This is a binomial situation.
We either get an even number or an odd number on any throw.
We don't need to examine the specific 6 numbers themselves.
There are ways to get 3 even outcomes in 8 throws (doesn't matter which even number) and ways to get 2 even outcomes.
since 3 even throws must go with 5 odd throws etc
and p=probability of an even result, 1-p=probability of an odd result.
Diving both sides by