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The formula for binomial distrubtion looks like that:

enter image description here

But is it possible to calculate proper results using slightly different formula?

The number of possible outcomes can be calculated using formula for variations with repetitions:

$m^N$

For example if I throw a coin 3 times, I get 8 possible outcomes. m=2 N=3 $2^3=8$

It looks to me like binomial distribution can be simply calculated by using this formula: $P(x)=\frac{{N\choose x}}{m^N}$

N = number of trials x = number of success m = number of outcomes from one trial

But I'm not sure. Maybe, it works only when there is a 50% probability of a success?

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  • $\begingroup$ Did you try it at any other value of $p$? Like $p=0.31$, say? It would seem the obvious thing to do when one particular value revealed a dramatic simplification -- to try another, less "special" value. $\endgroup$
    – Glen_b
    Sep 4 '17 at 23:18
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Yes this is correct assuming that $p=0.5$ because $m^N = 2^N$ and $p^x (1-p)^{N-x}=p^N= \frac{1}{2^N}$ and so the numerator and denominator cancel. This is obviously not true for other values of $p$.

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