statistical question about the even chances

[hanshuckebein]

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

[esoito]

4:0 = 12,5%
3:1 = 50%
2:2 = 37,5%

Sorry -- don't understand what you're saying here.

Please clarify.

[Bayes]

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

[Bayes]

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. 

[esoito]

Thanks to both of you.

Alles klar now. 

[hanshuckebein]

@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:

[hanshuckebein]

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

[hanshuckebein]

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

[Amiwrong]

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

[iggiv]

Welcome aboard bud. Or should I say: "Ð"обро пожаловаÑ,ÑŒ в Ñ,,орум"? 

[Steve]

Ð"а, spam мы все spamолжны выпspamÑ,ÑŒ воspamкspam. (правspamла Ñ,,орума)

[Skakus]

много воspamкspam

[Chris555p]

@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

[superman]

If 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.

What 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.

[Chris555p]

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