<s>[INST] Find Probability that even numbered face occurs odd number of times A Die  is tossed $2n+1$ times.  Find Probability that even numbered face occurs odd number of times
First i assumed let $a,b,c,d,e,f$ be number of times $1$ occurred, $2$ occurred and so on $6$ occurred respectively. Then we have
$$a+b+c+d+e+f=2n+1$$ is a linear equation where variables are non negative integers.
So number of non negative integer solutions of above is $\binom{2n+6}{5}$
But $b$, $d$ and $f$ should be odd. So let $b=2x+1$, $d=2y+1$ and $f=2z+1$ where $x,y,z$ are non negative integers.
so we get
$$a+2x+c+2y+e+2z=2n-2$$ which is possible in the following two cases
Case $1.$ $a,c,e$ are all even so let $a=2m+1$, $c=2q+1$ and $e=2r+1$ we get
$$m+x+q+y+r+z=n-1$$ so number of solutions is $\binom{n+4}{5}$
Case $2.$ Exactly two of $a,c,e$ are odd and other is even which can be possible in 3 ways.
So let $a=2m+1$, $c=2q+1$, $e=2r$ we get
$$m+x+q+y+r+z=n-2$$
Number of solutions is $3 \times \binom{n+3}{5}$
Hence Probability is
$$P(A)=\frac{\binom{n+4}{5}+3\binom{n+3}{5}}{\binom{2n+6}{5}}$$
Is my approach fine?
 [/INST] Even and odd faces both occur with a proabability of $\frac{1}{2}$. The probability of attaining an odd number of evens is
\begin{eqnarray*}
\sum_{i=0}^{n} \binom{2n+1}{2i+1} \left( \frac{1}{2} \right)^{2n+1}
\end{eqnarray*}
and an even number of odd faces is
\begin{eqnarray*}
\sum_{i=0}^{n} \binom{2n+1}{2i} \left( \frac{1}{2} \right)^{2n+1}
\end{eqnarray*}
Now note that these two sums are equal by the symmetry of the binomial coefficients so they are both $\color{red}{\frac{1}{2}}$.
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