How do i figure this out?
 

i posted this in the probability section, but not getting a response...
what is the probability of flopping a 4-flush or an open ended str8 draw on the flop with 10Js. i know that you are going to complete the draw about 33% of the time. Can some1 please explain to me how to perform this calculation.
thanks
--vitaly

Re: How do i figure this out?
 

so, let's say you have T[img]/forums/images/icons/heart.gif[/img] J[img]/forums/images/icons/heart.gif[/img] :

# of remaining cards : 50
# of remaining [img]/forums/images/icons/heart.gif[/img]s : 11

number of flops containing precisely two [img]/forums/images/icons/heart.gif[/img]s:
(11 choose 2) * 39 = 11 * 10 * 39

number of possible flops:
(50 choose 3) = 50 * 49 * 48

percent of flops containing precisely two [img]/forums/images/icons/heart.gif[/img]s if you hold two [img]/forums/images/icons/heart.gif[/img]s :
(11 * 10 * 39) / (50 * 49 * 48) = 3.64%

FYI, the percent of flops containing AT LEAST two [img]/forums/images/icons/heart.gif[/img]s if you hold two [img]/forums/images/icons/heart.gif[/img]s (this counts the times you flop a flush)
(11 * 10 * 48) / (50 * 49 * 48) = 4.49%

now, we need to count the open-enders you will flop. take for example, 89 (but not 789 or 89Q, since those aren't open-enders, they're straights...)

number of 8's : 4
number of 9's : 4

number of open-enders with 89: 4 * 4 * 40 (40 = 48 remaining cards minus the 4 7's and 4 Q's)

the same logic applies to the other two ways to flop open-ended:

number of open-enders with Q9: 4 * 4 * 40 (40 = 48 remaining cards minus the 4 8's and 4 K's)

number of open-enders with KQ: 4 * 4 * 40 (40 = 48 remaining cards minus the 4 9's and 4 A's)

percentage of open-ender flops:
3 ( 4 * 4 * 40) / (50 * 49 * 48) = 1.63%

percentage of flopped straights (four ways, 789, 89Q, 9QK, QKA):

4 ( 4 * 4 * 4 ) / (50 * 49 * 48) = 0.22%

now, we can't just sum up the percentages of flopped 4 flushes and flopped 4 straights, because there is overlap between those possibilities...

for 89, for example, we must subtract out:
8[img]/forums/images/icons/heart.gif[/img]9[img]/forums/images/icons/heart.gif[/img]Xo = 1 * 1 * 33
8[img]/forums/images/icons/heart.gif[/img]9oX[img]/forums/images/icons/heart.gif[/img] = 1 * 3 * 8
8o9[img]/forums/images/icons/heart.gif[/img]X[img]/forums/images/icons/heart.gif[/img] = 3 * 1 * 8

and, if we are counting flopped flushes, we must subtract out the four-straights that are flushes...

8[img]/forums/images/icons/heart.gif[/img]9[img]/forums/images/icons/heart.gif[/img]X[img]/forums/images/icons/heart.gif[/img] = 1 * 1 * 7


and for the 256 times you flop a straight, we need to subtract out the 4 straight flushes [img]/forums/images/icons/wink.gif[/img] and the
12 times you flop a straight and a four-flush...

so, the grand total is...

percent of time you will flop precicely a four-flush or four-straight:
[ (11 * 10 * 39) + (3 * 4 * 4 * 40) - (3 * (33 + 24 + 24)) ] / (50 * 49 * 48) =
[ 4290 + 1920 - 243 ] / 117600 = 5967/117600 = 5.07%

percent of time you will flop AT LEAST a four-flush or four straight (inludes flopping straights and flushes, but doesn't count flopping trips, boats, or quads, for example):
[ (11 * 10 * 48) + (3 * 4 * 4 * 40) + (4 * 4 * 4 * 4) - (3 * (33 + 24 + 24 + 7)) - 4 - 12) ] / (50 * 49 * 48)
[ 5280 + 1920 + 256 - 264 - 4 - 12 ] / 117600 = 7176/117600 = 6.10%

it's quite possible that, in my rush, i've made a small error in the double-counting subtractions, but this is basically the right process for answering probability questions like this, and the actual numbers I calculated should be very close to correct, probably within 0.25%...

