A bag has marbles of 4 colors: red, white, blue and green. Assume that if we take four marbles out at random (without replacement), each of the following is equally likely:

- One marble of each color is chosen
- One white, one blue and two reds are chosen
- One blue and three reds are chosen
- All four are red

What is the smallest number of possible marbles in the bag?

The answer is 21.

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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

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