Two Plus Two Older Archives A 7 Card Stud probability question
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#1
10-16-2002, 02:49 PM
 JTG51 Senior Member Join Date: Sep 2002 Location: CT Posts: 3,746
A 7 Card Stud probability question

I'm just starting to play a little 7CS, and don't really know how to go about figuring this one.

What are the chances that someone with a single pair showing on their board has a full house or better by 7th street?

Thanks.
#2
10-16-2002, 07:56 PM
 Dynasty Senior Member Join Date: Sep 2002 Location: Las Vegas Posts: 4,044
Re: A 7 Card Stud probability question

I don't know how to figure this out. However, I can tell you this: If somebody gives you the answer, it will not help you make better decisions at the table.

#3
10-16-2002, 11:21 PM
 JTG51 Senior Member Join Date: Sep 2002 Location: CT Posts: 3,746
How did I know you\'d say that?

I've heard you say that several times now. Isn't it OK to just ask questions out of curiosity sometimes?

I think that 1% at best of what's said on this particular forum will help any of us at the table. I'm sure that figuring out which light switch controls which bulb won't help me at the table at all. [img]/forums/images/icons/smile.gif[/img]
#4
10-17-2002, 12:09 AM
 Guest Posts: n/a
You must forgive Dynasty............

about half of his responses are dubious at best. He has some large ego that seems to be fed by criticizing others but I do believe he is generally harmless.

#5
10-17-2002, 12:27 AM
 JTG51 Senior Member Join Date: Sep 2002 Location: CT Posts: 3,746
I don\'t think so

I think that 99% of Dynasty's posts are helpful. A much higher percentage than most others.
#6
10-17-2002, 02:27 AM
 Dynasty Senior Member Join Date: Sep 2002 Location: Las Vegas Posts: 4,044
Re: How did I know you\'d say that?

Of course, you can ask just for the sake of knowledge. But, it's also important to know that certain information shouldn't be used to make your decisions at the table.

I'd hate to see you fold Jacks-up on the river because your opponent has a pair of 7's on his board and somebody calculated that there is a 23.8% chance that he has a full house.

BTW, if you want an Excel spreadsheet which calculates the probability of making a flush taking into account the exposed 3rd street cards, e-mail me and I'll send it to you.

#7
10-17-2002, 03:13 AM
 Bozeman Senior Member Join Date: Sep 2002 Location: On the road again Posts: 1,213
Re: How did I know you\'d say that?

You'll probably have trouble at the tables if you can't turn the lights on [img]/forums/images/icons/smile.gif[/img]
#8
10-17-2002, 01:34 PM
 JTG51 Senior Member Join Date: Sep 2002 Location: CT Posts: 3,746
LOL, good point. *NM*

#9
10-18-2002, 05:22 AM
 baggins Senior Member Join Date: Sep 2002 Location: chicago, il Posts: 605
Re: A 7 Card Stud probability question

while it may not help you directly with decisions at the table, pondering questions like this help make your mind agile, quick, creative, precise, and generally get in the habit of figuring out how to solve problems. this is all quite pertinent in regards to poker.
#10
10-17-2002, 08:18 AM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Re: A 7 Card Stud probability question

I will take a shot at this one. I will assume that the 3 face down cards are random. In a real game, this assumption is very wrong so the final probability is merely a guide. We will also have to ignore the possibility of straightflushes. Computations like this one often have many cases so they are frequently wrong. Don't believe the result until someone has confirmed it.

Suppose the opponents up cards are p p x1 x2. There are 3 down cards. There are C(52-4, 3) = 17296 possible sets of downcards. If he has a full house or four of a kind, then the downcards would have to be px-, xx-, pp-, ppx, pxx, or xxx. The size of each of these sets is

px- 2*6*40 = 480 ways
xx- 2*3*40 = 240 ways
pp- 1*40 = 40 ways
ppx 1*6 = 6 ways
pxx 2*2*3 = 12 ways
px1x2 2*3*3 = 18 ways
xxx 2 ways.
x1x1x2 3*3 = 9 ways
x1x2x2 3*3 = 9 ways

The resulting probability is (480+240+40+6+12+18+2+9+9)/17296 ~= 4.7 %.

Another way to derive the probability is p1 + p2 - p3 where

p1 is the probability that exactly 2 of the down cards have values of p, x1, or x2,
p2 is the probability that all of the down cards have values of p, x1, or x2, and
p3 is the probability that the opponent has 3 pair.

p1 = 3*(40/48*8/47*7/46)
p2 = 8*7*6/(48*47*46)
p3 = 6* (40/48*3/47*3/46.)

p1+p2-p3 ~= 4.7 %.

Dynasty might point out that in a real poker game, the opponent is more likely to have a pocket pair or paired door card. Also, the betting and cards exposed by the other players add a lot of information.

Hope that helps.

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