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View Full Version : Odds of flopping a set and catching a set (HE)


Caruso329
02-13-2005, 09:00 PM
Not sure if my equation is correct on this...

What are the odds of flopping a set?
If you don't flop a set, what are the odds that you will catch it on fourth street and fifth street?

My equation:
(52-2)/(2*3) = 8.3 to 1
(52-5)/2 = 23.5 to 1 (turn)
(52-6)/2 = 23 to 1 (river)
(52-5)/(2*2) = 11.75 to 1 (turn and river)

Caruso329
02-13-2005, 10:53 PM
I know this is simple, but could someone at least tell me if I'm right or not?

slickpoppa
02-13-2005, 11:04 PM
You are incorect. I will post the correct answer shortly

slickpoppa
02-13-2005, 11:10 PM
The correct equation for the flop is:

(2/50)*(48/49)*(47/48)*3 = .115, or in terms of odds 7.68:1

If you miss the flop, then for the turn it is 45/47 or 22.5:1

River:
44/46 or 22:1

Turn or River:
(45/47)*(2/46)*2 = .0832 or 11:1

Caruso329
02-13-2005, 11:29 PM
Thanks a lot. /images/graemlins/laugh.gif

d1sterbd
02-26-2005, 10:21 AM
Curious... why wouldn't this formula be correct:

(2/50) + (2/49) + (2/48) = .122

gaming_mouse
02-26-2005, 12:34 PM
[ QUOTE ]
The correct equation for the flop is:

(2/50)*(48/49)*(47/48)*3 = .115, or in terms of odds 7.68:1

If you miss the flop, then for the turn it is 45/47 or 22.5:1

River:
44/46 or 22:1

Turn or River:
(45/47)*(2/46)*2 = .0832 or 11:1

[/ QUOTE ]

I don't follow this method, but it seems to diverge slightly from the correct answer:

1 - ((45 / 47) * (44 / 46)) = 0.0841813136

elitegimp
02-26-2005, 02:18 PM
[ QUOTE ]
Curious... why wouldn't this formula be correct:

(2/50) + (2/49) + (2/48) = .122

[/ QUOTE ]

Because the 2/50 times the first card off the deck gives you your set, the second two cards don't matter.

Think of it this way -- 2/50 chance the first card is your set. If it's not (48/50), then 2/49 chance the next card is your set. So after two cards on the flop, you have a set 2/50+(48/50)*(2/49) times. Call this value 'p'... then there is a 1-p chance that you don't have a set after two cards on the flop, in that situation there is a (1-p)*2/48 chance you hit your set on the last card. Hence your total probability is

2/50 + 48/50*2/49 + (1-2/50-48/50*2/49)*2/48 = 0.1176

In fact, this is an over-estimate as well, because it doesn't take into account the times you hit quads (i.e. when your first card gives you a set, and either the second or third card on the flop gives you quads). Notice, though, that everyone else is giving the odds that you hit a set or full-house, so including quads shouldn't get me flamed too badly /images/graemlins/smile.gif

top-spin
02-26-2005, 03:50 PM
I've used the (1-method) as well. It seems to work when you're looking to hit 1 out of a specific number of outs, like making a set, or pairing something on the flop, or making your flush or straight draw post flop.

I've come up with a formula sheet that might be instructive:

http://www.vhot.com/poker/poker.pdf

On there are the specific formulas with the outs listed. I'd welcome any comments, esp if i'm wrong!! And alternate formulas would be cool. I'm still learning.

For example you'll make a set by the river 20% of the time starting with just your hole cards, so preflop u've got 5:1 by the river. You'll flop a set 11.76% of the time (yes includes quads).

If you miss ur set on the flop, you'll make it by the river about 8.4% of the time, and just over 4% for the turn and river seperately.

I have a seperate question, and as u can see, some of my formulas have blanks and question marks.

I dont know how to calculate when you HAVE to hit TWO cards.

Eg, i've got 78. I want to make a straight, let's keep it simple, i want only open ended straight draws.

i need to flop 69,56,9T. Then after that i have 4 outs, ez to for me to pickup. I understand the calculations will probably include a "don't care" scenario for the 3rd card on the flop, so any percentage computed will include actually hitting a straight on the flop.

How to i calcluate hitting BOTH of those on the flop? How about hitting BOTH by the turn? I have to hit both by the turn or i'm folding. I'm really at a loss. If i'm on a draw by the river, I know i have an 8.7% chance to hit one of my 4 outs on the river.

And if u include INSIDE straight draws, you have to hit 45,59,46,6T,TJ,9J. That's like 24 cards. Forget inside draws at this point, or maybe any advice I get will help me compute the inside draw.

Help? /images/graemlins/smile.gif

Thanks!!

gaming_mouse
02-26-2005, 03:59 PM
top spin,

See this thread. (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=probability&Number=1239471 &Forum=f11&Words=straight&Searchpage=0&Limit=25&Ma in=1239215&Search=true&where=bodysub&Name=20330&da terange=1&newerval=5&newertype=w&olderval=&olderty pe=&bodyprev=#Post1239471) Let me know if you have any quesions.

gm

top-spin
02-26-2005, 05:43 PM
thanks! I've never figured out combinations like that but it really makes sense.. i'll study on it !

ty

slickpoppa
02-26-2005, 07:12 PM
[ QUOTE ]
[ QUOTE ]
The correct equation for the flop is:

(2/50)*(48/49)*(47/48)*3 = .115, or in terms of odds 7.68:1

If you miss the flop, then for the turn it is 45/47 or 22.5:1

River:
44/46 or 22:1

Turn or River:
(45/47)*(2/46)*2 = .0832 or 11:1

[/ QUOTE ]

I don't follow this method, but it seems to diverge slightly from the correct answer:

1 - ((45 / 47) * (44 / 46)) = 0.0841813136

[/ QUOTE ]

GM, your answer is incorrect because it includes the probability of hitting quads. It's a good number to know, but technically the OP's question asked about hitting a set, not a set or better.

gaming_mouse
02-26-2005, 07:14 PM
[ QUOTE ]
GM, your answer is incorrect because it includes the probability of hitting quads. It's a good number to know, but technically the OP's question asked about hitting a set, not a set or better.

[/ QUOTE ]

yeah, technically, but that was intentional. i assumed he meant "at least a set," since that is what you want to know in practice usually.

gm

slickpoppa
02-26-2005, 07:19 PM
yeah, well i'm a stickler for the technicalities, lol. Anyway to get my answer like yours i would add:
(45/47)*(2/46)*2 + (2/47)*(1/46) = .08418

I don't understand what is hard to follow about my notation

gaming_mouse
02-26-2005, 07:25 PM
[ QUOTE ]
I don't understand what is hard to follow about my notation

[/ QUOTE ]

nothing at all. i was just feeling lazy, saw that the 1- method was simpler, and didn't want to think about what you were doing. sorry /images/graemlins/smile.gif