I have some difficulties to find out the probability of winning 3 spins minimum out of 7 spin.
It's an interesting money management:bkr 99 units,win 29 units, if at least 3 W out of 7 spins.
Who has the answer?
I'll illustrate some details about it later on.
I think it's not possible to tell without knowing the chance of winning for each individual spin. If all bets are EC, the chance of hitting 3 out of 7 would be 57.15% I believe, but I'm not a math expert. :)
Thanks ATI,but I'm afraid that your figure is wrong.
I'm speaking about EC,european roulette,and I'd like to know the probability of winning 3 spins ,AT LEAST,out of seven spins.
I think that is somewhere between 75% and 85%.
Other answers?Thanks in advance
beretta28 - the probability on any & all spins is always the same. On the 37 pocket Live dealer wheel for EC bets the win probablity remains the same on any spin @ 18/37 = 48.65% of a win & 19/37 = 51.35% of a loss. So certainly the answer Isn't 57.15% or 75% - 85% ?
For maths each spin should be written as 48.65% x1, x1, x1 repeating & always = 48.65% & the same with 51.35% x1, x1, x1 repeating & always 51.35%.
Asking what the probability is of winning at least 3 out of 7 spins is flawed as the probabiliy remains the same 48/51% for each spin. It's little different to asking "what are my chances of backing RED for 100 spins & coming out a winner".
If it was 57% or 80% then with a reasonable progression lots of us would be always in the winners circle & as most of us know by our losses this isn't the case.
How often do players say back only RED or ODD & due to randoms variances a string of BLACKS or EVENS appear or 5 BLACKS OR 5 EVENS out of 7 results spin up leaving you with only 2/7 REDS or 2/7 ODDS. Then the pendulum swings your way & sometimes 10 REDS OR 10 ODDS back to back
may spin up. So on that information the strike rate for 2/7 spins is as low as 28.57% but later in the game it was 142.86% for 10/7 spins.
This 3/7 probability question is asking for a fixed answer for values that do fluctuate, accordingly there can't be any valid answer & so the question is largely invalid ? It's something akin to asking "How long's a piece of string" ?
ausguy,thanks,but I'm afraid there is a misunderstanding.
Some examples:
Probability that in 20 spins there are exactly 10 Red and 10 Black: 17,62%
Probability that in 10 spins there are exactly 5 Red and 5 Black: 24,62%
Probability that in 100 spins there are exactly 50 Red and 50 Black: 2,63 %
These calculations need "binomial calculator"!
Probability that in 7 spins on Red and Black,for instance,there are at least 3 Red:??? binomial calculator is not enough or,better,I don't use it correctly.
Sorry but your answer is not pertinent or I'm mistaken
Beretta
Binomial.
First you are saying 7C3
Which is 7!/3!/4! (4 is 7-3) =35
Then x by 18/37^3 x by 19/37^4
=28.02%
Quote from: Turner on Jul 09, 06:10 AM 2014
Beretta
Binomial.
First you are saying 7C3
Which is 7!/3!/4! (4 is 7-3) =35
Then x by 18/37^3 x by 19/37^4
=28.02%
So......
(How many combos of 3 in 7) x (3 wins) x (4 losses) x 100%
Turner,thank you very much.
I have already received 4 different answers: 83,7%, 74,9%, 57,15% and now 28,02%.
First of all,just with common sense,I'd say that the probability of winning in 7 spins on EC,at least 3 spins is more than 50%.
So I don't understand your 28,02%:it could be the probability of winning less than 3 spins in 7 spins.May be?
I think that the good figure is 74,9%,but I'm not so sure and I'm still confused.
I would like to have confirmation of the good figure and then I'll post the related money management(BKR 99 units and ,if 3 W at least in 7 spins,29 units won)
Beretta...well thats my understanding of binomial formula. Theres a calculator on saliu.com/roulette2.html
I havnt used it
I found this very similar example on the web.
Same method.......what would you think the odds of 6 wins in 10?
The probability of getting exactly x success in n trials, with the probability of success on a single trial being p is:
P(X=x) = nCx * p^x * q^(n-x)
Example:
A coin is tossed 10 times. What is the probability that exactly 6 heads will occur.
Success = "A head is flipped on a single coin"
p = 0.5
q = 0.5
n = 10
x = 6
P(x=6) = 10C6 * 0.5^6 * 0.5^4 = 210 * 0.015625 * 0.0625 = 0.205078125....
Or...20.5%
......... So... This is me again....forget HE (zero)
3 Ec wins in 7 spins...for red....say RBBRBRB
7C3 x0.5^3 x 0.5^4
35 x 0.125 x 0.0625 = 27.3%
..................
7C3 is how many 3s in 7 = 7!/3!/4!
7! means factorial 7 or 7x6x5x4x3x2x1
With no zero...odds of 3wins is 0.5x 0.5 x 0.5 (0.5^3)
With zero, odds of 3 wins is 18/37^3
4 losses is 0.5^4 without zero
With zero is 19/37^4
Beretta....
Your ausguy reply examples (5 reds in 10 spins...etc) are what I get using the binomial equasion I mentioned....only your probabilities seem to be without zero
5 reds in 10
10C5 x 0.5^5 x 0.5^5 = 24.6 %
Im certain your answer is 28%
Thanks again Turner.
