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Walkinshaw30t
Walkinshaw30t
Walkinshaw30t
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Joined: Apr 11, 2013
June 24th, 2013 at 2:32:03 AM permalink
Hi,
Lately I have become very interested in probabilities, variance and standard deviation (as you may notice from my other threads!)

My current thoughts are regarding 'streaks' in even money games.
If a streak of say 7 or more blacks in roulette happens approx 0.0064 of the time.
- because this is quite a low number is there a high variance of this event occuring?
- how would I calculate this?

Thanks in advance for anyone who may be able to help!
Time will tell
EvenBob
EvenBob
EvenBob
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June 24th, 2013 at 6:51:11 AM permalink
What difference does it make. The chance of it going
to 8 in a row is just the same as it staying at 7.
"It's not called gambling if the math is on your side."
7craps
7craps
7craps
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Joined: Jan 23, 2010
June 24th, 2013 at 8:54:43 AM permalink
Quote: Walkinshaw30t

Hi,
Lately I have become very interested in probabilities, variance and standard deviation (as you may notice from my other threads!)

From many of your other posts your understanding of the basics is very limited IMO. (Do not fear, you are in the majority)
Start here
http://stattrek.com/probability/set-and-subset.aspx?Tutorial=Stat
or here
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat

Teach yourself statistics

Quote: Walkinshaw30t

My current thoughts are regarding 'streaks' in even money games.
If a streak of say 7 or more blacks in roulette happens approx 0.0064 of the time.
- because this is quite a low number is there a high variance of this event occuring?
- how would I calculate this?

You are asking an advanced question and need to understand more basics (like what is variance and how is it calculated and what does that result mean...)
to learn how to solve this.

for 18/38 = 0.005350844897313
18/37 = 0.006449045889425 (1 in 155.061 trials but each TRIAL is NOT 1 spin in length for your example)

I gather you want 18/37 which you did not even specify.
That figure is only for the next 7 spins.
Not in 10 spins: 0.016384062529891
or
in 100 spins: 0.275468908341602
# of trials increase so does the probability of seeing your streak (run)
start here
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html
(there are also a few WoV threads about this linked at that page. I think it still is accurate.)
https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/#post119117
average # of spins = 300.01
variance = 86649.92307
standard deviation = 294.4
Var(N) = 1−(p^(1+2r)−qpr(1+2r)/q^2p^2r
where p = probability of success (18/37 in your example)
q = 1-p
r = length of the streak (run)

The distribution (for this "wait" time) is NOT a normal distribution
It is not the same as the # of reds in 300 spins



Good Luck.

The one mistake I see you always making in your other threads is
mixing up the standard deviation for the number of successes in N trials (say # of reds in 100 spins)
and the standard deviation of the wagers over N trials (say # of bets on red in 100 spins).
They are totally two different animals but it appears to me (I may be 100% wrong)
you consider them both the same.
Maybe others will reply there too
winsome johnny (not Win some johnny)
Walkinshaw30t
Walkinshaw30t
Walkinshaw30t
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Joined: Apr 11, 2013
June 24th, 2013 at 6:40:43 PM permalink
Quote: EvenBob

What difference does it make. The chance of it going
to 8 in a row is just the same as it staying at 7.



Yes I understand that- that is not what Im interested in. Im interested in standard deviation- how much it can be relied on and how many trials would be required to see the results fall within the expected standard deviations eg 68% with in 1, 95% within 2 etc
Time will tell
Walkinshaw30t
Walkinshaw30t
Walkinshaw30t
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Joined: Apr 11, 2013
June 24th, 2013 at 6:58:49 PM permalink
Quote: 7craps

From many of your other posts your understanding of the basics is very limited IMO. (Do not fear, you are in the majority)
Start here
http://stattrek.com/probability/set-and-subset.aspx?Tutorial=Stat
or here
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat

Teach yourself statistics

You are asking an advanced question and need to understand more basics (like what is variance and how is it calculated and what does that result mean...)
to learn how to solve this.

for 18/38 = 0.005350844897313
18/37 = 0.006449045889425 (1 in 155.061 trials but each TRIAL is NOT 1 spin in length for your example)

I gather you want 18/37 which you did not even specify.
That figure is only for the next 7 spins.
Not in 10 spins: 0.016384062529891
or
in 100 spins: 0.275468908341602
# of trials increase so does the probability of seeing your streak (run)
start here
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html
(there are also a few WoV threads about this linked at that page. I think it still is accurate.)
https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/#post119117
average # of spins = 300.01
variance = 86649.92307
standard deviation = 294.4

Var(N) = 1−(p^(1+2r)−qpr(1+2r)/q^2p^2r
where p = probability of success (18/37 in your example)
q = 1-p
r = length of the streak (run)

The distribution (for this "wait" time) is NOT a normal distribution
It is not the same as the # of reds in 300 spins



Good Luck.

