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>> Wikipedia claims that for large N, the binomial distribution
>> approximates the normal distribution.
>
> Yes, it seems my parameter values fit in ok for this approximation to be
> pretty accurate.
Central Limit Theorum FTW! :-D
>> I presume it's relatively easy to work out the inverse function for a
>> normal distribution...
>
> Excel has a built in function to do this, so this looks like the way to go.
Cool.
So are you *really* trying to estimate coin flips? Or is there some more
complex problem you're trying to solve here?
Also... I'm impressed that I managed to find an actual solution to this
using nothing but Wipikedia and an incomplete understanding of probability.
PS. Apparently Wolfram Alpha can plot all these distributions for you if
you type in something suggestive like "binomial distribution, n=100, p=0.2".
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> So are you *really* trying to estimate coin flips? Or is there some more
> complex problem you're trying to solve here?
Not really more complex, just a game that is played repeatedly and you have
a certain % chance of winning (say 57%). I just want to know if you play
10,100, 1000 times what is the *likely* number of wins in some kind of worst
case and best case scenario. ie it's obviously possible to win every single
game, but that is *really* unlikely, I want to know how many you'd win with
90% or 99% confidence.
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scott wrote:
>> So are you *really* trying to estimate coin flips? Or is there some
>> more complex problem you're trying to solve here?
>
> Not really more complex, just a game that is played repeatedly and you
> have a certain % chance of winning (say 57%). I just want to know if
> you play 10,100, 1000 times what is the *likely* number of wins in some
> kind of worst case and best case scenario. ie it's obviously possible
> to win every single game, but that is *really* unlikely, I want to know
> how many you'd win with 90% or 99% confidence.
Right. So it's ye olde "2SD = 95% confidence" rule then.
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Invisible wrote:
> I presume it's relatively easy to work out the inverse function for a
> normal distribution...
Actually it's impossible to analytically compute. Fortunately many
pieces of software have a built in function to numerically compute it.
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>> I presume it's relatively easy to work out the inverse function for a
>> normal distribution...
>
> Actually it's impossible to analytically compute.
Define "analytically".
Last time I checked, it's impossible to compute the square root of a
number "exactly" - and yet it's trivially easy to approximate it to any
desired accuracy.
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Invisible wrote:
>>> I presume it's relatively easy to work out the inverse function for a
>>> normal distribution...
>>
>> Actually it's impossible to analytically compute.
>
> Define "analytically".
As a closed-form expression. As I said, it's still possible to compute
numerically.
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Kevin Wampler wrote:
> Invisible wrote:
>>>> I presume it's relatively easy to work out the inverse function for
>>>> a normal distribution...
>>>
>>> Actually it's impossible to analytically compute.
>>
>> Define "analytically".
>
> As a closed-form expression. As I said, it's still possible to compute
> numerically.
And Sqrt[2] erf^-1 (2z - 1) doesn't count because...?
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Invisible wrote:
> Kevin Wampler wrote:
>> Invisible wrote:
>>>>> I presume it's relatively easy to work out the inverse function for
>>>>> a normal distribution...
>>>>
>>>> Actually it's impossible to analytically compute.
>>>
>>> Define "analytically".
>>
>> As a closed-form expression. As I said, it's still possible to
>> compute numerically.
>
> And Sqrt[2] erf^-1 (2z - 1) doesn't count because...?
erf isn't an elementary function. See:
http://en.wikipedia.org/wiki/Closed-form_expression
http://en.wikipedia.org/wiki/Error_function#Taylor_series
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>>>>>> I presume it's relatively easy to work out the inverse function
>>>>>> for a normal distribution...
>>>>>
>>>>> Actually it's impossible to analytically compute.
>>>>
>>>> Define "analytically".
>>>
>>> As a closed-form expression. As I said, it's still possible to
>>> compute numerically.
>>
>> And Sqrt[2] erf^-1 (2z - 1) doesn't count because...?
>
> erf isn't an elementary function. See:
>
> http://en.wikipedia.org/wiki/Closed-form_expression
>
> http://en.wikipedia.org/wiki/Error_function#Taylor_series
Right. So what you're actually saying is that you can't compute the
confidence interval exactly using a finite number of applications of a
particular arbitrarily chosen set of functions - the elementary functions.
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Invisible wrote:
>>>> As a closed-form expression.
>
> Right. So what you're actually saying is that you can't compute the
> confidence interval exactly using a finite number of applications of a
> particular arbitrarily chosen set of functions - the elementary functions.
That is indeed the definition.
This obviously doesn't matter for scott's case since he just wants a
numeric answer and Excel seems to have erf built in, I just wanted to
point out that far from being "relatively easy" to work out, the
cumulative distribution function of a Gaussian is generally the first
function people encounter which is impossible to actually compute in
closed form (normally this will come up in first or second year calculus
at some point). I'll agree that at first glance it certainly *looks*
like it would be simple to compute though, which is part of why it's
interesting.
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