A 7 Card Stud probability question

[JTG51]

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.

[Dynasty]

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.

[JTG51]

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]

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.

[JTG51]

I think that 99% of Dynasty's posts are helpful. A much higher percentage than most others.

[Dynasty]

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.

[Bozeman]

You'll probably have trouble at the tables if you can't turn the lights on [img]/forums/images/icons/smile.gif[/img]

[irchans]

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.

[SittingBull]

and pairs his doorcard on 4th,it's reasonable to assume he has a set.
Hence, the odds of his filling up by 7th Str. is about 2 1/2 to 1 against.
Of course,if the player plays many 3rd str. hands,this assumption is usually incorrect.
If U are locking horns with a typical player,U need to determine whether or not he has paired his doorcard on the latter Strs. If the single pair is not a doorcard match,then he probably has two pairs.
The odds against filling on 5th S [img]/forums/images/icons/laugh.gif[/img] tr. is about 5 to 1 against,assuming two pair on 5th and 10 to 1 against filling by 7th.

Happy pokering, JTG!
Sitting Bull

[JTG51]

Those are some good points, all of which would have to be considered in a real game. Thanks.