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Beating The Lines
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Wondering if you guys can help me out. I've been tracking something that has gone 25-16 the past two weeks and need to know the % of the time that this would happen having a 50-50 chance on each game. I used to have the formula and used it a lot last year but I can't find it anymore.

If you know what I'm talking about please post the formula for me. Thanks!
 

EV Whore
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I have to be missing something, but wouldn't it just be (25/41) = 61%?

I assume by 50/50 chance you mean it has an equal probability of occuring on any given play...
 

Beating The Lines
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I have to be missing something, but wouldn't it just be (25/41) = 61%?

I assume by 50/50 chance you mean it has an equal probability of occuring on any given play...

I mean the odds of something that has a 50/50 chance hitting at 61% after doing it 41 times. So like flipping a coin and landing on heads 25 out of 41 times. Theres a formula to determine how often this will happen by chance. Like 5 times out of 100 or something like that. Once I have enough data to where it will happen at less than 1 in 100 times by chance than I usually feel pretty good about the system and start betting the games that fit it.

So just trying to see how much bigger I need my sample size to be before I can feel comfortable betting these.
 

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If you have a TI-83 calculator, click on DISTR (which is 2nd function VARS) scroll down to binompdf and enter:

binompdf(41, 0.5, 25)
... this is the probability that something will hit EXACTLY 25 times in 41 trials when the probability of each individual event is 0.5 (i.e. 50%) = the answer I got was 4%

there is also a command called BINOMCDF

binomcdf (41, 0.5, 24)
... is the probability that something will hit up to 24 times in 41 trials when the probability of each individual event is 0.5 (i.e. 50%) ... which is the cumulative probability ... This is more useful because it is the probability that it will hit no more than 24 times = the answer I got was 92%

... which means there is a 8% probability that you will hit more than 24/41 plays when you're just flipping coins.
 

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I'm not sure of the formula anymore, but if I ran this program right I came up with a 10.551% chance of getting at least 25 wins out of 41, assuming chances of winning are 50%

maybe someone else can verify, or maybe I'll be able to find the formula
 

Rx. Senior
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http://faculty.vassar.edu/lowry/binomialX.html

plug in:

n= 26
k= 41
p= .5

use the link to determine probabilities. Here's the formula in case you want to see it.


This pagewill calculate and/or estimate binomial probabilities for situations of the general "k out of n" type, where k is the number of times a binomial outcome is observed or stipulated to occur, p is the probability that the outcome will occur on any particular occasion, q is the complementary probability (1-p) that the outcome will not occur on any particular occasion, and n is the number of occasions.

For example: In 100 tosses of a coin, with 60 "heads" outcomes observed or stipulated to occur among the 100 tosses,
n = 100 [the number of opportunities for a head to occur]
k = 60 [the stipulated number of heads]
p = .5 [the probability that a head will occur on any particular toss]
q = .5 [the probability that a head will not occur on any particular toss]


--------------------------------------------------------------------------------

Method 1: If n1000, exact binomial probabilities will be calculated through repeated applications of the standard binomial formulaQ
P(k out of n) =
n!
--------------------------------------------------------------------------------
k!(n-k)!
(pk)(qn-k)



In principle, Method 1 is preferable in all cases, since it involves direct calculation of exact binomial probabilities. Its limitation is that it is not computationally feasible with very large samples. The programming on this page is capable of performing the calculation up through n=1000.

--------------------------------------------------------------------------------

Method 2: If np5 and nq5, binomial probabilities will be estimated by way of the binomial approximation of the normal distribution, according to the formulaQ
z =
(k— )±.5
--------------------------------------------------------------------------------
-

where: - = np
[the mean of the binomial sampling distribution]
- = sqrt[npq]
[the standard deviation of the binomial sampling distribution]


--------------------------------------------------------------------------------

Method 3: If n≥150 and the mean (np) and variance (npq) of the binomial sampling distribution are within 10% of each other, binomial probabilities will be estimated through repeated applications of the Poisson probability function

TP(k out of n) =
(e-np)(npk)
--------------------------------------------------------------------------------
k!

where e = the base of the natural logarithms.

The defining characteristic of a Poisson distribution is that its mean and variance are identical. In a binomial sampling distribution, this condition is approximated as p becomes very small, providing that n is relatively large. The programming on this page permits the Poisson procedure to be performed whenever np and npq are within 10% of each other, providing that n≥150. Do keep in mind, however, that the results of the Poisson procedure are only approximations of the true binomial probabilities, valid only in the degree that the binomial mean and variance are very close.



