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?

First let us simplify the problem: the probability of an even face is 1/2, so is the probability of an odd face. The problem is therefore equivalent to a number of coin tosses of a fair coin, with even/odd faces corresponding to heads/tails.
Define the events $E$: the number of heads is even and $F$: the number of tails is even. On the one hand, heads and tails are perfectly interchangeable, therefore
$$
P(E)=P(F).
$$
Moreover, given an odd number of trials, either the number of heads is even or the number of tails is even, which implies
$$
P(E)+P(F)=1.
$$
You may then conclude that
$$
P(E)=P(F)=1/2.
$$
Therefore, the probability that even numbered face occurs odd number of times is 1/2.
