hi, hoping people can help....

what are the odds that by the river a flush is possible? same thing with straight?......... and i assume it's ok to multiply the 2 together to get "no flush" AND "no straight" possible.

important numbers in hold-em...............absolutely vital in omaha.

and i make no assumptions regarding players holdings and/or cards folded.

thanks in advance ........ i should know how to do this but i don't ......... i'll see if i can figure it out on my own too

I’ll do the easiest one:

Flush by the river: Hero has two suited cards:

Pr = (C(11,3)*C(39,2) + C(11,4)*39 +C(11,5) + 3*C(13,3))/C(50,5)
= 6.44%

So, no flush is 93.56%. Note that the last term is the board flushing a suit different than yours.

Unsuited is a bit more difficult but very doable. I got 2.0% for a flush again including the flush consisting of 5 board cards of one of the two suits you don’t have.

Then Pr No Flush = Pr(suited) Pr(no flush|suited) + Pr(off suit)*Pr(no flush|off suit)

Straights are more tricky as the starting hand possibilities are more varied; no gap, 1 gap, … n gap; middle cards or end cards, ace or no ace, pair or no pair.

You can’t simply multiply the no-straight and no-flush probabilities together because they are not independent. For example, if you have a flush, then you know there are at least 5 cards with different ranks which is favorable for a straight whereas if you don’t have a flush there’s the possibility of several ranks having more than 1 card dealt which reduces the straight possibility. Thus, for complete accuracy you would have to treat each type of flush separately and then combine. Also, there’s the possibility of a straight flush or a straight and flush on the same deal.