-switters

Re: How do i figure this out?
 

</font><blockquote><font class="small">In reply to:</font><hr />
percent of flops containing precisely two [img]/forums/images/icons/heart.gif[/img] if you hold two [img]/forums/images/icons/heart.gif[/img] :
(11 * 10 * 39) / (50 * 49 * 48) = 3.64%

FYI, the percent of flops containing AT LEAST two [img]/forums/images/icons/heart.gif[/img] if you hold two [img]/forums/images/icons/heart.gif[/img] (this counts the times you flop a flush)
(11 * 10 * 48) / (50 * 49 * 48) = 4.49%


[/ QUOTE ]

Don't we need to multiply these percentages by 3, since we don't care what order the cards come in?

Re: How do i figure this out?
 

no. in both cases we are simply counting how many hands fulfill the given criteria, regardless of order...

so, the number of possible two-[img]/forums/images/icons/heart.gif[/img] combinations from the 11 remaining [img]/forums/images/icons/heart.gif[/img]s is (11 choose 2), which is 11 * 10 = 110.

since a flop is three cards, there are 39 possible ways that each of those combinations could come with a non-[img]/forums/images/icons/heart.gif[/img] as the third card, or 48 ways that the card could come, with or without a [img]/forums/images/icons/heart.gif[/img]. this is regardless of order, this only speaks to the actual combination of cards, so the count is (110 * 39) or (110 * 48), depending on whether you want to count flopped flushes. 6[img]/forums/images/icons/heart.gif[/img]3[img]/forums/images/icons/heart.gif[/img]K[img]/forums/images/icons/spade.gif[/img] and K[img]/forums/images/icons/spade.gif[/img]6[img]/forums/images/icons/heart.gif[/img]3[img]/forums/images/icons/heart.gif[/img] are only one combination, and should only be counted once.

-switters

how can this be?
 

If you are going to flop a made flush/str or 4 flush open ended 6% of the time, and you are going to complete about 1/3 of these, that leaves you with making a flush or srt8 only 2 % of the time. So how can it be correct to play medium suited connectors late? The pot odds just dont justify it. You arent gonna win w 1 pair, and the prob of getting 2 pair or trips isnt very good either. Also, is it ever correct to play 10Jo if you know that you can only win w a str8? If 6 players limp to you, do you throw J10o away? how bout 78o?
-vitaly

Re: how can this be?
 

you can win more ways with these hands than just by making str8s and flushes. that also factors in somewhat

b

Re: how can this be?
 

in addition to the str8's and flushes, there are many other hands you like to flop with 10[img]/forums/images/icons/heart.gif[/img]J[img]/forums/images/icons/heart.gif[/img]

(using (x C y) to represent the choose function)

quads : 2 ways
boats : 10's full ((3 C 2) * 3 = 9) ways + J's full ((3 C 2) * 3 = 9) ways = 18 ways
trips : 2 * ((3 C 2) * 44) = 264 ways
two pair : 3 * 3 * 44 = 396 ways
top pair : 2 * (3 * (32 C 2)) = 5952 ways ( some of these are double-counted with the flush and str8 draws from yesterday...)

unless you have other reason to be very scared, you will usually like all of these hands with a number of limpers before you, and these add another:

18 + 264 + 396 + 5952 = 6630 ways to win (minus a few for the double counting... call it an even 6000)
6000 / 117600 = 5.10%

and, if there are many limpers, you may even like a hand like two overcards, and a gut shot to the nut straight... boards like x78, x79, where x &lt; 7... these would add:

2 * (4 * 4 * 20) = 640 ways (once again, a few of these are double counted...)

bottom line, you're only going to flop one of the hands we've discussed here about 10 - 11% of the time, which isn't enough direct odds to call, but when you do flop a monster (&lt; 2%) or make a huge draw (9 / 3 = 3%) of the time, you figure to get paid off huge against a large field, which is why you play these hands into 5 or 6 way pots. Playing them into 4 or less players is probably not generally a profitable play, except in rare circumstances... which we all already knew [img]/forums/images/icons/wink.gif[/img]

hope this helps!

