I have one dice and I roll it six times, recording the number I get each time. What is the probability that the number of 2s I get equals the number of 5s.
I try solve this using the trinomial distribution
$P(X_1 = x_1, X_2=x_2,X_3=3)=\frac{n!}{x_1!x_2!x_3!}p_1^{x_1}p_2^{x_2}p_3^{x_3} $ , or
$P(X_1 = x_1, X_2=x_2)=\frac{n!}{x_1!x_2!(n-x_1-x_2)!}p_1^{x_1}p_2^{x_2}(1-p_1-p_2)^{(n-x_1-x_2)}$
with $X_1 =$ number of faces equaling 2, $X_2 =$ number of faces equaling 5,
then we have to find
$P(X_1 = 1, X_2=1)+P(X_1 = 2, X_2=2)+P(X_1 = 3, X_2=3), n=6$
and
$P(X_1 = 1, X_2=1)=\frac{6!}{4!}*\frac{1}{6^2}(\frac{2}{3})^{4}=\frac{40}{243}, p_1=p_2=1/6$ $P(X_1 = 2, X_2=2)=\frac{6!}{2!2!2!}*\frac{2^22^2}{6^2*6^2}(\frac{1}{3})^{2}=\frac{10}{81}, p_1=p_2=2/6$ $P(X_1 = 3, X_2=3)=\frac{6!}{3!3!}*\frac{3^33^3}{6^3*6^3}=\frac{5}{16}, p_1=p_2=3/6$
so $P(X_1 = 1, X_2=1)+P(X_1 = 2, X_2=2)+P(X_1 = 3, X_2=3)=\frac{2335}{3888}=0.60057...$
Did I solve it right? A probability of about 60% seems a bit too big for this scenario. Is the problem with how I have defined my $p_1$ and $p_2$. I assumed that in this case it would be the proportion of which the event happens, so $\frac{1}{6}$ of the n faces shows a certain number.etc Other wise, I would think that still with $p_1=p_2$ but they are the probability that $x_1$ or $x_2$ faces of a dice are a certain number which then I would find with the binomial distribution.
The final answer looks too big because the calculations for $P(X_1=X_2=2,3)$ are wrong; $p_1,p_2$ remain at $\frac16$ for these cases. $$P(X_1=X_2=2)=\frac{6!}{2!2!2!}(1/6)^2(1/6)^2(2/3)^2=5/162$$ $$P(X_1=X_2=3)=\frac{6!}{3!3!0!}(1/6)^3(1/6)^3(2/3)^0=5/11664$$ Then there is the case where no 2s and 5s are rolled, which is still an equal number of each: $$P(X_1=X_2=0)=\frac{6!}{0!0!6!}(1/6)^0(1/6)^0(2/3)^6=\frac{64}{729}$$ Adding these up gives a final probability of around $0.2837$.