--------------------------------------------------------------------------------

To proceed, enter the values for n, k, and p into the designated cells below, and then click the «Calculate» button. (The value of q will be calculated and entered automatically). The value entered for p can be either a decimal fraction such as .25 or a common fraction such as 1/4. Whenever possible, it is better to enter the common fraction rather than a rounded decimal fraction: 1/3 rather than .3333; 1/6 rather than .1667; and so forth.

--------------------------------------------------------------------------------
n k p q








--------------------------------------------------------------------------------
Parametersof binomial sampling distribution:
--------------------------------------------------------------------------------

mean


variance


standard deviation



--------------------------------------------------------------------------------

binomial z-ratio

(if applicable)




--------------------------------------------------------------------------------


Method 1. exact binomial calculation


Method 2. approximation via normal


Method 3. approximation via Poisson




Method 1. exact binomial calculation


Method 2. approximation via normal


Method 3. approximation via Poisson




Method 1. exact binomial calculation


Method 2. approximation via normal


Method 3. approximation via Poisson



--------------------------------------------------------------------------------
For hypothesis testing


One-Tail
Two-Tail

Method 1. exact binomial calculation



Method 2. approximation via normal



Method 3. approximation via Poisson





--------------------------------------------------------------------------------
 

Beating The Lines
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If you have a TI-83 calculator, click on DISTR (which is 2nd function VARS) scroll down to binompdf and enter:

binompdf(41, 0.5, 25)
... this is the probability that something will hit EXACTLY 25 times in 41 trials when the probability of each individual event is 0.5 (i.e. 50%) = the answer I got was 4%

there is also a command called BINOMCDF

binomcdf (41, 0.5, 24)
... is the probability that something will hit up to 24 times in 41 trials when the probability of each individual event is 0.5 (i.e. 50%) ... which is the cumulative probability ... This is more useful because it is the probability that it will hit no more than 24 times = the answer I got was 92%

... which means there is a 8% probability that you will hit more than 24/41 plays when you're just flipping coins.

Thanks thats what I'm looking for. Is there anything on there to tell the probability of hitting 24 times or more out of 41?

Thanks again.
 

New member
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If you want to do it manually, the formula is:

(nCx)*(p^x)*(1-p)^(n-x)

so in this case,

41C25*0.5^41 ...

and 41C25 is 41!/(25!*16!)

and it's closer to 4.69%

If you want to do the same thing for 0,1,2,.... all the way up to 24 and add them up, you get binomcdf(41,0.5,24)
 

Beating The Lines
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I'm not sure of the formula anymore, but if I ran this program right I came up with a 10.551% chance of getting at least 25 wins out of 41, assuming chances of winning are 50%

maybe someone else can verify, or maybe I'll be able to find the formula

Ok that looks about right. Thank guys.
 

New member
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Thanks thats what I'm looking for. Is there anything on there to tell the probability of hitting 24 times or more out of 41?

Thanks again.

The probability of hitting it EXACTLY 24 times is 0.069 (or 6.9% or 7%), so the probability of hitting it up to 23 times would be 92% - 7% = 85%, so the probability of hitting 24 times or more would be 15%
 

New member
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I'm not sure of the formula anymore, but if I ran this program right I came up with a 10.551% chance of getting at least 25 wins out of 41, assuming chances of winning are 50%

maybe someone else can verify, or maybe I'll be able to find the formula

You are right, goodcall ... I re-did on my calculator it and I mistakenly wrote 92% when I should have written 94.13%

... and thanks for the site, raiders
 

Rx. Senior
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sorry Cincy, plugged in wrong numbers first time. Went with 25 or more. I edited post above.
 

New member
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Well raiders, he wanted 25 out of 41 at first and then he wanted 24 out of 41 ...

.. but thanks for the site. That is pretty neat ... I was doing it on my calculator and it was a pain.
 

Beating The Lines
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Well raiders, he wanted 25 out of 41 at first and then he wanted 24 out of 41 ...

.. but thanks for the site. That is pretty neat ... I was doing it on my calculator and it was a pain.

25 out of 41. It's 25-16 since I've been tracking.

Sorry guys I typed it wrong about half way through the thread.
 

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