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statistical question about the even chances

Started by hanshuckebein, Oct 11, 10:20 AM 2010

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0 Members and 4 Guests are viewing this topic.

GLC

Quote from: Chris555p on Dec 05, 10:26 AM 2013
Hi Superman

Thanks for replying; In terms of light progression, I mean for ex increasing bet size by a certain percentage
on a Loss, reducing it by a certain percentage on a win. Using a 10$ base bet this would be….


Level 1 - $10-$16-$22-$28-$34 = $110 (Up $6 on a lost, down $4 on win.)
Level 2 - $15-$24-$33-$42-$51 = $165 (Up $9 on a lost, down $6 on win.)
Level 3 - $30-$48-$66-$84-$102 = $330 ( Up $18 on a lost, down $12 on win.)
Level 4 - $60-$96-$132-$168-$204 = $660 ( Up $36 on a lost, down $24 on win.)  Total : 1265$

You mentioned that the probality of losing….. is very slim, but you would AND it would happen often enough
for you to back at where you started or less that you started with….

Playing on real Wheel, is there a way of calculating or quantifying approximately this very slim probability of
missing 5 bets in a row; at 4 different levels of progression  starting at level 1, base bet  10$ …....?

Thanks

Cheers


Chris

Chris,  Are you just playing a level until you lose 5 bets in a row?  I don't see the value in that because at the end of the 5 bets you will be a substantial amount in the hole.

The only thing I can think of is that you are hoping to have enough wins early in the progression at each level to recover all losses.  Those odds are pretty slim indeed.  If you don't win on bet 1 or 2 at each level, you'll be losing and going into the hole without even losing 5 in a row at each level.

Maybe I'm not having a "meeting of the minds" with you.  Can you explain further?

GLC
In my case it doesn't matter.  I'm both!

GLC

What about using this progression:

Level 1:  1-1-1-1-1
Level 2:  2-2-2-2-2
Level 3:  3-3-3-3-3
Level 4:  4-4-4-4-4

or

1-2-3-5-9
2-4-6-10-18
3-6-9-15-27
4-8-12-20-36

In my case it doesn't matter.  I'm both!

Chris555p

Hi GLC

Thanks for replying.To answer your question, No I do not play until I loose 5 times in a row. The way it works is
I wait for a trigger then I start betting 10$ at level 1; If I win, I stop betting, wait for the next trigger and do the
same thing again.

If I loose, I move the next bet to the right betting $16; if I win, I stop betting; and reduce my bet by 4$.
The goal is to stay at level 1 as long as possible.

If for ex I loose all the 5 bets in a level in a row, I usually take a break and come back later on to play next level.
The procedure is the same as level 1 except the bet size for level 2 is bigger to recoup
level 1 losses. The probability for losing 5 bets in a row is very low, 1/32.

If I lose all 5 bets at level 2, I move to level 3 after taking a break.

If I lose all 5 bets at level 3 I move to level 4 again after taking a break….

The bet selection for the ec is very important; win goal has to be realistic, to avoid getting into deep hole.

From experience it is very rare for me to go beyond level 2. losing 2 or more levels in a row one after
another is extremely low.

Therefore, I was wondering what is the probability for losing 5 bets in a row at 4 different levels of betting.
I would greatly appreciate your thought on this calculation of probability. Thanks.


Cheers

Chris

GLC

I don't consider myself a math guru, but I have thought about this kind of thing a lot.

One of the methods I tested for a while was to play a 5 step capped martingale.

1-2-4-8-16
2-4-8-16-32
4-8-16-32-64
8-16-32-64-128

While playing the 1st level, any units won went into profits and couldn't be lost in that session.  That means I would have to lose 5 bets in a row which happens about once ever 32 attacks, mathematically.

Upon losing 5 bets in a row, I would move to the 2nd level and play at the 2nd level until I recovered 21 of the 31 lost units from level 1.  Then I'd drop back down to 1-2-4-8-16 and continue trying to recover the remaining 10 units at level 1.

If I lost at level 2 before recovering my 21 units from level 1, I'd move to level 3 to recover 42 of the 62 lost units from level 2.  I'd continue in this way until I reached my win target of 50 units or lost at the 4th level.  I never lost all 465 units, but it was still a pretty wild ride and I don't know if I could play it for real money unless I was betting small units.  Then I would have to deal with the possibility of getting bored because I couldn't afford to risk enough to make it interesting.