The one mistake I see you always making in your other threads is
mixing up the standard deviation for the number of successes in N trials (say # of reds in 100 spins)
and the standard deviation of the wagers over N trials (say # of bets on red in 100 spins).
They are totally two different animals but it appears to me (I may be 100% wrong)
you consider them both the same.
Maybe others will reply there too




Cheers thanks for reply and sending through the links,
Im probably not explaining myself clear enough but what I want to know is: if a streak of 7 reds happens on average 1/155. What is the variance associated with this happening? Would it be possible to go 300 spins without seeing a 7 streak? What about 2500?
And how many standard deviations away from average these would be?
I understand this is quite a difficult question.
Thanks
Time will tell
7craps
7craps
7craps
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Joined: Jan 23, 2010
June 24th, 2013 at 7:53:21 PM permalink
Quote: Walkinshaw30t

Im interested in standard deviation- how much it can be relied on and how many trials would be required to see the results fall within the expected standard deviations eg 68% with in 1, 95% within 2 etc

This shows your lack of understanding the concept of standard deviation.
standard deviation estimates (quite well with a good sample size) the Binomial Probability Distribution a range of possible outcomes.
It really is as simple as that. 1SD is always 1SD when we talk about a normal distribution.
Your streak question is NOT a normal distribution as seen from the photo I posted.

Let us start with a simple example.
Copy and Paste here

Say Bet RED 18/37 for 74 spins

How many REDS in 74 spins
EV = 36
Is this correct??
I could be wrong here.
But we know we will not always get exactly 36 reds in 74 spins.
There is variance.

We should just calculate a table using the Binomial Probability Distribution and you can add all the values up and see what is what.
There is math to that unless you use a program to do it for you.
Here we are
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat

 x    prob[X=x]   prob[X<x]  prob[X>=x]  prob[X<=x]   prob[X>x]

21    0.0001854   0.0001207   0.9998793   0.0003061   0.9996939
22    0.0004230   0.0003061   0.9996939   0.0007291   0.9992709
23    0.0009061   0.0007291   0.9992709   0.0016353   0.9983647
24    0.0018242   0.0016353   0.9983647   0.0034594   0.9965406
25    0.0034563   0.0034594   0.9965406   0.0069157   0.9930843
26    0.0061710   0.0069157   0.9930843   0.0130867   0.9869133
27    0.0103932   0.0130867   0.9869133   0.0234799   0.9765201
28    0.0165275   0.0234799   0.9765201   0.0400074   0.9599926
29    0.0248363   0.0400074   0.9599926   0.0648437   0.9351563
30    0.0352937   0.0648437   0.9351563   0.1001374   0.8998626
31    0.0474577   0.1001374   0.8998626   0.1475951   0.8524049
32    0.0604149   0.1475951   0.8524049   0.2080101   0.7919899
33    0.0728448   0.2080101   0.7919899   0.2808549   0.7191451
34    0.0832190   0.2808549   0.7191451   0.3640739   0.6359261
35    0.0901018   0.3640739   0.6359261   0.4541756   0.5458244
36    0.0924729   0.4541756   0.5458244   0.5466485   0.4533515
37    0.0899736   0.5466485   0.4533515   0.6366221   0.3633779
38    0.0829950   0.6366221   0.3633779   0.7196171   0.2803829
39    0.0725787   0.7196171   0.2803829   0.7921958   0.2078042
40    0.0601639   0.7921958   0.2078042   0.8523596   0.1476404
41    0.0472661   0.8523596   0.1476404   0.8996257   0.1003743
42    0.0351830   0.8996257   0.1003743   0.9348088   0.0651912
43    0.0248047   0.9348088   0.0651912   0.9596135   0.0403865
44    0.0165562   0.9596135   0.0403865   0.9761697   0.0238303
45    0.0104566   0.9761697   0.0238303   0.9866263   0.0133737
46    0.0062452   0.9866263   0.0133737   0.9928715   0.0071285
47    0.0035247   0.9928715   0.0071285   0.9963963   0.0036037
48    0.0018783   0.9963963   0.0036037   0.9982746   0.0017254
49    0.0009442   0.9982746   0.0017254   0.9992188   0.0007812
50    0.0004473   0.9992188   0.0007812   0.9996660   0.0003340
51    0.0001994   0.9996660   0.0003340   0.9998654   0.0001346


Here is an easier way
Binomial Standard Deviation = for 1 trial = square root of (P * (1-P)) = 0.49981
This is in your studies

