Author Topic: Probabilities before and after the event  (Read 5724 times)

kav

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Probabilities before and after the event
« on: July 20, 2016, 02:57:19 PM »
This might be a stupid question, or a mind game or a smart question or it may just help us clarify some things...

Let's say I have a system that uses the Paroli progression (bet more up as you win) on the Even chances (Red Black)
I have considered the chances of 2 consecutive wins is 0,25 or 25% and for 3 consecutive wins is 0,125 or 12,5%.
I have decided that I will stop after 2 consecutive wins and start over.

The question is this:
Since AFTER the 2 consecutive wins the chance for a 3rd win is 50%, now my chance for a series of 3 wins is no longer 12,5%, but 50%. Would it be smart to "change plans" and go for a 3rd win?
« Last Edit: July 20, 2016, 03:47:58 PM by kav »


 

Jesper

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Re: Probabilities before and after the event
« Reply #1 on: July 20, 2016, 03:43:03 PM »
We allways play with the words as well. We have near 50% of double our money at an EC every time we try. This next bet is higher, and have very little to do with the previous bets, they finance the trial, that is the only connection we should see.
 
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Dane

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Re: Probabilities before and after the event
« Reply #2 on: July 20, 2016, 05:06:26 PM »
PLAY WITH WORDS? Well, maybe! Take it rolig (Danish word mening easy).
ROLIG in Swedish is something funny; and we surely is having funny questions here.

In general paroli players might find it hard to take it rolig.
They never know when to stop. Of course the probabiltiy for winning  next TIME
is 0.5 if you have no Zero. Rena rama sanningen. The whole truth. Basic knowledge.

If I choose to play Paroli, I'll bet both RED and EVEN at the same time .
I only double up after winning both.
In all other cases I start anew. Only eight of the 18 red numbers are Even. This irregular distribution means that
I'll (in most cases) come out from the paroli without losing everything. Winning here and losing there in most cases means that I can take it rolig, no matter how many times I just won!
 
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palestis

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Re: Probabilities before and after the event
« Reply #3 on: July 21, 2016, 11:16:48 PM »
This might be a stupid question, or a mind game or a smart question or it may just help us clarify some things...

Let's say I have a system that uses the Paroli progression (bet more up as you win) on the Even chances (Red Black)
I have considered the chances of 2 consecutive wins is 0,25 or 25% and for 3 consecutive wins is 0,125 or 12,5%.
I have decided that I will stop after 2 consecutive wins and start over.

The question is this:
Since AFTER the 2 consecutive wins the chance for a 3rd win is 50%, now my chance for a series of 3 wins is no longer 12,5%, but 50%. Would it be smart to "change plans" and go for a 3rd win?
In purely statistical  terms, the 3rd bet has a 50% chance to win. Once you already know the first results (which happened to be wins), you revert to the original probability.
But in practical terms you may not get the benefit of the 50%, for the following reasons.
First you will find that reaching the 3rd step it's not that easy. You will lose more times in the first 2 steps, than winning both. Which means whenever you are lucky enough to reach the 3rd step, you probably lag behind in terms of B/R, due to more frequent losses in the first 2 steps.
Secondly, regardless of probability, you can't expect to be consecutively lucky, and likewise you can't be unlucky.
So even if it seems that you have an strong 50% probability in each bet, the original 12.5% still works in the background. That is y the 50% seems to be so weak, when it shouldn't be.

But for all practical purposes, I find the notion of at least one win in a series of 3 (or more) bets, far more important than aiming to hit  many consecutive bets.
First of all, the probability of hitting an EC in at least one of 3 bets is 87.5%.
Secondly, nobody says that you can't wait to lose the first two bets virtually.
Will the 3rd bet have an 87.50% probability to win after losing the first two bets?
Statistically is still 50%, but in reality because of the principle of non consecutively losing (or winning ) events, I find that you have a better than 50% chance TO BREAK A SERIALLY LOSING PATTERN.
In this case the 87.5% works in the background, in your favor this time.
I just don't accept that a 87.50% probability keeps getting shot and reduced to 50% each and every time. Then y bother have the probability of series theorem?
As far as I know it is a valid part of probability theory.
The picture below says it all:
Yes they say that each time you toss a coin the chance of one side is 50%.
But since this choice is RESTRICTED by the fact of life "that many same things in a row" are very unlikely, then I take that to mean that at some point the other side is LONG OVERDUE. NOT NECCESSARILY enough times to close the imbalance gab, but at least for one time only.
In roulette you just have to wait to bet for that one time only. Then the cycle starts all over again.
« Last Edit: July 21, 2016, 11:21:49 PM by palestis »
 
