Find the distribution of Y given by the number of balls received by the first urn

Question

Four urns are to be filled by two indistinguishable balls. Find the distribution of the random variable $Y =$ "# balls in the first urn".

Solution. Clearly, $Y = \{0,1,2\}$. Let $E_i$ be the event "the i-th ball goes to urn 1", $i=1,2$. Then $$\Pr(Y=0) = \Pr(\bar E_1\cap\bar E_2) = \frac{3}{4}\frac{3}{4} = \frac{9}{4^2}\,,$$ $$\Pr(Y=1) = \Pr(\{\bar E_1\cap E_2\}\cup\{E_1\cap \bar E_2\}) = \frac{3}{4}\frac{1}{4}+\frac{1}{4}\frac{3}{4} = \frac{6}{4^2}\,,$$ and $$\Pr(Y=2) = 1-\frac{15}{4^2} = \frac{1}{4^2}\,.$$

Now I'd like to compute such probabilities in a different way, using the classical definition, that is "# events in favour of E over the overall # events". The sample space is $$\mathcal{S} = \{2000, 0200, 0020, 0002,1100, 1010, 1001, 0110, 0101, 0011\}\,,$$ where "2000" means that the first urn receives 2 balls and the others receive no balls. Clearly, $\#\mathcal{S} = 10$. Now we see that the event $Y = 0$ realises whenever one of $\{0200, 0020, 0002, 0110, 0101, 0011\}$ realises. Hence $$\Pr(Y=0) = \Pr(\{0200\})+\Pr(\{0020\})+\Pr(\{0002\})+\Pr(\{0110\})+\Pr(\{0101\})+\Pr(\{0011\})\,.$$

But here I'm stuck. While I see that $$\Pr(\{0200\}) = \Pr(\{0020\})=\Pr(\{0002\}) = 1/4^2\,,$$ I don't see why $$\Pr(\{0110\})=\Pr(\{0101\})=\Pr(\{0011\}) = \frac{1}{8}\,.$$

My guess is that I'd get those probabilities only if I "force" $\mathcal{S}$ to include two further events such as $0000$ and $1111$, which in this experiment receive probability zero any way. But his seems a little bit artificial. I'm I correct ? Any hint is highly appreciated.

Update While the previous $\mathcal{S}$ is valid, is not directly useful for solving the problem (using the classical definition) since it is not symmetric, i.e. its simple events are not equiprobable; see also the comments and answers below. The key point in constructing a useful sample space is to distinguish between the two balls, say one is red and one is black, although latter on we will forget about the colour. Then this new sample space $\mathcal{S}^*$ (see Graham Kemp's answer), has 16 equiprobable simple events.

From this it can be easily seen that $$\Pr(Y=0) = \frac{9}{16}\,,$$ because there are 9 favourable cases to the event "first urn receives no balls" over 16 overall cases.

Answer 1

Colour the balls blue and green, so a blue-green colour-blind observer cannot distinguish them.

Now your sample set is of the 16 equally probable events: $$\def\b{\color{navy}{\overline{ 1}}}\def\g{\color{darkgreen}{\underline{ 1}}} \left\{\raise{3.5ex} {2000, 0200,0020,0002, \\\b\g 00, \b 0\g 0, \b 00 \g, 0\b\g 0, 0\b 0\g, 00\b\g, \\ \g\b 00, \g 0\b 0, \g 00\b, 0\g\b 0, 0\g 0\b, 00\g\b}\right\} $$

Note: $\{2000\}$ is the event of "both balls go into the first urn", $\{\b\g 00\}$ is the event "the blue ball goes into the first urn, the green ball goes into the second, and $\{\g\b 00\}$ is the event "the green ball goes into the first urn, the blue ball goes into the second."   Are you convinced these are equally probable?

This you get the odds $9:6:1$

Answer 2

Let $F_i$ denote the event that the first ball is placed in urn $i$ and let $S_i$ denote the event that the second ball is placed in urn $i$.

Then e.g.:$$\{0110\}=(F_2\cap S_3)\cup (F_3\cap S_2)$$and: $$\{2000\}=F_1\cap S_1$$

Now realize that events like $F_2\cap S_3$, $F_3\cap S_2$ and $F_1\cap S_1$ have equal probability so that: $$P(\{0110\})=2P(\{2000\})=\frac18$$