-switters

ERROR CORRECTION
 

oops. the math from the previous posts was off - in most cases by a factor of three. The process I used was sound, but I was miscomputing the choose function throughout.

apologies for the confusion -- I knew I shouldn't have rushed through that post without a good calculator so many years since my last probability class... [img]/forums/images/icons/frown.gif[/img]


anyways -- here are the important numbers, FIXED:

number of flops containing precisely two [img]/forums/images/icons/heart.gif[/img]s:
(11 choose 2) * 39 = (11 * 10 / 2) * 39 = 2145

number of possible flops:
(50 choose 3) = (50 * 49 * 48 / 6) = 19600

percent of flops containing precisely two [img]/forums/images/icons/heart.gif[/img]s if you hold two [img]/forums/images/icons/heart.gif[/img]s :
2145 / 19600 = 10.94%

number of flops containing AT LEAST two [img]/forums/images/icons/heart.gif[/img]s:
(11 choose 2) * 39 = (11 * 10 / 2) * 48 = 2640

the percent of flops containing AT LEAST two [img]/forums/images/icons/heart.gif[/img]s if you hold two [img]/forums/images/icons/heart.gif[/img]s (this counts the times you flop a flush)
2640 / 19600 = 13.47%

number of open-ender flops:
3 ( 4 * 4 * 40) + 2 (4 * 4 * 4) = 2048

percentage of open-ender flops:
2048 / 19600 = 10.45%

percentage of flopped straights (four ways, 789, 89Q, 9QK, QKA):
(4 ( 4 * 4 * 4 ) = 256) / 19600 = 1.31%

now, we can't just sum up the percentages of flopped 4 flushes and flopped 4 straights, because there is overlap between those possibilities...

percent of time you will flop precisely a four-flush or four-straight:
[ 2145 + 2048 - (3 * (33 + 24 + 24)) ] / 19600 =
[ 2145 + 2048 - 243 ] / 19600 = 3950/19600 = 20.15%

percent of time you will flop AT LEAST a four-flush or four straight (inludes flopping straights and flushes, but doesn't count flopping trips, boats, or quads, for example):
[ 2640 + 2048 + 256 - 264 - 4 - 12) ] / 19600 = 4664/19600 = 23.80%

-switters

sorry, pudley...
 

Pudley -- it turns out that you were right -- we DO need to multiply them by three...

the reason is: I was computing the choose function wrong...

I computed (11 choose 2) as 11 * 10, when it should have been (11 * 10)/2.

I computed (50 choose 3) as 50 * 49 * 48 when it should have been (50 * 49 * 48)/6.

so, instead of:
(11 * 10 * 39) / (50 * 49 * 48)

I shoulda had:
(11 * 10 * 39 / 2) / (50 * 49 * 48 / 6)
= 6(11 * 10 * 39) / 2(50 * 49 * 48)
= 3 * (11 * 10 * 39) / (50 * 40 * 48)

or, exactly what you said. [img]/forums/images/icons/smile.gif[/img]

sorry bout that, everyone...

-Bryon

ERROR CORRECTION
 

turns out I was computing the choose function wrong... correcting the math from the previous post:

quads : 2 ways
boats : 10's full ((3 C 2) * 3 = 9) ways + J's full ((3 C 2) * 3 = 9) ways = 18 ways
trips : 2 * ((3 C 2) * 44) = 264 ways
two pair : 3 * 3 * 44 = 396 ways
top pair : 2 * (3 * (32 C 2)) = 2976 ways ( some of these are double-counted with the flush and str8 draws )

18 + 264 + 396 + 2976 = 3654 ways to win (minus a few for the double counting... call it an even 3000)
3000 / 19600 = 15.31%


add that to the str8 draws and flushes, and you will get a flop you find very favorable about 1/3 of the time... so maybe you WILL want to be playing JTs into four and five way fields...

apologies for the confusion...

-switters