You know how to use that value?
For trials less than about 30 we do not even think of using it.
But for 74 spins we do (the data is very close to being normally distributed, right?? There still will be a small error)

what is the square root of 74?
Multiply that by the SD of 1 trial = 0.49981
about 4.3?
So a 1SD range = about 4.3(about 68% or look here: nice table about half way down)
http://en.wikipedia.org/wiki/Standard_deviation

EV = 36
SD = 4.3
That 1SD range of 31.7 to 40.3 IS a 68% probability. (68.2689492%)
It does not matter how many trials there are, 500, 5000 or 5 million
A 1 SD range = 68.2689492%. That means about a 32% probability to be outside of 1SD, equally too high or too low.
A 2SD range is wider, as is a 3SD and a 4SD range. There are probabilities related to each range.

of course it is IMPOSSIBLE to get exactly 31.7 Reds in 74 spins, using my example
so we can see we have approximated the Binomial distribution.

How close are we to the 1SD range??

try 32 to 40 = 70.5% so we gots a small error.
But with a small sample size, the error is to be expected. A larger sample size and the smaller the error.
There is a formula for that too but you need to really understand the Binomial Probability Distribution
and variance and standard deviation first.
Then see how to use the central limit theorem to solve common math related problems.

So, there must be a very specific question you want answered other than the streak question above.
Fire away and see where your understanding now lies.
Continued Good Luck
winsome johnny (not Win some johnny)
7craps
7craps
7craps
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Joined: Jan 23, 2010
June 24th, 2013 at 8:08:24 PM permalink
Quote: Walkinshaw30t

Would it be possible to go 300 spins without seeing a 7 streak?

yes. Does this not seem right??
only a 0.632729561221891 probability of seeing at least 1 such streak.
That is about a 36.8% chance (probability) of NOT seeing such a streak.

a handy streak calculator here
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

Quote: Walkinshaw30t

What about 2500?

expected # of runs of length 7 or more in 2500 spins = about 8.26
This is just the average. Here is part of that distribution using a computer program in Excel
Event	                                 Run Probability        1 in
0 runs of length 7 or more	            0.000208547	    4,795.08 
at least 1 run of length 7 or more	0.999791452787651	1.00 
at least 2 runs of length 7 or more	0.997977803412864	1.00 
at least 3 runs of length 7 or more	0.990139483734160	1.01 
at least 4 runs of length 7 or more	0.967693431018414	1.03 
at least 5 runs of length 7 or more	0.919781386396954	1.09 
at least 6 runs of length 7 or more	0.838469580627413	1.19 
at least 7 runs of length 7 or more	0.724186199048575	1.38 
at least 8 runs of length 7 or more	0.587364455037691	1.70 
at least 9 runs of length 7 or more	0.444931004198057	2.25 
at least 10 runs of length 7 or more	0.313958401980447	3.19 
at least 11 runs of length 7 or more	0.206251693272468	4.85 
at least 12 runs of length 7 or more	0.126240487898944	7.92

Quote: Walkinshaw30t

And how many standard deviations away from average these would be?
I understand this is quite a difficult question.
Thanks

Now you go from asking about the chance (probability) of an event to the standard deviation over many trials
I gave the formula for that in my first post. I guess it went past you too quickly.
winsome johnny (not Win some johnny)
Walkinshaw30t
Walkinshaw30t
Walkinshaw30t
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Joined: Apr 11, 2013
June 24th, 2013 at 8:16:27 PM permalink
Deleted
Time will tell
Walkinshaw30t
Walkinshaw30t
Walkinshaw30t
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Joined: Apr 11, 2013
June 24th, 2013 at 8:52:06 PM permalink
For the example you gave of 74 spins- if i was to make 74 spins and write down results of which sd bracket the results fell into and repeat this 100 times, can i guarantee that 68% will be within 1 standard deviation, 95% within 2 SD etc
If not what sample size would be required to achieve the expected SD results? What would the variance be for 100 samples of 74 spins?


