I'm looking for an equation that will work for any positive integer n. I know that multiple coin flips approaches a normal distribution, but I'm looking for answers that deal specifically with random bits or coin flips. Basically, you can't have 4.2 coins land on heads.
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It sounds like you're asking for an exact closed form for the sum $\sum_{k=\lceil 3n/4 \rceil}^{n} {{n}\choose{k}}$. I don't think there is one. – mjqxxxx May 25 '15 at 22:47
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By definition it is: $$\begin{align} \Bbb P \quad & = \quad \sum_{k=\lceil 3n/4\rceil}^n \binom{n}{k}p^k(1-p)^{n-k} \\[1ex] & = \quad \tfrac 1{2^n}\sum_{k=\lceil 3n/4\rceil}^n \binom{n}{k} & : p=1/2 \end{align}$$
And that's why Normal approximations get used.
Graham Kemp
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Yes, I'm aware that that would work, but is there any way of doing it without a sigma function? – Finn Flaherty May 25 '15 at 22:13
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Is there any way I can answer my original question without a sigma function? – Finn Flaherty May 25 '15 at 22:15
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btw, if you want to estimate the probability of getting exactly 1/4 heads, the answer is here: http://math.stackexchange.com/questions/1208016/show-that-r-kn-n-le-binomknn-r-kn-where-r-k-dfrackkk-1k It turns out that $\binom{4n}{n} \approx \sqrt{\dfrac{2}{3\pi n}}\left(\dfrac{4^4} {3^3}\right)^{n}$ so that $\frac1{2^{4n}}\binom{4n}{n} \approx \sqrt{\dfrac{2}{3\pi n}}\left(\dfrac{2^4} {3^3}\right)^{n}$. – marty cohen May 25 '15 at 22:19
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Yes, I see why the approximation works, but in this instance, I don't need and can't use an approximation, which brings us back to the point that this is not the problem I need help with. – Finn Flaherty May 25 '15 at 22:25
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In other words, $p(n/4) \approx \sqrt{c/n} d^n$ where $c = 8/(3\pi)$ and $d = 2/(3^{3/4}) \approx 0.877$. – marty cohen May 25 '15 at 22:26
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The answer to this question will not be a number, the answer will be a function. – Finn Flaherty May 25 '15 at 22:32