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I need to calculate the probabilities associated with the next experiment: I have to roll a dice a certain amount of times. Each time I roll the dice I get a certain amount of points but, and this is the tricky part, it can be a combination of different kind of points. Let´s say that if the roll is:
6 I get 2 red points
5 I get 1 red point and 1 blue point
4 I get 2 blue points
3, 2 or 1 I don´t get any point.
I want to know what´s the probability after 10 rolls of getting al least 2 red points and 2 blue points.
I can calculate the probability of getting at least 2 red points as the inverse of the probability of getting 0 or 1 red points. The possible ways this could happen would be (please correct me if I´m wrong):

Rolling 6 zero times and 5 zero times, that is $$(4/6)^{10}=0.01734$$
Rolling 6 zero times and 5 once, that is $$(1/6)*(4/6)^9=0.004335$$

So $1-0.01734-0.004335=0.9783$ is the probability of obtaining at least 2 red points after throwing the dice ten times. I can calculate the probability of getting at least 2 blue points the same way but, and here is where I´m stuck, how can I calculate the probability of both events at the same time? They are not independent, cause the roll of the dice applies to red points and blue points at the same time, so for example, the more red points I get the smaller my chances of getting the required blue ones.
In the problem I actually have to solve this combination of points can change and I have even more different "kinds" of points, and the number of turns is variable, so I would appreciate a method to solve this problem rather than a solution, but any help would be welcomed. Help! Thank you!

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Define the polynomial $p$ by $$p=\frac{1}{6}x^2+\frac{1}{6}xy + \frac{1}{6}y^2+ \frac{1}{2}.$$ Then, the coefficient on $x^iy^j$ in $p^n$ gives the probability of having exactly $i$ red points and $j$ blue points after $n$ rolls.

Using a computer algebra system, we can find $p^{10}$ to be $$ \frac{1}{60466176} x^{20} + \frac{5}{30233088} y x^{19} + ... + \left(\frac{5}{30233088} y^{19} + \frac{5}{1119744} y^{17} + \frac{5}{93312} y^{15} + \frac{35}{93312} y^{13} + \frac{35}{20736} y^{11} + \frac{35}{6912} y^9 + \frac{35}{3456} y^7 + \frac{5}{384} y^5 + \frac{5}{512} y^3 + \frac{5}{1536} y\right) x + \left(\frac{1}{60466176} y^{20} + \frac{5}{10077696} y^{18} + \frac{5}{746496} y^{16} + \frac{5}{93312} y^{14} + \frac{35}{124416} y^{12} + \frac{7}{6912} y^{10} + \frac{35}{13824} y^8 + \frac{5}{1152} y^6 + \frac{5}{1024} y^4 + \frac{5}{1536} y^2 + \frac{1}{1024}\right) = \sum_{i=0}^{20} \sum_{j=0}^{20} c_{i,j} x^i y^j. $$ Then $$ 1-\sum_{\substack{ 0 \le i,j \le 20 \\ i<2 \,\,\mbox{or}\,\, j<2}} c_{i,j} = \frac{53382023}{60466176}$$ gives the probability of having at least 2 red points and 2 blue points after 10 rolls.

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  • $\begingroup$ Thanks for answering. Where does that polynomial come from? $\endgroup$ – Rodrigo Nov 29 '13 at 20:37
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    $\begingroup$ We create it from the probabilities you are given: the coefficient of $x^i y^j$ is the probability of having $i$ red points and $j$ blue points after one roll. The 1/2 is there since there is a 1/2 chance of zero red and zero blue (1/2 = 1/2 x^0 y^0). $\endgroup$ – Matthew Conroy Nov 29 '13 at 20:42
  • $\begingroup$ Sorry for my ignorance, but I think I still don´t get it. Does that mean that the first term in p is the probability of getting 2 red points (hence the $x^{2}$), the second is the probability of getting 1 red and 1 blue (hence the $xy$), the third the probability of getting 2 blue points and the fourth the probability of getting none of them? Which is the mathematical theory that lies behind that formula? As I said, the problem I actually have to solve "resembles" this one, but I need to apply the solution to a different problem. $\endgroup$ – Rodrigo Dec 4 '13 at 12:21
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    $\begingroup$ I like to think of this method as essentially an accounting method: there is nothing deep about it, it is just a good way to handle tedious computations with convenient notation. Yes, your understanding of $p$ is correct. As a start to understanding it, work out, by hand, what $p^2$ is. Do you see why this yields the appropriate probabilities when rolling the die twice? $\endgroup$ – Matthew Conroy Dec 4 '13 at 20:15
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    $\begingroup$ I use PARI/GP. It is free, and I've been using it for years. It does the job well. pari.math.u-bordeaux.fr $\endgroup$ – Matthew Conroy Dec 4 '13 at 20:16

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