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3 spins out of 7/probability

Started by beretta28, Jul 07, 05:04 PM 2014

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beretta28

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.

ati

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.  :)

beretta28

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

ausguy

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" ?

beretta28

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

Turner

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%

Turner

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%

beretta28

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)

Turner

Beretta...well thats my understanding of binomial formula. Theres a calculator on saliu.com/roulette2.html
I havnt used it

Turner

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

Turner

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%

beretta28

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.

Turner

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

Turner


beretta28

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,...%

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