[spoiler]
 x    prob[X=x]   prob[X<x]  prob[X>=x]  prob[X<=x]   prob[X>x]

21    0.0001854   0.0001207   0.9998793   0.0003061   0.9996939
22    0.0004230   0.0003061   0.9996939   0.0007291   0.9992709
23    0.0009061   0.0007291   0.9992709   0.0016353   0.9983647
24    0.0018242   0.0016353   0.9983647   0.0034594   0.9965406
25    0.0034563   0.0034594   0.9965406   0.0069157   0.9930843
26    0.0061710   0.0069157   0.9930843   0.0130867   0.9869133
27    0.0103932   0.0130867   0.9869133   0.0234799   0.9765201
28    0.0165275   0.0234799   0.9765201   0.0400074   0.9599926
29    0.0248363   0.0400074   0.9599926   0.0648437   0.9351563
30    0.0352937   0.0648437   0.9351563   0.1001374   0.8998626
31    0.0474577   0.1001374   0.8998626   0.1475951   0.8524049
32    0.0604149   0.1475951   0.8524049   0.2080101   0.7919899
33    0.0728448   0.2080101   0.7919899   0.2808549   0.7191451
34    0.0832190   0.2808549   0.7191451   0.3640739   0.6359261
35    0.0901018   0.3640739   0.6359261   0.4541756   0.5458244
36    0.0924729   0.4541756   0.5458244   0.5466485   0.4533515
37    0.0899736   0.5466485   0.4533515   0.6366221   0.3633779
38    0.0829950   0.6366221
Time will tell
7craps
7craps
7craps
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  • Posts: 1977
Joined: Jan 23, 2010
June 25th, 2013 at 2:26:25 PM permalink
Quote: Walkinshaw30t

For the example you gave of 74 spins-
if i was to make 74 spins and write down results of which sd bracket the results fell into and repeat this 100 times,
can i guarantee that 68% will be within 1 standard deviation, 95% within 2 SD etc

Nice question.
No guarantee when we deal with
the probability of a random event or many random events
unless the probability for success or failure equals 1 or 0.

Quote: Walkinshaw30t

If not what sample size would be required to achieve the expected SD results?

This is an advanced question.
The answer depends on what 'error' one is willing to accept because we are dealing, in these gambling examples,
with Independent and identically distributed random variables
http://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables

68% +/- 10% error (58% - 78%)
68% +/- 1% error (67% - 69%)
requires a 100 times larger sample size than the 10% error
68% +/- .1% error (67.9% - 68.1%)
requires a 100 times larger sample size than the 1% error
Quote: Walkinshaw30t

What would the variance be for 100 samples of 74 spins?

100 sessions of 74 spins each.
We track red at 18/37

You want to know how many sessions would fall in that 1SD range.
We keep it simple and use 68% as the probability for 1SD for 1 event.
This has been proven in many prob and stats book. Used here without any proof.

Again, this is a basic binomial probability distribution type question.
where
N = # of trials
P = Prob of success (is a constant) = 68%

N*P = EV = 68 (sessions expected to fall into the 1SD range)
Variance = N*P*(1-P)
I get 21.76
SD = square root of the variance I get 4.664761

Your 1SD range over 100 sessions would be 68 + and - 4.664761
63.335238 to 72.664761
I even ran a simulation using WinStats
Thanks again Rick Parris
 I used 68.2% as the probability because I can      
group         middle     freq  freq/100
----------------------------------------------
45.50 <= x < 46.50    46.00        2     0.00%
46.50 <= x < 47.50    47.00        5     0.00%
47.50 <= x < 48.50    48.00       11     0.00%
48.50 <= x < 49.50    49.00       30     0.00%
49.50 <= x < 50.50    50.00       57     0.01%
50.50 <= x < 51.50    51.00      134     0.01%
51.50 <= x < 52.50    52.00      273     0.03%
52.50 <= x < 53.50    53.00      535     0.05%
53.50 <= x < 54.50    54.00     1012     0.10%
54.50 <= x < 55.50    55.00     1831     0.18%
55.50 <= x < 56.50    56.00     3085     0.31%
56.50 <= x < 57.50    57.00     5197     0.52%
57.50 <= x < 58.50    58.00     7891     0.79%
58.50 <= x < 59.50    59.00    12500     1.25%
59.50 <= x < 60.50    60.00    18344     1.83%
60.50 <= x < 61.50    61.00    25693     2.57%
61.50 <= x < 62.50    62.00    34605     3.46%
62.50 <= x < 63.50    63.00    44998     4.50%
63.50 <= x < 64.50    64.00    55828     5.58%
64.50 <= x < 65.50    65.00    65865     6.59%
65.50 <= x < 66.50    66.00    75514     7.55%
66.50 <= x < 67.50    67.00    81316     8.13%
67.50 <= x < 68.50    68.00    85331     8.53%
68.50 <= x < 69.50    69.00    84564     8.46%
69.50 <= x < 70.50    70.00    80606     8.06%
70.50 <= x < 71.50    71.00    72816     7.28%
71.50 <= x < 72.50    72.00    63378     6.34%
72.50 <= x < 73.50    73.00    51571     5.16%
73.50 <= x < 74.50    74.00    40655     4.07%
74.50 <= x < 75.50    75.00    30246     3.02%
75.50 <= x < 76.50    76.00    21284     2.13%
76.50 <= x < 77.50    77.00    14459     1.45%
77.50 <= x < 78.50    78.00     9049     0.90%
78.50 <= x < 79.50    79.00     5382     0.54%
79.50 <= x < 80.50    80.00     2954     0.30%
80.50 <= x < 81.50    81.00     1605     0.16%
81.50 <= x < 82.50    82.00      767     0.08%
82.50 <= x < 83.50    83.00      353     0.04%
83.50 <= x < 84.50    84.00      148     0.01%
84.50 <= x < 85.50    85.00       61     0.01%
85.50 <= x < 86.50    86.00       29     0.00%
86.50 <= x < 87.50    87.00       10     0.00%
87.50 <= x < 88.50    88.00        4     0.00%
88.50 <= x < 89.50    89.00        2     0.00%

