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Consider an urn containing $c$ balls, $\alpha$ of which are red, $\beta$ of which are blue, and $\gamma$ of which are green, and $\alpha+\beta+\gamma=c$.

We perform $n$ trials with replacement of one element at a time from the urn.

Since $\alpha,\beta,\gamma,c$ and $n$ are integer numbers, at each trial corresponds one and only one extracted ball. Therefore, I guess we can represent the possible sequences of balls extracted in $n$ trials as follows:

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On the $x$-axis we report the consecutive $n$ trials and on the $y$-axis we report the balls that can be extracted at each trial (in this example, $\alpha=3, \beta=2, \gamma=3$, and $c=8$, $n=18$).

In particular, the solid green line represents a sequence of trials in which only green balls were extracted, whereas the dashed line represents a sequence of trials in which at least one red ball and at least one blue balls were extracted.

I would say that all the paths related to the extraction of only green balls have the same chance to occur, but this does not seem to be true for the paths related to the extractions of at least one red ball and at least one blue ball (unless $\alpha=\beta=\gamma$).

In the former case, we can write, that, given the $n$ trials,

$$ P(\text{"to get only green balls"})=\left(\frac{\gamma}{c}\right)^n=\frac{N_{\gamma,c}^n}{N_{c}^n}, $$

where $N_{\gamma,c}^n=\gamma^n$ is the number of possible paths (or functions of $n$) that we can build with $\gamma$ balls, and $N_{c}^n=c^n$ is the number of possible paths that we can build with $c$ balls. This result was shown here by Henry.

My question is: Given $n$ independent trials,

How can we write the probability $$ P(\text{"to get at least one red ball and at least one blue ball"}) $$ in terms of a ratio between numbers of paths (analogously to what we have done for the extraction of all green balls), since the "favorable cases" are clearly not equiprobable?

The question rises from the fact that it is easy to show (e.g. by means of Bayes' theorem) that the probability to extract at least one red ball and at least one blue ball in $n$ independent trials is $1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n$.

But this probability can be written as $\frac{c^n-(\alpha+\gamma)^n-(\beta+\gamma)^n +\gamma^n}{c^n}$, where the denominator is the same total number of paths $N_c^n=c^n$ we used to write the expression of $P(\text{"to get only green balls"})$ in terms of ratio between number of paths.

Therefore, it seems that the numerator $c^n-(\alpha+\gamma)^n-(\beta+\gamma)^n +\gamma^n$ should represent the number of paths related to the sequences of extractions with at least one red ball and at least one blue ball. If we divide this number by $N_c^n$ we get the Bayes' probability, exactly as the favorable sequences for this event were equiprobable!

I apologize for naivety and incorrectness, and I thank you for any comment or suggestion!

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  • $\begingroup$ Of course all $c^n$ outcomes are equiprobable. What makes you think they are not? $\endgroup$ – bof Sep 24 '18 at 6:01
  • $\begingroup$ @bof My problem are not the $c^n$ cases (possible cases), but the favorable cases of the event "to get at least one red ball and at least one blue ball". These are for sure not equiprobable. $\endgroup$ – user559615 Sep 24 '18 at 6:07
  • $\begingroup$ I don't understand what you are trying to say. There are $c^n$ possible cases, and they are all equiprobable, each of them having probability $c^{-n}.$ Included among those $c^n$ possible cases are the favorable cases, $c^n-(\alpha+\gamma)^n-(\beta+\gamma)^n+\gamma^n$ in number, and they are equiprobable, each having probability $c^{-n}.$ What am I missing? $\endgroup$ – bof Sep 24 '18 at 6:20
  • $\begingroup$ @bof I think you pointed out exactly my problem! I believe that I am the one missing something here! Do you assess that the $c^n-(\alpha+\gamma)^n-(\beta+\gamma)^n+\gamma^n$ paths representing the sequences related to the extraction of at least one red ball and at least one blue ball are equiprobable, right? $\endgroup$ – user559615 Sep 24 '18 at 6:28
  • $\begingroup$ Yes. Maybe it would help to consider a small case, say $c=6$, $\alpha=1$, $\beta=1$, $\gamma=3$, $n=2$. The experiment is just like throwing a pair of dice with one side painted red, two sides painted blue, and three sides painted green. Each of the $36$ possible outcomes has probability $1/36$ (under the usual assumptions of independent fair dice), included among them are the four favorable cases $R_1B_1,\ R_1B_2,\ B_1R_1,\ B_2\ R_1$. $\endgroup$ – bof Sep 24 '18 at 6:34

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