The answer to my first question is 76,6%(Zero taken into account) that in 7 spins a player wins at least 3 spins.
The profit(29 units) is too low compared with bkr (99 units) and with the risk(23,4%) of losing it all.
A bit more interesting is what you mention, that is:
10 spins:bkr 97 units,profits 32 units,only if you won't have exactly 5 Reds and 5 Blacks.
Probability that perfect equilibrium occurs(bkr 97 units lost, in this case) is 25,5%(Zero taken into account).
Of course you win 32 units as soon as you reach 6 W out of 10 spins(probability 74,5%).
The probability is lower than first example,bur ratio win/bkr is more favorable.
Beretta....
I think your figure is cumulative binomial probability
This means at most 3 wins in 7, which includes 0 wins, 1 win, 2 wins, 3 wins.
The Binomial probability, which is 3 out of 7 wins is 28%
RRRBBBB is 28%
betting red ....RRR BBBB or RR BBBBB or R BBBBBB or BBBBBBB is 75%
OR....
A) 3 wins in 7 = 28%
B) 0,1,2 or 3 wins in 7 is 75%
Is your question A or B
Cheers
Turner
link:://stattrek.com/online-calculator/binomial.aspx (link:://stattrek.com/online-calculator/binomial.aspx)
.
Betting Red
RRR win ,RRBR win,RBBBRR win,BBRRBBR win, BBBRRR win,RBRBRB, win RBRR ,and so on
Goal:to win 3 RED in 7 spins at the most= 76,6% ,because of Zero,otherwise 75,...%
Beretta,
sometimes, you see clever arses in forums who persist about how they are right and the other is wrong, and I really dont want to come across like that....or I will look like some arrogant Maths guy....when I am neither of those 2 (well I am a guy lol)
I dont think its 75%....but I could always be wrong.
I will leave you with it mate. Good luck.
I don't want to be arrogant neither.
May be it's difficult to explain the problem on the Forum
I obtained the figure I mentioned by four different sources,two of them very qualified and two of them Ph.D in math/statistiques very soon.
Interesting topic.but how do you win 29 to 99br?
It's a very old algorithm used for betting(horses,football....)
Now is available for roulette.
Here the bets, in units:
15 20 24 24 16
10 16 24 32 32
4 8 16 32 64
Start from the first figure up on the left(15).
In case of W you go below,in case of L you go right.
Starting from 15,if W,next bet 10 or 20 if L and so on:the figure below if W,the figure at the right if L.
3 wins in a row(it happens):15 + 10 + 4= +29
3 wins at least in 7 spins : + 29 units.
If you lose 5 spins in a row or 5 L and 2 W ,in 7 spins at the most: -99 units
Is it clear?
A better money management(ratio yield/bkr related to % of W or L) doesn't exist.
Very boring and hit and run method.
Clear as diamonds thanks.i will give it a go .off to the casino.
x warrior
what about your results at the Casino,with this approach?
x Turner
I have understood your mistake or my very confused question.
Your 28,02% is the probability to win EXACTLY 3 spins in,at the most,7 spins.
I asked the probabilty to win AT LEAST 3 spins in 7 spins,that is 75%,because of Zero,otherwise,without Zero,is almost 77%
I was only there for 37 spins and won them all ,I will try this weekend again .
hi, just for fun 3 sessions on BV, 3 on dublin. I was betting color before the last. All 6 won.
beretta28, maybe you have any better bet selection?
Congratulations Biagle!
I prefer play for single....BBR I'd play B. RRRRRRB I'd play R,if L RRRRRBB I wait for change of colour RRRRRRBBBBBR I'd play B.
This money management is very interesting because if you lose - 99 units,but if you win + 29 units,almost 30% of BKR.
Very easy and fast ricovery.
Quote from: beretta28 on Jul 10, 09:31 AM 2014
It's a very old algorithm used for betting(horses,football....)
Now is available for roulette.
Here the bets, in units:
15 20 24 24 16
10 16 24 32 32
4 8 16 32 64
Start from the first figure up on the left(15).
In case of W you go below,in case of L you go right.
Starting from 15,if W,next bet 10 or 20 if L and so on:the figure below if W,the figure at the right if L.
3 wins in a row(it happens):15 + 10 + 4= +29
3 wins at least in 7 spins : + 29 units.
If you lose 5 spins in a row or 5 L and 2 W ,in 7 spins at the most: -99 units
Is it clear?
A better money management(ratio yield/bkr related to % of W or L) doesn't exist.
Very boring and hit and run method.
question if you lose 3 in a row and then win what unit do bet so if I lose 15 20 24 then win 24 now what? Anyone.
Quote from: warrior on Jul 23, 11:09 AM 2014
question if you lose 3 in a row and then win what unit do bet so if I lose 15 20 24 then win 24 now what? Anyone.
Hi Warrior,
My understanding is you would move down and bet 16.
Jim
.
Perfect,it wasn't complicated....
Quote from: JimmieB on Jul 23, 11:20 AM 2014
Hi Warrior,
My understanding is you would move down and bet 16.
Jim
are you saying line 2 where the bet starts at 10 and then( 16)
Sorry Warrior, my mistake that's what I get for trying to do work and check on the forum at the same time!!
The next bet you would have had after you won the 24 units on line 1 (column 4) would be 32 (line 2, column 4).
Jim