Q: Help to understand simple probability problem from Kahneman 'Thinking fast and slow' I'm new to statistics and I'm trying to solve this easy problem. Unfortunately I can't find a solution that would confirm that scenario with 4 same coloured marbles is six times more likely than the other one.
Here's the problem from Kahneman's book:

Imagine a large urn filled with marbles. 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, records the number of red
  balls in the sample, throws the balls back into the urn, and then 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.”

My solution (I assumed that there are 10k marbles in total):
probability of drawing 4 marbles with same colour:
5000/10000 * 4999/9999 * 4998/9998 * 4997/9997 = 6,25%

there are two combinations (rrrr, wwww), so:
6,25% * 2 = 12,49%

probability of drawing 2 marbles of each colour:
5000/10000 * 4999/9999 * 5000/9998 * 4999/9997 = 6,25%

here we have six combinations (wwrr, rrww, wrrw, rwwr, rwrw, wrwr):
6,25% * 6 = 37,51%

 A: Note that the author says

the outcome “4 red” or “4 white.”

and not

the outcome “4 red or 4 white.”

My interpretation of this is that when you get to the final step of your calculation, it is instead

there is one combination (either rrrr or wwww), so:
6,25% = 6,25%


which is very close to one-sixth of the other probability.
The reason it's not exactly one-sixth is basically because $10\,000\neq 9\,999$. The urn has a finite number of balls, and therefore the probability distribution for the second, third and fourth balls depend on the balls before them.
This error becomes smaller and smaller the more balls the urn contains. Try with a million instead, and see.
A: Firstly, keep it general. Let the number of red and white marbles be $n$ each (making a total of $2n$ marbles).
The number of ways of drawing $4$ marbles from the urn is $\binom {2n} {4}$. This is the denominator for both the probabilities we are going to compute next (and therefore we can ignore it when considering the ratio between the two probabilities).
Now the number of ways of getting $2$ red and $2$ white marbles when drawing $4$ marbles is
$$N(2R2W) = \binom n 2 \binom n 2 = \dfrac{n^2(n - 1)^2}{4}.$$
The number of ways of getting $4$ red (or, alternatively, $4$ white) marbles is $$N(4R) = N(4W) = \binom n 4 = \dfrac{n(n - 1)(n - 2)(n - 3)}{24}.$$
The ratio of the probabilities is the same as the ratio of the above numbers (since the denominator will be the same for both probabilities). Thus,
$$\dfrac{N(2R2W)}{N(4R)} = 6 \dfrac{n(n - 1)}{(n - 2)(n - 3)} \approx 6\ \text{for large n}.$$
In particular, your interpretation of "as often as the outcome “4 red” or “4 white”" is "$4$ marbles of the same colour", but apparently that is not what the author meant (and it's his fault, because he should've at least said "the outcomes", indicating that he means each alternative possibility).
