Help with poker math

[ClaudCunningham]

will someone please take the time to show me how to calculate odds from the
following example....


hero has 4,5 hearts

villian 8,9 clubs

flop
6 clubs, 7clubs, 3 diamonds
so villian is 53 to 47 percent favorite, but i cant figure out how to do the
math!!

I've had it explained to me that villian has 14 outs and is favorite, but doesnt
that leave hero the rest of the deck as outs ???

Can someone please walk me through the math to arrive at this . I think it will
help me better "conceptualize" things

thanks

The key is villian has 14 outs twice

So, chance that the 45 holds up is:

(chance that villian misses on the turn)*(chance that villian misses on the river given that he missed on the turn)

=

(31/45)*(30/44) ~ .47

[AaronBrown]

There are 45*44/2 = 990 possible combinations of turn and river cards.

Any turn + river with a club, 5 or 10 in it wins for villain. That's 14 cards (9 clubs, 2 5's since 5/images/graemlins/heart.gif is in hero's hand and we already counted 5/images/graemlins/club.gif, 3 10's since we already counted 10/images/graemlins/club.gif). There are 14*13/2 = 91 ways to get two of those cards and 14*31 = 434 ways to get one of them. That's a total of 525. Any other turn and river combination win for hero.

525/990 = 35/66 = 0.53 for villain.

I use a shorthand method to figure out the percentages in my head. It is as follows:
With two cards to come take his number of outs and multiply by 4 ie)flush draw has 9 outs or 9x4=36% chance with two cards to come.
With one card to come take the number of outs and multiply by 2 and add 1 ie)Flush draw with 9 outs 9x2 + 1 =19%.
The actual percents for a flush draw with two cards to come is 35% and with one card to come it is ~20%. Usually you won't have to face situations with more than 10 outs, but when you do this shorthand methodology begins to over estimate with two cards to come and be about 1% short with one card to come. Btw, I learned this shorthand tool from Chip Reese on a tv interview.