plz help....i cant figure this out

[Cyrus]

"What is the probability of flopping a 4-flush or an open ended str8 draw on the flop with 10Js?"

It comes down to a 7% probability of flopping a 4-flush draw (and nothing else) and to a 8.5% probability of flopping an open-ended (8-outs) straight draw (and nothing else).

You have a 16% probability of flopping a 4-flush draw or an open-ended straight draw (but nothing else).

If you have flopped the 4-flush draw, the probability of completing to a flush with 2 cards to come is 35%; if you have flopped the open-ended straight draw, the probability of completing with 2 cards to come is 31.5%.

[switters]

I think these numbers are too high. I posted a response to vkotlyar's similar post in the mid-high stakes section:
link to other post...

if you hold T[img]/forums/images/icons/spade.gif[/img]J[img]/forums/images/icons/spade.gif[/img], there are 11 [img]/forums/images/icons/spade.gif[/img]s left.

the number of flops that contain two of them are:

number of two card [img]/forums/images/icons/spade.gif[/img] combos = (11 choose 2) = 110
*TIMES*
the number of non-[img]/forums/images/icons/spade.gif[/img] cards remaining = 39
= 4290 ways to flop a four-flush.

there are (50 choose 3) possible boards, given whatever hand you have = 117600

which means you flop a four-flush 3.64% of the time.

I see now that I made a small error in my other post, in that there are more 8-out straight draws than I gave credit for...

ways to flop an 8-out straight:
true "open-enders": 89x, Q9x, KQx, where x doesn't complete the straight.
there are 4 * 4 * 40 of each of these (4 8's, 4 9's, and 40 remaining cards that don't complete the straight)
double-gut shots: 79K, 8QA
there are 4 * 4 * 4 of each of these

so:
3 * (4 * 4 * 40) + 2 * (4 * 4 * 4) = 1920 + 128 = 2048

dividing by the number of flops: 2048 / 117600 = 1.74%

also -- you can't just add these up to get

3.64% + 1.74% = 5.38%, because many of the hands are being "double-counted" (i.e., we have counted 8[img]/forums/images/icons/spade.gif[/img]9[img]/forums/images/icons/spade.gif[/img]2[img]/forums/images/icons/diamond.gif[/img] as both a flush draw and a straight draw). The real number works out to more like 5.10%

I'm pretty sure my math is solid here, but I'd love to hear about it if there are disagreements -- Cyrus, where did you get your numbers?

-switters

[irchans]

50 Choose 3 is 19600.

[irchans]

switters,

I think you reasoning is correct, but your calculator is broke. The computations involving the choose function seem to be using the permutation function instead. When we fix that, your probabilities become:

4- flush = c[11, 2]* (3*13) /c[50, 3.] = 10.9%
4-straight = 2048./19600. = 10.4%

[switters]

Thanks!

you're absolutely right -- I didn't have a calculator with choose[] handy, and I was accidentaly computing the permutation function ((50! / (50-3)!) = 50 * 49 * 48) rather than the choose function ((50! / ((50-3)! * 3!)) = 50 * 49 * 48 / 6) which is why I was counting 117600 flops instead of the real number, 117600 / 6 = 19600!

thanks, irchans! apologies, Cyrus!

-switters

[BruceZ]

4-straight = 2048./19600. = 10.4%

This is 3*16*40 + 4*4*4*2 = 2048, but that overcounts cases where we pair the board. It must be

[ 3(16*34 + 6*4*2) + 4*4*4*2 ]/19,600 = 9.7%.

BTW, if we ignore the double belly busters, the number of 4-straight flops is 1776, a nice number.

[Cyrus]

"Number of two card combos = (11 choose 2) = 110 ... There are (50 choose 3) possible boards, given whatever hand you have = 117,600"

C(11,2)=55 ...You forgot to divide by (2*1) [img]/forums/images/icons/wink.gif[/img]
C(50,3)=19,600 ...117,600 is the # of possible Permutations

The archived thread also has wrong figures.