----------------------------------------------
                   grouped data
items:                  1000000   

minimum value:               46.00
first quartile:              65.00
median:                      68.00
third quartile:              71.00
maximum value:               89.00

mean value:                  68.21
midrange:                    67.50

range:                       43.00
interquartile range:          6.00
mean abs deviation:           3.72

sample variance (n):         21.68
sample variance (n-1):       21.68
sample std dev (n):           4.66
sample std dev (n-1):         4.66

----------------------------------------------
                  cumulative
----------------------------------------------
45.50 <= x < 46.50    46.00        2     0.00%
46.50 <= x < 47.50    47.00        7     0.00%
47.50 <= x < 48.50    48.00       18     0.00%
48.50 <= x < 49.50    49.00       48     0.00%
49.50 <= x < 50.50    50.00      105     0.01%
50.50 <= x < 51.50    51.00      239     0.02%
51.50 <= x < 52.50    52.00      512     0.05%
52.50 <= x < 53.50    53.00     1047     0.10%
53.50 <= x < 54.50    54.00     2059     0.21%
54.50 <= x < 55.50    55.00     3890     0.39%
55.50 <= x < 56.50    56.00     6975     0.70%
56.50 <= x < 57.50    57.00    12172     1.22%
57.50 <= x < 58.50    58.00    20063     2.01%
58.50 <= x < 59.50    59.00    32563     3.26%
59.50 <= x < 60.50    60.00    50907     5.09%
60.50 <= x < 61.50    61.00    76600     7.66%
61.50 <= x < 62.50    62.00   111205    11.12%
62.50 <= x < 63.50    63.00   156203    15.62%
63.50 <= x < 64.50    64.00   212031    21.20%
64.50 <= x < 65.50    65.00   277896    27.79%
65.50 <= x < 66.50    66.00   353410    35.34%
66.50 <= x < 67.50    67.00   434726    43.47%
67.50 <= x < 68.50    68.00   520057    52.01%
68.50 <= x < 69.50    69.00   604621    60.46%
69.50 <= x < 70.50    70.00   685227    68.52%
70.50 <= x < 71.50    71.00   758043    75.80%
71.50 <= x < 72.50    72.00   821421    82.14%
72.50 <= x < 73.50    73.00   872992    87.30%
73.50 <= x < 74.50    74.00   913647    91.36%
74.50 <= x < 75.50    75.00   943893    94.39%
75.50 <= x < 76.50    76.00   965177    96.52%
76.50 <= x < 77.50    77.00   979636    97.96%
77.50 <= x < 78.50    78.00   988685    98.87%
78.50 <= x < 79.50    79.00   994067    99.41%
79.50 <= x < 80.50    80.00   997021    99.70%
80.50 <= x < 81.50    81.00   998626    99.86%
81.50 <= x < 82.50    82.00   999393    99.94%
82.50 <= x < 83.50    83.00   999746    99.97%
83.50 <= x < 84.50    84.00   999894    99.99%
84.50 <= x < 85.50    85.00   999955   100.00%
85.50 <= x < 86.50    86.00   999984   100.00%
86.50 <= x < 87.50    87.00   999994   100.00%
87.50 <= x < 88.50    88.00   999998   100.00%
88.50 <= x < 89.50    89.00  1000000   100.00%


Say one flips a fair coin 10 times.
We expect 5 Heads and 5 Tails, most would agree.
But if 1 million people did this experiment we would find that less than 25% of them actually got exactly 5 Heads.
More got 6 or higher or 4 or less
Distributions of random events

"Many events can't be predicted with total certainty.
The best we can say is how likely they are to happen, using the idea of probability."
http://www.mathsisfun.com/data/probability.html

Excellent
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winsome johnny (not Win some johnny)
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