Problem with urn I'm reading Thinking fast and slow and there's this problem I cannot get a correct answer. Here it is: 

Half the marbles are red, half are white. Next imagine a very patient person (or a robot) who blindly draws 4 marbles from the urn, and does it all again, many times. If you summarize the results, you will find that the outcome "2 red, 2 white" occurs (almost exactly) 6 times as often as the outcome "4 red" or "4 white". 

Could you please explain and show why it's 6 times? 
 A: Imagine there are $r$ red and $w$ white marbles in the urn. Let $\text{R}$ be the event "a red marble is drawn" and $\text{W}$ be the event "a white marble is drawn". 4 balls are drawn, without replacement between the draws.
Then $P(\text{RRRR}) = \frac{r}{r+w}\cdot \frac{r-1}{r+w-1}\cdot \frac{r-2}{r+w-2}\cdot \frac{r-3}{r+w-3}$.
However, if $r$ and $w$ are both very large (much, much larger than 4 ... perhaps something more like 400 each, say), then each of those fractions is very close to $\frac{1}{2}$, and the product is about $\frac{1}{16}$. 
By symmetry, the calculation for $\text{WWWW}$ is similar.
Indeed, so would it be for WWWR, or RWWR or any other specific order of R and W over 4 marbles.
e.g. $P(RRRW)=\frac{r}{r+w}\cdot \frac{r-1}{r+w-1}\cdot \frac{r-2}{r+w-2}\cdot \frac{w}{r+w-3}$.
But $P(\text{two R, two W, in any order}) = P(RRWW)+P(RWRW)+ P(RWWR)\\
\hspace{6.5cm}+P(WRRW)+P(WRWR)+P(WWRR)$
When $r$ and $w$ are large, each of those 6 arrangements has about the same chance as $P(RRRR)$ and $P(WWWW)$, close to $\frac{1}{16}$.
So - roughly speaking, because there are lots of both kinds, in equal numbers, and there are 6 ways of arranging two red and two white marbles in order, there's about six times the chance of observing two red and two white (about 6/16) in any order as four red (about 1/16) or four white (about 1/16).
