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I can't find this exact question. The specific question I have is this:

If you have a bag with one red, one blue, and one green ball, the chance of picking any particular one is 1/3. Simple enough. Let's say you do it eight times in a row, though. Are the chances of picking the same color eight times in a row 1/6561, or about 0.015%? I'm not sure if you would triple that, since it could happen for any particular color, or if that's already factored in somehow. Or maybe it's both, depending on if the color matters.

The more general question is what would the formula for figuring out any particular combination of colors, e.g. three reds, three blues, and two greens, if the order of selection doesn't matter. I'd also like to see the most generic formula for any number of balls and any number of selections.

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The probability of picking red $8$ times in a row is $\frac{1}{3^8}$.

The probability of picking blue $8$ times in a row is the same, as is the probability of green $8$ times in a row. So the probability of some colour $8$ times in a row is $3\cdot \frac{1}{3^8}$.

For the somewhat more general problem that you mention, suppose we pick a ball $8$ times in a row, with replacement.

The probability of any particular sequence of colours is $\frac{1}{3^8}$. We now ask how many sequences there are that have $3$ red, $3$ blue, and $2$ green. This is given by the multinomial coefficient $\binom{8}{3,3,2}$, which is $\frac{8!}{3!3!2!}$. So our probability is $\frac{8!}{3!3!2!}\cdot \frac{1}{3^8}$.

Remark: More generally, suppose that we repeat an experiment independently $n$ times, with possible outcomes $A_1,A_2,\dots,A_r$, that have probabilities $p_1,p_2,\dots,p_r$. Then the probability that $A_1$ will occur $a_1$ times, and $A_2$ will occur $a_2$ times, and so on, where $a_1+a_2+\cdots+a_r=n$ is given by $$\binom{n}{a_1,a_2,\dots,a_r} p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}.$$

The multinomial coefficient in front is $\frac{n!}{a_1!a_2!\cdots a_r!}$. It is the number of "words" of length $n$, over the alphabet $\{A_1,\dots,A_r\}$, that have exactly $a_1$ occurrences of $A_1$, $a_2$ occurrences of $A_2$, and so on up to $A_r$.

For more details, you might want to search under multinomial distribution.

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  • $\begingroup$ Cool, that matches what I was thinking. Can you answer the more general question? $\endgroup$
    – Joe M
    Jul 9, 2014 at 17:53
  • $\begingroup$ Okay, so the chances of any combination of three, three, two would then be triple that value? $\endgroup$
    – Joe M
    Jul 9, 2014 at 21:48
  • $\begingroup$ Yes, as you have seen, there are $3$ cases, $2$ red, $2$ blue, $2$ green, and the others evenly split. $\endgroup$ Jul 9, 2014 at 21:50
  • $\begingroup$ That's fun. I just did the math for all ten combinations and they added up correctly. $\endgroup$
    – Joe M
    Jul 9, 2014 at 21:57
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    $\begingroup$ Nice that you used the word "fun." I am (probably) much older, have done these things forever. It is still fun. $\endgroup$ Jul 9, 2014 at 22:06

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