Alternative to the Binomial PMF? Since hypergeometric distribution is the without replacement version of the Binomial distribution why can't we replace the combinations in the Hypergeometric PMF with combinations with replacement and expect the same result with the standard Binomial PMF ?
To elaborate this is the regular combination formula counting unordered without replacement :
$$C(n,r)= \frac{n!}{r!(n−r)!}$$
And this is the combination with replacement formula counting unordered with replacement:
$$C^R(n,r)= \frac{(n+r−1)!}{r!(n−1)!}$$
and this is the hypergeometric distribution PMF:
For $X\sim\operatorname{HGeom}(w,b,n)$
$$P(X=k)= \frac{C(w,k)C(b,n-k)}{C(w+b,n)}$$
My question is, since binomial and hypergeometric are both fixed number of trials where Binomial is counting with replacement and Hypergeometric is counting without replacement, why doesn't altering the Hypergeometric formula to be
$$P(X=k)= \frac{C^R(w,k)C^R(b,n-k)}{C^R(w+b,n)}$$
give us the PMF of the Binomial Distribution ? What am I missing ?
 A: If you have for example a bag with $6$ red balls and $4$ blues balls and you want the probability that drawing $3$ balls gives you $2$ red and $1$ blue, then

*

*without replacement this is $\dfrac{C(6,2)C(4,1)}{C(10,3)} = \dfrac{60}{120}$ or $\dfrac{360}{720} =0.5$

*with replacement this is $C(3,2)\dfrac{C(6,1)}{C(10,1)}\dfrac{C(6,1)}{C(10,1)}\dfrac{C(4,1)}{C(10,1)} = \dfrac{432}{1000}=0.432$
which are very different calculations and there is no real correspondence between the two.  The "with replacement" binomial version has a succession of products of small fractions which are either all the same or $1 $ minus the same term and this really cannot be consolidated into ratio of single binomial coefficients.
A: One thing you're missing is that you need a logical reason why something should be true before wondering why it isn't.
The binomial distribution is that of the number of successes in a non-random number of independent trials with the same probability of success on each trial. Suppose we want the probability of exactly two successes in six trials, with probability $0.3$ of success on each trial.
\begin{align}
ssffff & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
sfsfff & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
sffsff & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
sfffsf & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
sffffs & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
fssfff & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
fsfsff & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
fsffsf & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
fsfffs & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
ffssff & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
ffsfsf & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
ffsffs & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
fffssf & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
fffsfs & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4 \\
ffffss & \quad \longleftarrow \quad \text{The probability of this is } 0.3^2\times0.7^4
\end{align}
Their are $15$ of these: $\binom 6 2 = 15.$ The probability that one of these happens is
$$
\overbrace{0.3^2\times0.7^4 + \cdots\cdots + 0.3^2\times0.7^4}^{15 \text{ terms}} = 15 \times 0.3^2\times0.7^4
$$
The reason $15$ appears is that it is the number of ways to choose $2$ out of $6$ without replacement. Those $15$ ways appear in the list above.
