Probability of even number of dice result Suppose we roll a dice (with 6 sides) 21 times.
What are the odds of getting either 1 or 2 an even amount of times?
I tried to calculate it by representing the number of times we get 1 or two as a binomial variable $B(n=21, p=\frac{1}{3})$, And summing the probability of getting $0, 2,\ldots, 20$ and I get the probability is $0.5$, which from what I saw in a different way of calculation is wrong.
Where did my method go wrong?

Here's a potential hint:
For any given number to appear an even amount of times, there are
$$
{21\choose2}(5^{21-2})+{21\choose4}(5^{21-4})+{21\choose6}(5^{21-6})+\cdots+{21\choose20}(5^{21-20})=\sum_{n=1}^{10}{21\choose2n}(5^{21-2n})
$$
possibilities.