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scepticus

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Re: Probabilities before and after the event
« Reply #4 on: July 22, 2016, 12:33:59 AM »
Doesn't the Reverend Bayes Theorem state that the Probability at the start is the same as at the end ?
I agree with Palestis  that no realistic gambler seeks to win EVERY bet but seeks only to win sufficient to show a profit.
 

mr j

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Re: Probabilities before and after the event
« Reply #5 on: July 22, 2016, 12:36:46 AM »
Some of it can be a play on words, no doubt.

If the 12 hits three times in a row....the goofs at the table yell out...WHAT ARE THE ODDS? Its simply 1/38 (or 37) per.

but if someone BEFORE the fact says 12 will hit three times in a row (and it does), then its 1/54,872.

Ken
 
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palestis

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Re: Probabilities before and after the event
« Reply #6 on: July 22, 2016, 02:43:23 AM »
Doesn't the Reverend Bayes Theorem state that the Probability at the start is the same as at the end ?
I agree with Palestis  that no realistic gambler seeks to win EVERY bet but seeks only to win sufficient to show a profit.
This is a controversial issue, that I believe deserves more serious discussion. Because winning
consistently may depend on it.
We can easily compute the probability of winning at least one bet in a predetermined series of bets.
That's what many system players do whether they know it or not.
They wait for a trigger and then they start betting a predetermined series of bets with a specific target in their sight.
Since a series of bets aims only at winning one bet, normally the betting cycle (trigger plus bets), ends right there, and it only restarts after a new trigger appears. For example the probability to hit one EC in a series of 3 predetermined bets is 87.5%. Whether you hit it in the 1st, or 2nd  or 3rd bet  doesn't matter. What is certain is that you have an 87.5% chance to hit it in one of 3 bets. Once you hit it,  it is very risky to continue for more wins. Because that 87.5% no longer applies.
Now the question is if the 87.5% still applies in the 3rd bet after the first 2 bets failed.
I'm am not sure about that, but what I am certain about is the fact that is higher that 50%.
That's what my long time tests over the years have proven.
Betting only the 3rd bet in a 3 bets series, after the first 2 failed in my presence, has won many more times than it lost, even if the probability supposedly is 50%. Not only I have tested it with EC's but with dozens, columns, 3 quads, 2 DS's and many other combinations of groups.
And in all cases, betting the later stages of a series (provided the earlier stages lost virtually of course), won far more frequently than lost.
And the simple reason to avoid betting the earlier stages of a series is to save money, and use them in bets that have higher winning certainty. 
Could it be that by purposely waiting to lose the early stages, in a way we ride out a variance that is possibly present? Where betting from the beginning, we may be penalized by a variance?
I have used Ellison's system in the past. It's ok, but it's nothing special if played as described.
However waiting to lose virtually all bets after a trigger, when the second trigger appeared the results changed dramatically. There were a lot more wins in a new trigger after the first failed.
And I don't remember seeing 3 triggers lose in a row. Which means after 2 failed back to back triggers, you better bet heavy in the 3rd. The waiting time is worth its weight in gold. 

« Last Edit: July 22, 2016, 02:54:40 AM by palestis »
 
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MrPerfect.

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Re: Probabilities before and after the event
« Reply #7 on: July 31, 2016, 11:42:03 PM »
Waiting for a trigger start to make sense if the ods are disrupt.  For example, if you bet 18 numbers that are best for that wheel instead of black/ red...  if you have positive expectatation, means you have extra hits for your numbers. These extra hits will eventually have to form some patterns for their distrebution...  that will be jenuine patterns wich you can relay on. In such a situation you will have more consequtive hits then expected. So hit to your numbers may become trigger itself.
 I do not advise to use skips and consequtive hits as a pattern in your bets decision, it's not an optimal way. Just point the fact that they can be used to maximise profit per spin , if you do not mind higher risk/ reward ratio.
 

dobbelsteen

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Re: Probabilities before and after the event
« Reply #8 on: August 01, 2016, 08:54:15 AM »
My method is start betting after a virtual loss of  a 10 event trigger.Every random sequence is a trigger , that means you have never to wait for a trigger.
You can use every betting selection and built in a stop loss moment.An stop loss after 10 no-hits is sufficient. Noting of the results of the spins is not nessesary and the betting is very simple. After all hits a new trial start directly
 

Reyth

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Re: Probabilities before and after the event
« Reply #9 on: August 01, 2016, 04:42:06 PM »
This might be a stupid question, or a mind game or a smart question or it may just help us clarify some things...