With your method, anytime you lose 5 more bets than you win, you will have to move to the next level.  That means that you must reach a point where you have had 20 more losses than wins in the session.  When that happens, you will be down your $1265.  This won't happen very often, but when it does that's a big chunk to give back in one session.

This should happen about once in approximately 400 bets.  Not spins, bets.  That's a long session because of all the spins that you're tracking plus the 400 bets.

GLC
In my case it doesn't matter.  I'm both!

Chris555p

Hi GLC

Thanks for the response; Assuming if I were to loose the 4 levels (- 1265$) playing with 10 $ base bet,
the only thing I would hope is that it happens after after several weeks or months of playing so that the
profits made during these periods would be far Superior than the losses. Therefore, the bet selection
must be very solid.

You mentioned that the probability for loosing all five bets at the four different levels... should happen about once
in approximately 400 bets.  Not spins, bets..... Is it possible to please explain how the probability of once every 400 bets
was calculated, thanks.


Cheers


Chris



GLC

Remember that I prefaced my comments with the caveat that I'm not a math guy. 

I was thinking that on average we lose 5 more bets than we win in every 100 placed bets.  So at 5 per 100, in 400 it would be 20.  One error I made is that I was basing this on playing double zero roulette which is what I have to play.  Playing single zero roulette it would double that to 800 placed bets, roughly speaking.

I may be overlooking something here.  But the above is my logic.

Granted the wins for each win/loss combination may keep you ahead of a major loss.  That's the unknown factor.

GLC
In my case it doesn't matter.  I'm both!

Chris555p

HI GLC

Thanks for the explanation. I totally agree with your explanation. I have made hundreds of bets using this
sort of progression over a period of several weeks and up to now the results are great.

Given the win/loss combination as u mentioned, I have never up to now suffered suffered any major loss;
I realize that eventually this may happen....but I hope that when this happens, the gain accummulated in the
mean time will easily absorb such losses.

Thanks again my friend.


Cheers

Chris

GLC

Chris,  One point I will make just for your consideration.  Since you will lose more times than you win in a session unless you are just lucky for a particular session, why not adjust your bet progression per following logic.

What you are playing is a modified D'Alembert type progression.  The basic D'A is +1 on a loss, -1 on a win.  Every win will result in a 1 unit win.  If we have an equal number of losses and wins, we will be up exactly 1 unit for each win.  The time we get into trouble is when we have more losses than wins by enough to overpower the number of wins we have.  Since we expect to lose more than we win, many designers try to compensate for this tendency to drift toward ever higher bet sizes.

One such way is called half peak.  The author of this method suggests that anytime you recover to half of your highest draw down, you take the loss and re-set.  You have to decide how deep into the hole you must go before half peak kicks in.  In other words, -14 is too soon in my mind.  It would mean that you must stop when you get back to -7.  Something like -50 is more realistic.  So we would reset if we got back to -25.

Another way to do it is to drop back more on a win than you increase on a loss.  What you do with your progression is the opposite. You increase more on a loss than you draw back on a win.  This has the tendency to cause your bets to grow in size even if you have 50/50 win loss ratio.  You could increase by 4 on a loss and decrease by 5 on a win.

Or you could increase by 4 and draw back by 4 on the 1st 3 bets in each level and then increase by 4 and draw back by 5 for the next 2 bets at each level.

Of course, instead of 4 it would be proportionate for the larger levels.

Just a thought you may find interesting.

Good luck with this method.  I think I'm on to something else.

GLC
In my case it doesn't matter.  I'm both!

Chris555p

Hi GLC

Thanks for your inetersting suggestion. Would it possible to provide some examples using
the LW registry showing how to use the half peak method on a sample of say 20 bets.  Thanks

Cheers

Chris

GLC

Here's an attempt to attach the original half-peak system.

Here's another twist on D'Alembert

link:://:.rouletteforum.cc/index.php?topic=13450.0

I'm not saying that either of these is a sure way to win.  Maybe the way you're playing is better.  I don't have time to compare them.  You must decide.  Sorry if I've just wasted you time.
In my case it doesn't matter.  I'm both!

Chris555p

Hi GLC

Thanks for the explanation and for the links; I will definitely study them
and do comparaisons.

Cheers


Chris

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