I' ve recently read a short article that states the following:
in 4 consecutive spins you have a probability for these ratios to appear of:
4:0 = 12,5%
3:1 = 50%
2:2 = 37,5%
is this true and how is the calculation done?
cheers and thanks
hans
4:0 = 12,5%
3:1 = 50%
2:2 = 37,5%
Sorry -- don't understand what you're saying here.
Please clarify.
The simple way to prove this is just to use the basic probability formula:
P = (number of favourable outcomes)/(total number of outcomes)
But note that the outcomes should be equally likely. Generate a list of the total number of outcomes (the sample space):
R,R,R,R
R,R,R,B
R,R,B,R
R,R,B,B
R,B,R,R
R,B,R,B
R,B,B,R
R,B,B,B
B,R,R,R
B,R,R,B
B,R,B,R
B,R,B,B
B,B,R,R
B,B,R,B
B,B,B,R
B,B,B,B
Then count the sequences which are "favourable". e.g. for 4:0, there are 2 (RRRR & BBBB) which is 2 out of 16 (1 out of 8 ), so the probability is 1/8 = 12.5%. Do the same for 3:1 and 2:2 to get the stated percentages.
This is impractical if you want to know the percentages for longer sequences. It would be tedious to work out the figures for a sequence which is 20 spins long, because there are 220 = 1,048,576 patterns, but the principle is the same. In that case you would use some more complex maths (the binomial distribution), which can give you probabilities of ratios for any length of sequence.
link:://stattrek.com/Lesson2/Binomial.aspx (link:://stattrek.com/Lesson2/Binomial.aspx)
Nice going. Just keep an eye on the variance, it has a nasty habit of biting you in the arse just when you least expect it. :'(
Thanks to both of you.
Alles klar now. ;)
@esoito
I mean that if you look at 4 consecutive spins you have a probability of for example
B 4 R 0 = 12,5%
B 3 R 1 = 50%
B 2 R 2 = 37,5%
@bayes
thank you very much for taking the time to explain things. :thumbsup:
I checked the results of my my former post an had to do a few corrections. so here are the correct results. (sorry guys that things look a little mixed up now. )
I did a 30 x 100 spins test session (wiesbaden spins, 1st 15 sessions dealer spun, 2nd 15 sessions airball). I regarded 4 consecutive spins as 1 "complete game". but I only bet after the 3rd spin where there is either a 3:0 or 2:1 situation. I then bet that the 4th spin would bring the 3:1 situation - all just flatbetting.
out of my 30 sessions I lost 9 and ended up with a total profit of 44,5 units.
you can find the exact results in the attachement.
what I'm going to do next is to continue playing the sessions I lost to see if and when they would recover.
cheers
hans
so I continued to play the sessions I lost within 100 spins. I played them until the total net recovered to 500.
doing this I lost 1 out of the total 30 sessions and ended up with a profit of 73 units.
the excact results can be found in the attachment.
any of your comments or ideas are welcome. :)
cheers
hans
Hollo there! I'm a new one here at the forum, but 've been reading it for quite a long time. Would like to say that i admire with many thoughts that i've found here! You Guys are really awesome!
Have encountered on this topic and was a bit confused. It sounds really reasonable strategy for flatbetting. Or it is not? It certanly has some logic in it. Can please sombody expierienced and sophisticated in such things give me a clue? I just can't assume why it shouldn't win in a long term?
Thank you.
P.s. and once again, Thank You all for what you doing here. You are giving many people food for their brain.
Philipp
Welcome aboard bud. Or should I say: "Ð"обро пожаловаÑ,ÑŒ в Ñ,,орум"? :)
Ð"а, spam мы вÑе spamолжны выпspamÑ,ÑŒ воspamкspam. (правspamла Ñ,,орума)
много воspamкspam
@Bayes and everyone
Is it possible to assist me with the calculation of a probability. Say if I play a certain B/R
strategy where I play R 5 spins in a row. The probability of R not coming
5 times in a row is 1/32 ( 1/2x1/2x1/2x1/2x1/2 = 1/32 = 3.1%).
Say If I were to lose and I play the same strategy with a light negative progression
at 4 different levels. What is the probability that I would lose the 4 different levels, 4 times
in a row......? Thanks.
Cheers
Chris
QuoteIf I were to lose and I play the same strategy with a light negative progression
at 4 different levels
It would depend on what you say is light, you could do
1,1,1,1
2,2,2,2
3,3,3,3
etc
And depending on how much you are down each set could dictate what you use the nexy level, would you want one win to put you ahead? or 2 wins or would you want the whole 4 bets to win to end the challenge, not an easy question to answer but I know what you mean.
QuoteWhat is the probability that I would lose the 4 different levels, 4 times
in a row......?
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.
I know I have tried all different combinations of EC's RRBBRRBB set patterns, but eventually what you think wont happen happens, I'm not saying you wont get away with it for a while, even a long time but you don't know WHEN it will happen therefore pointless staying rigid
Footnote: sounds a good one for HIT n RUN some would say, even in that respect, it will happen but could take weeks/months, I don't agree with hit n run.
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
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
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
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
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 $1265. This won't happen very often, but when it does that's a big chunk to give back in one session.
This should happen about once in approximately 400 bets. Not spins, bets. That's a long session because of all the spins that you're tracking plus the 400 bets.
GLC
Hi GLC
Thanks for the response; Assuming if I were to loose the 4 levels (- 1265$) 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
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
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
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
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
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.
Hi GLC
Thanks for the explanation and for the links; I will definitely study them
and do comparaisons.
Cheers
Chris