At a local tournament, me and a friend were having silly £1 bets on flops, all cards over 8, 2 cards the same suit, really trivial stuff like that. I offered a bet at evens, that an 8, 4 or 2 would be on the flop. I know I'm getting the best of it, I was thinking probably 55/45 favourite, but he seemed to think I was 75% favourite-ish. We didn't really talk about it, but it just got me wondering what the actual odds were?

Any chance of some help please guys?

That's 12 cards total. You need at least one.
Pr(at least 1) = Pr(1) + Pr(2) + Pr(3)

= [(12C1)(40C2) + (12C2)*40 + 12C3] / 52C3
= 47/85 = 55.294%

I have no idea what any of that means, but you seem to know your stuff.

Random internet man, I trust you implicitly. Cheers for the help

Or

1 - (40/52 * 39/51 * 38/50)

=~ 55.294%

Gahh, I usually would have taken that approach. Something about this forum decreases my brain's efficiency :@

fwiw, heehaw, you answered it in the way that Bruce answers most questions. But that's because Bruce usually has to.

When the solution with fractions is simple enough, and the solution with combinations is simpler, I usually try to give both since more people understand the solution with fractions. In this case the solution with fractions is much simpler, so I'd just give that one. Using fractions is the same as using permutations instead of combinations.

Then there's the inclusion-exclusion solution:

3*(12/52) -
C(3,2)*(12/52 * 11/51) +
C(3,3)*(12/52 * 11/51 * 10/50)

=~ 55.294%