50 Followers Problem!

Let there be $r$ red, $w$ white, $b$ blue and $g$ green marbles in the bag. For each of the given scenarios to be equally likely to occur, they need to each be represented by the same number of outcomes. These outcome totals are:

1.) $\dbinom{r}{1} \dbinom{w}{1} \dbinom{b}{1} \dbinom{g}{1} = rwbg,$

2.) $\dbinom{r}{2} \dbinom{w}{1} \dbinom{b}{1} = \dfrac{r(r - 1)wb}{2},$

3.) $\dbinom{r}{3} \dbinom{b}{1} = \dfrac{r(r - 1)(r - 2)b}{6},$ and

4.) $\dbinom{r}{4} = \dfrac{r(r - 1)(r - 2)(r - 3)}{24}.$

Equating 1.) and 2.) gives us that $g = \dfrac{r - 1}{2}.$

Equating 2.) and 3.) gives us that $w = \dfrac{r - 2}{3}.$

Equating 3.) and 4.) gives us that $b = \dfrac{r - 3}{4}.$

Thus $r$ must be each of $1 \pmod{2}, 2 \pmod{3}$ and $3 \pmod{4}.$ The first of these equivalences states that $r$ is odd. The second equivalence gives us that $r$ can be $2, 5, 8, 11, 14, 17, .....$ And the third equivalence gives us that $r$ can be $3, 7, 11, 15, 19, .....$ The least value of $r$ consistent with all these requirements is $r = 11,$ and this will in turn lead to the least possible values for $w,b,g.$

For $r = 11$ we thus have $w = 3, b = 2$ and $g = 5,$ giving us

$r + w + b + g = 11 + 3 + 2 + 5 = \boxed{21}.$