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Classic and simple problem: Three indistinguishable dice (6-sided) are rolled at the same time. What's the probability that we have rolled three of a kind?

Now I know that the answer is $\frac{1}{36}$, as explained for example in a thread on this site here.

But when I think about it from the following way, I get another solution. Where's my error?

Order does not count, so the space of possible outcomes is $\Omega = \{(x_1 \leq x_2 \leq x_3)\ \lvert\ x_i \in \{1, \cdots, 6\} \}$ with $\lvert\Omega\rvert = \binom{6+3-1}{3} = 56$.
The set of outcomes we want is $\{(1,1,1),\cdots(6,6,6)\}$ with cardinality 6. So the probability of rolling three of a kind is $\frac{6}{56}$.

Can you point me to my thinking error? Thanks in advance!

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johnnycrab
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    Under the "reasonable" model of dice throwing (results independent, $1$ to $6$ equally likely for each die) your sample space does not consist of equally likely outcomes. The same argument would show that when you toss two coins, the probability of two heads is $1/3$. – André Nicolas Apr 11 '16 at 21:42
  • Can you explain why not? My intuition was: Take ${1, \cdots, 6}^3$, but here we distinguish the dice. So I tried to identify the problem with an urn model where one pulls three balls (balls numbered 1 to 6) with putting the ball back after each pull and order irrelevant. Why can't I draw this parallel? – johnnycrab Apr 11 '16 at 21:50
  • @AndréNicolas Okay, so in order to make the outcomes equally likely, we have to "distinguish" the dice in the form of the sample space ${1, \cdots, 6}^3$? But wouldn't this say "this is the first one, this is the second one, this is the third one", when only the numbers showing up are important? – johnnycrab Apr 11 '16 at 21:59
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    If "first, second, third" makes you think there might be a built-in conceptual error, suppose the dice are coloured blue, white, and red. The probability should not depend on the colouring. – André Nicolas Apr 11 '16 at 22:03
  • @AndréNicolas Got it. Thank you. :) – johnnycrab Apr 11 '16 at 22:08
  • You are welcome. Variants of the question have been asked and answered a number of times on MSE, so I was reluctant to write yet another answer. – André Nicolas Apr 11 '16 at 22:12

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