Overpair likelihood

[Zapped]

As an engineer and a new Hold'Em player, I enjoy re-learning the probability & stats I've mostly forgotten! Permit me, if you will, to offer up a bad-beat story (aka bad play on my part) to motivate a question.

In a no-limit game, I'd limped in pre-flop with a medium pocket pair (99), and saw an 843 flop (rainbow) with two other callers. I was in late position where I called some good-sized bets from the other two players, and then called thru the turn and the river. At the end the board read 84362 with no flush possibilities. I knew I was safe from flushes or higher, and for whatever reason I wasn't too worried about folks with potential straights (from pocket 75, A5, 5x).

Much to my chagrin, the two other players revealed pocket cowboys and rockets at showdown - AA *and* KK. Rockets took the pot, and the spectators all went "whoa".

So here are my questions, with some attempted answers. Please correct me where I'm wrong, or add an answer where I have nothin' but rags ...


Q1) I hold 99 against one other player, with a board showing a high-card 8. Ignoring trips or better for a moment, what is the likelihood of a pocket overpair in this situation?

A1) My calculation says there are 45 unseen cards, so my opponent could have 45C2 possible hands, of which there are five overpairs (AKQJT) each of which can be made 4C2 ways. 5 * 4C2= 30 ways that can happen. [ 5 * 4C2 ]/ 45C2 = 3.03% or about 32-to-1 against.


Q2) I hold 99 against one other player, with a board showing 84362. What is the likelihood of a pocket overpair OR a pocket pair matching one card on the board for trips?

A2) In addition to the overpairs from (A1), I've got to add in the possibiliies of 88, 44, 33, 66, or 22. Since one of the cards is already on the board, we're choosing two cards from three of the same rank, five different ways = 5 * 3C2 = 15. So the total ways I can be beat by an overpair or trips are 30 (from A1) plus 15 = 45. 45 / (45C2) = 4.55% or 21-to-1 against.


Q3) How do I deal with three players total (two opponents), if the question is EITHER player beating me with overpair or trips as in (Q2)?

A3) This is where I get fuzzy, because the two other players share the same deck. I suspect I can't just add 45 ways + 45 ways = 90 ways = 90/990 = 9.09% = 10-to-1 against. Maybe it's [ 1 - (not-p)*(not-p) ] = [ 1 - ((990-45)/990)*((990-45)/990) ] = 8.88% = 10.25-to-1 against? I'm just not sure.


Q4) How do I extend this to N+1 total players (N opponents), again if the question is ANY player beating me with overpair or trips as in (Q2)?

A4) My brain has imploded, I can speculate no more. /images/graemlins/smile.gif

[Indiana]

gotta love those engineers. Well, your calculations look right and if you don't mind me saying, I think you are asking the right types of questions. By the way, last night the flop came J high and I held QQ. Using your numbers it looks like I did the right thing in calling down and then putting my last few bucks in on the river....

With respect to your last question, you are right, you cannot just add the probabilities. Do you remember this equation: P(A U B) =P(A) + P(B) -P(A and B)? This is what you need to use. you already calculated that Pr(A) + Pr(B) = .0455*2=.091. Now you just need to calculate P(A and B) where A is the event that you are beat by one player and B is you being beat by the other player...
To do this, you use the formula P(A and B)=P(A and B beats
you with the set)+P(A and B beats you with the overpair)=P(A|B has the set)*P(A)+P(A|B has the overpair)*P(A) where
P(A|B) is the prob. that A beats you 'given that' B beats you. Now, P(A|B has the set)=(5*4C2 +4*3C2)/43C2 =.047
because there are still 5 different overpairs A can beat you with and now 4 sets left to get you with of the 43 cards left(cause B has two of the "45" that were left). Next, P(A|B has the overpair)=(5*3C2 +4*4C2))/43C2= .043
because there are now 4 overpairs A could have and still 5 sets. Finally, we put it all together and have P(AUB)=.091 -.0455*{.047 +.043}=.0869. As you can see, in general just add the two probabilities and subtract a little because of the intersection.

*Conclusion: Stack those guys because you will usually have 2 of them together beat!!!

Indiana


Hope this helps,

Indiana