If a weighted coin is tossed N times, count the times it lands heads as H. 
The probability of landing heads is P(H).

I assume the expected value of H is N*P(H).

But how do I find out what range of H is X% likely to happen?  For example, 
I want to be able to say that in 90% of cases H will be in the range (A,B).

What I've done so far is for 0<=H<=N calculated (N choose H)/P, where P is 
the total permutations of N items in N.  This should give me the probability 
of exactly H heads coming up, yes?  I can keep a cumulative total of these 
probabilites for each H, and then figure out the range of H between (say) 
10% and 90%.  But it seems a bit long-winded, I can't figure out how to 
calculate the range directly.  Also Excel can't figure out the permutations 
when N is bigger than 150 or so, I want it to work for higher values of N.

Any ideas?  Hopefully it's something obvious!

Post a reply to this message

scott wrote:
> If a weighted coin is tossed N times, count the times it lands heads as 
> H. The probability of landing heads is P(H).

I presume by this you explicitly mean that P(H) /= 50%?

> I assume the expected value of H is N*P(H).

Yes. That would be the definition of "probability". (Assuming all the 
trails are independent.)

> But how do I find out what range of H is X% likely to happen?  For 
> example, I want to be able to say that in 90% of cases H will be in the 
> range (A,B).

Now you're talking about probability distributions. If you can decide 
what probability distribution this experiment has, you can [potentially] 
directly compute the result you're after.

Wikipedia informs me that what you're looking for is a "binomial 
distribution". According to this, we have the probability of getting 
exactly K heads as

   (N choose K) * P(H)^K * (1 - P(H))^(N-K)

More usefully for you, the cumulative probability - the probability of K 
heads OR LESS, is given by

   I[1 - P(H)] (N - K, N + K)

where I[x](a, b) represents the regularised incomplete beta function,

http://en.wikipedia.org/wiki/Regularized_incomplete_beta_function

Unless Excel has a built-in function to compute I(), I would suggest it 
would be simpler and easier to directly compute the sum over the 
individual probabilities. (Especially since you want the probability for 
a value between K1 and K2, not between K1 and 0.)

You can also compute the cumulative probability by the integral

   (N - K) (N choose K) integral[0 .. 1-P(H)] t^(N-K-1) (1 - t)^K dt

But I'm guessing Excel can't do that.

Apparently you can also estimate the maximum probability by

   exp( -2 * (N * P(H) - K)^2 / N )

(The true cumulative probability is less than or equal to this. 
Apparently for P(H) = 50% it's quite a good approximation, but I'm not 
sure for other values.)

Post a reply to this message

Invisible wrote:

> More usefully for you, the cumulative probability - the probability of K 
> heads OR LESS, is given by
> 
>   I[1 - P(H)] (N - K, N + K)
> 
> where I[x](a, b) represents the regularised incomplete beta function,
> 
> http://en.wikipedia.org/wiki/Regularized_incomplete_beta_function

http://reference.wolfram.com/mathematica/ref/BetaRegularized.html

So Wolfram Alpha will compute it, for example.

But that doesn't matter:

http://office.microsoft.com/en-us/excel/HP052090051033.aspx

Post a reply to this message

> Wikipedia informs me that what you're looking for is a "binomial 
> distribution".

Thanks - that was the key I needed!

> According to this, we have the probability of getting exactly K heads as
>
>   (N choose K) * P(H)^K * (1 - P(H))^(N-K)

That *is* useful, at least using this formula I get correct results, 
although it's still a bit clumsy.

I will continue with Excel now trying to make it a bit neater.

Post a reply to this message

scott wrote:
>> Wikipedia informs me that what you're looking for is a "binomial 
>> distribution".
> 
> Thanks - that was the key I needed!
> 
>> According to this, we have the probability of getting exactly K heads as
>>
>>   (N choose K) * P(H)^K * (1 - P(H))^(N-K)
> 
> That *is* useful, at least using this formula I get correct results, 
> although it's still a bit clumsy.
> 
> I will continue with Excel now trying to make it a bit neater.

See my other post; Excel has a function to directly compute a binomial 
distribution, apparently. (And cumulatively, if you wish.)

Post a reply to this message

> See my other post; Excel has a function to directly compute a binomial 
> distribution, apparently. (And cumulatively, if you wish.)

I need the inverse function though, I want to tell it the cumulative 
probability (eg 10% and 90%) and it tell me how many successful trials that 
relates to.  There seems to be no BINOMDISTINV function.

Post a reply to this message

scott wrote:
>> See my other post; Excel has a function to directly compute a binomial 
>> distribution, apparently. (And cumulatively, if you wish.)
> 
> I need the inverse function though, I want to tell it the cumulative 
> probability (eg 10% and 90%) and it tell me how many successful trials 
> that relates to.  There seems to be no BINOMDISTINV function.

Wikipedia claims that for large N, the binomial distribution 
approximates the normal distribution. (If you plot the normal 
distribution with expected value N * P(H) and varience N * P(H) * (1 - 
P(H)).)

This approximation works best for large N and for P(H) near to 50%.

I presume it's relatively easy to work out the inverse function for a 
normal distribution...

Post a reply to this message

> 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.

> 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.

Post a reply to this message

>> 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".

Post a reply to this message

> 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.

Post a reply to this message