Hi ipokeyou,

This is a binomial situation.

We either get an even numberoran 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 matterwhicheven number) and ways to get 2 even outcomes.

Therefore

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.

Hence,

Diving both sides by