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I believe this is a duplicate of this post, but I think someone can easily clarify my misunderstanding of the Beta pdf:

$f(x)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}(x)^{a-1}(1-x)^{b-1}$ for $x\in[0,1]$ and $a,b>0$

where $\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}=\frac{(a+b-1)!}{(a-1)!*(b-1)!*1!}$ if $a$ and $b$ are integers. I'm not quite sure what the lingering $1!$ in the denominator signifies; I noticed that $(a-1)+(b-1)\ne (a+b-1)$ so we include $1!$ in the denominator to reconcile the left hand side.

Here's an example to illustrate my confusion: if my friend flips a coin five times in another room and then tells me there were three heads and two tails, the Beta pdf tells me I only need to know the results for three of his coin flips in order to deduce the results of the remaining two coin flips because the reciprocal of the Beta function's output is $\frac{4!}{2!*1!*1!}$.

This doesn't seem quite right to me. Unless my friend tells me which three coin flips were heads, it would seem I actually need to know the results of four of his coin flips to deduce the result of his last remaining coin flip.

Could someone explain why my friend only needs to share the results for three coin flips and not four? What does the $1!$ represent?

E. Kaufman
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    Re "The Beta pdf tells me I only need to know the results for three of his coin flips:" Are you perhaps confounding the Beta distribution with the Binomial distribution?? There's no "1!" in the formula, anyway: it doesn't represent anything. Are you perhaps trying to interpret the Beta function as if it were a multinomial coefficient? – whuber Jul 03 '22 at 20:28
  • @whuber Not really. It just feels as if we should multiply $x^2(1-x)^1$ by some other* coefficient instead of $\binom{4}{2,1,1}$ in order to compute $f(x)$ but I know that's not true.

    Re "Are you perhaps trying to interpret the Beta function as if it were a multinomial coefficient?" Yes; I thought we had to if $(a-1)+(b-1)\ne(a+b-1)$

     – E. Kaufman Jul 03 '22 at 20:41
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    The essence of the Beta distribution is that its density is proportional to $x^{\alpha-1}(1-x)^{\beta-1}$ over $[0,1]$. Since that does not integrate to $1$ over $[0,1]$, you need to divide by what the integral is equal to, namely the Beta function $B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$. In the same way that the Beta density looks vaguely like the binomial probability mass function even though the distributions are very different, the Beta function looks vaguely like the reciprocal of the binomial coefficient but is in fact different – Henry Jul 03 '22 at 22:45
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    $1!=1$ so I don't see why adding it changes anything. –  Jul 04 '22 at 05:25

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