Let's say I have a system that uses the Paroli progression (bet more up as you win) on the Even chances (Red Black)
I have considered the chances of 2 consecutive wins is 0,25 or 25% and for 3 consecutive wins is 0,125 or 12,5%.
I have decided that I will stop after 2 consecutive wins and start over.

The question is this:
Since AFTER the 2 consecutive wins the chance for a 3rd win is 50%, now my chance for a series of 3 wins is no longer 12,5%, but 50%. Would it be smart to "change plans" and go for a 3rd win?
In purely statistical  terms, the 3rd bet has a 50% chance to win. Once you already know the first results (which happened to be wins), you revert to the original probability.
But in practical terms you may not get the benefit of the 50%, for the following reasons.
First you will find that reaching the 3rd step it's not that easy. You will lose more times in the first 2 steps, than winning both. Which means whenever you are lucky enough to reach the 3rd step, you probably lag behind in terms of B/R, due to more frequent losses in the first 2 steps.
Secondly, regardless of probability, you can't expect to be consecutively lucky, and likewise you can't be unlucky.
So even if it seems that you have an strong 50% probability in each bet, the original 12.5% still works in the background. That is y the 50% seems to be so weak, when it shouldn't be.

But for all practical purposes, I find the notion of at least one win in a series of 3 (or more) bets, far more important than aiming to hit  many consecutive bets.
First of all, the probability of hitting an EC in at least one of 3 bets is 87.5%.
Secondly, nobody says that you can't wait to lose the first two bets virtually.
Will the 3rd bet have an 87.50% probability to win after losing the first two bets?
Statistically is still 50%, but in reality because of the principle of non consecutively losing (or winning ) events, I find that you have a better than 50% chance TO BREAK A SERIALLY LOSING PATTERN.
In this case the 87.5% works in the background, in your favor this time.
I just don't accept that a 87.50% probability keeps getting shot and reduced to 50% each and every time. Then y bother have the probability of series theorem?
As far as I know it is a valid part of probability theory.
The picture below says it all:
Yes they say that each time you toss a coin the chance of one side is 50%.
But since this choice is RESTRICTED by the fact of life "that many same things in a row" are very unlikely, then I take that to mean that at some point the other side is LONG OVERDUE. NOT NECCESSARILY enough times to close the imbalance gab, but at least for one time only.
In roulette you just have to wait to bet for that one time only. Then the cycle starts all over again.

LOL sorry I wanted to like this post again.

 
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palestis

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Re: Probabilities before and after the event
« Reply #10 on: August 01, 2016, 08:26:11 PM »
Great minds think alike.
 

Real

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Re: Probabilities before and after the event
« Reply #11 on: August 10, 2016, 06:08:40 PM »
A question for you guys. 

In just four spins, no zero, which do you feel is the most likely to hit?

RRRR or RBRRB ?
« Last Edit: August 11, 2016, 12:51:14 AM by kav »
 

kav

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Re: Probabilities before and after the event
« Reply #12 on: August 11, 2016, 12:49:45 AM »
You mean RRRR vs RBRB I guess.
If we leave "feelings" aside, mathematically both sequences have the same probability.
« Last Edit: August 11, 2016, 01:06:41 AM by kav »
 

Bayes

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Re: Probabilities before and after the event
« Reply #13 on: August 11, 2016, 07:41:24 AM »
I agree with Real. The post should have been left in its original place.

As regards the question, sorry to point out the bleedin' obvious but it has to be RRRR because you only have 4 spins!
 

kav

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Re: Probabilities before and after the event
« Reply #14 on: August 11, 2016, 01:17:15 PM »
After popular demand I re merged the posts.
Now can someone please enlighten me?

You mean that RRRR is more probable than RBRB?
 
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