Probability of someone flopping a set.

[GooperMC]

Ironmans post has got me thinking. I know that 1:3 someone will be dealt AXs of his suit what I am wondering is what is the probability that someone was dealt something that would flop a set on a non-paired board.

I ran the calculations but this is the first time I have done a calculation like this so I was hoping that someone could validate my methods.

Hypothetical flop of 78J.

P1: the probably of someone having a 7 = 1 – the probability that the player doesn’t have any 7s.

P1 = 1- (C(3,3) * C(42,33) / C(45, 36)) = .9845

P2: the probability that person has a second 7 too

This person can have
XXX: C(42, 3) = 11480
7XX: C(42, 2) * C(2,1) = 1722
77X: C(42, 1) * C(2,2) = 42

P2 = (1722 + 42) / 13244 = .133

P3: the probability that someone has a set of 7’s = (.133)(.9845) = .13

The probability that some has a set of 8’s or J’s should be the same as 7’s

P4: the probability that someone has a set should be approx 1 – the probability that nobody has a set

P4 = 1 – (.87)^3 = .34

I am a little tired this morning but that number seems in the ballpark but I very probably (no pun intended) made an error somewhere along the way.

On a complete side note thanks to Buzz for showing me an easy way to do this kind of calculation.

[KJ o]

In practical play, I'd guess the probabilities of sets of 7 and 8 are lower as those are typically poor starting hands and folded preflop.

(I thought your calculations looked reasonable, but I'm not to good, so my word is worth precious little there.)

[gergery]

Here is the simple methodology:

Probability that someone has a pair of 7’s is C(3,2) *C(43,2) / C(45,4) = 1.8%

This is because there are 3 7’s out and you must choose 2 of them. Then you can have any other cards still remaining in the deck for the other 2 cards.

To find the chances they have one of three possible sets, you just multiply by 3. So now we have 5.4%. But this double counts slightly, because sometimes you could be dealt 77JJ, so need to count the number of times you’ll get 3377, 77JJ, etc. Those are pretty small in this instance, so I find it easier to just deduct a small amount, so let’s call this 5.3%.

So chances 1 opponent does not have a set is 1-5.3% or 94.7% Against 3 opponents that’s .947^3 = .85, so 1-.85 = .15 or a 15% chance that someone has a set. Again, however, that understates things slightly, as once you know that two opponents don’t have a set, your third opponents is very slightly more likely to have a set. So you need to adjust 15% upward to about 15.8%. Of course, this doesn't take into account the fact that no one in their right mind will play 77xx in O8 without very good xx cards, so the real chances that someone was dealt this AND did not fold is significantly less.

This is my crude, Excel based way of getting at the numbers quickly, without having to do something like this (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=probability&Number=1859270 &Forum=f11&Words=&Searchpage=0&Limit=25&Main=18592 70&Search=true&where=bodysub&Name=11039&daterange= 1&newerval=1&newertype=y&olderval=&oldertype=&body prev=#Post1859270) (Buzz's posts also show correct methodology), which is the methodology you need to use. You can also review some of these articles (http://www.math.sfu.ca/~alspach/pokerdigest.html) for a more in-depth look at how to think thru combinatorics in poker

--Greg