-5 we calculate the probability of losing spins 1-10 as
(1 ^ 5) * ((19 / 37) ^ 5).
So sequential probability really is useless for using the past to predict the future in a game of random and independent trials such as roulette.

By the way, the title of the John Solitude book from which you quoted is Roulette Fact And Fiction, available for free download from his web site. I could not find where the author tried to show that as an event fails occur in a series of independent trials the probability of it occurring increases. Another by the way, this fallacy is commonly called The Maturity of Chances.

I could not find where the author tried to show that as an event fails occur in a series of independent trials the probability of it occurring increases.

It sais that probability to loose diminishes… but never goes to 0. (as quoted from Sequential probability part of the book, begining from page 50)

Nevermind my friend, at least ve live in democray and have the right to belive and interpret various things in different ways

I could not find where the author tried to show that as an event fails occur in a series of independent trials the probability of it occurring increases.

It sais that probability to loose diminishes… but never goes to 0. (as quoted from Sequential probability part of the book, begining from page 50)

OK, I have read the entire section of the book on sequential probability theory, and I have to agree—sort of. I am not going to quote the book’s entire discussion, only what I feel is the summary of it on pages 61-62. (The question being addressed is whether or not there is an advantage to waiting for a dozen not to show for 15 spins before betting on it.)

The benefit however waiting for the extreme to happen before playing is:

a) A player will play far less spins, most of the time he’ll be observing so he’ll be less exposed to the house edge ... .
b) If we consider the mathematical odds for 15 non appearances of a dozen (after which the player starts betting on the 16th spin), the mathematical odds for a 15 in a row turning in for instance a 30 in a row event, are very remote if we consider the whole sequence.
c) On the downside: although the mathematical odds would be very low for a 15 non appearance of a dozen to turn into for instance a 30 non appearance sequence, it would be wrong to state it could never happen, in fact it WILL happen occasionally and the player is deliberately betting on out of the ordinary streaks which are already closer to turning into an extra ordinary event.

I have to admit I like the footwork. The author never explicitly states that the probability of an event occurring increases as it fails to occur, but (b) certainly implies that it does.

codekiddy - 05 April 2012 08:02 PM

Nevermind my friend, at least ve live in democray and have the right to belive and interpret various things in different ways

In my opinion it is wrong to apply the same probability calculation to a sequence of events once the sequence has started. To be consistent one would also have to consider 15 spins during which a bet won at least once, and then conclude there was still a chance of losing all 15 of those spins. As I showed in my previous post, once the result of a spin is known, the probability that a given bet, whether actually made or not, has won is either 1 if it did win or 0 if it lost. But as you say, to each his own.

OK, I have read the entire section of the book on sequential probability theory…..

You are joking wright? :D
I’m glad you read it dude,

I have to admit that since now I have change my mind in several ways, and why did I change my mind is the fact of pure matematics and knoweledge gained from some books which I finished reading.

I would like to share some of these books with you my friend and other people here (if you like learning a lot as I do of course).
I’m not trying to say anything else then learning and knowing more is good!

These are just some of most interesting about roulette and,
If you want more, or if anyone on this forum (who like roulette matematics and such) wants some specific book/s let me know and I’ll upload it for you :D (These are all free books of course!)

I have to say that after reading one or more of these book you’ll feal that spent time is much more worth then spending time trying systems and similar worthless ilusion crap.
You have to agree that:
Every roulette system has one thing that is the same in all roulette systems, and that is: worthless, lousy, yellow and sucking illusion of wining :D

ay you wait for a dozen not to appear for 15 spins. First of all you might be there most of the day and get 1 or 2 bets in. because of this alone you could never make decent profit. Off course every event is a new event and you might wait the whole day to get 2 losses….. It does not matter how you look at it, roulette is a negative expectancy game. You will loose by the house edge in the longrun.

It was my pleasure chating with you escher, now it’s time to sleep (+1 UTC here LOL)

Every spin is a new event. The odds do not change period. Someone once said you would never get this sequence. 1.2.3.4.5.6.7.8.9 in roulette. Why not? It has the same odds of coming up as any 9 number sequence. Playing roulette blindly….well yes, it is not beatable.

Dude!
If we put 100$ on the table each,
You say that number 5 will come10 times in a row!
I say that number 5 will come at least once.

Hwo has better oods in this bet? You or me?
You’re obviously going to lose in a long run!!
Why?

Now plese don’t tell me that we have the same chances in a long run ( not a chance even in a short run)

Lol…. Why would you have better odds? You would only wager on no. 5 until it comes witin the next 10 spins if it comes at all and then stop. I say it makes no difference if I bet all 10 spins on no.5 if it comes in at all. What if it comes in 3 times in 10 spins? Who made the better bet?

Whether you bet every spin on 1 number, or 10 spins and then wait for whatever reason… There is no difference at all. You will loose in the longrun. A couple of thousand spins I bet we are both 2,7 percent down. That is because you have random 37 numbers that could be spun. Someone said 37 pockets of freedom. And that is what it is. You have no way to beat he odds if you cannot have accurate bets in other words, you need to be able to win more than loose according to the numbers covered and the payout. Betting 5 numbers, if you can win 1/7, you have an edge albeit not a big one. There must be a reason why your bets will be 1/7 or better. No system player can give a reason because there is none. Only visual ballistics and bias have reasons why they win. Those are the facts. Which is better betting blindly 1 number for 35 spins, or betting blindly 35 numbers for 1 spin. Which is the better odds? There is no difference in the longrun.

Chances (in a short run) for one number coming out 10 times in a row are not the same as chances of one number coming out at least once in 10 spins. Isn’it?
Now In a long run The chances for some number coming out 10 times in a row and so multiple times are much lower then chances for some number coming once in each 10 spins and so multiple times. Isn’it.
The abowe 2 statements can be matematiclay prooved.

Can you proove the oposite?
I would like to see that proof.

What if it comes in 3 times in 10 spins? Who made the better bet?

In this case you’ll loose the initial 100$ wager because number 5 didn’t come 10 times but 3 times in a row (while my statement of coming out at least once holds, because it comed out 3 times, so I win 100 bucks)

Chances (in a short run) for one number coming out 10 times in a row are not the same as chances of one number coming out at least once in 10 spins. Isn’it?
Now In a long run The chances for some number coming out 10 times in a row and so multiple times are much lower then chances for some number coming once in each 10 spins and so multiple times. Isn’it.
The abowe 2 statements can be matematiclay prooved.

Can you proove the oposite?
I would like to see that proof.

What if it comes in 3 times in 10 spins? Who made the better bet?

In this case you’ll loose the initial 100$ wager because number 5 didn’t come 10 times but 3 times in a row (while my statement of coming out at least once holds, because it comed out 3 times, so I win 100 bucks)

Kinid regards.

lol, That is a ridiculous arguement. Next you are going to tell me that it is impossible for 1.2.3.4.5.6.7.8.9.10. to fall in that order right?

The thing is you are warping my words and twisting it to try and make a point. What I’m telling you is that every spins is a new event and that all 37 numbers are in play. There is no way in hell that you will ever gain an advantage because you won and then stopped and waited for whatever reason. Tell me, what makes the 1st and the 20th spins any different from each other?

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