1

calculating my own odds is pretty easy. But for me it's very challenging to calculate the opponent's odd yet. Got a very nice book which focuses at exactly this problem. The problem: It just shows the math, but not how to get there. I'm just gonna quote the page where 'Two pairs odds' are getting calculated during the Flop stage:

**Two pairs odds**
For a specific opponent, let us denote by A' the event: "That opponent will achieve two pairs (PPDDx) as his/her final hand" and by B' the event: "At least on opponent will achieve two pairs (PPDDx) as his/her final hand". We calculated the probabilities P(A') and P(B') at the moment when 3 community cards were dealt.
     The variables the probabilites depend on are:
     p" = number of viewed community P-cards
     d" = number of viewed community D-cards
     p  = number of viewed P-cards
     d  = number of viewed D-cards
     n  = number of your opponents
  We studied all the cases with respect to the possible values of variables p" and d". For each case, we calculated the probability P(A') as a function of *p* and *d* and the probability (B') as a function of *p*, *d*, and *n*. We grouped the cases with the same probabilities conveniently in nine larger cases and obtained the following results:

**1)** If p"=0 and d"=0, then:
P(A') = [(4-p)*(3-p)/2162]*[(3-d)*(4-d)/1980]+[(4-d)*(3-d)/2162]*[(4-p)*(3-p)/1980]+[(4-p)*(4-d)/1081]*[(3-p)*(3-d)/990]

P(B') = [(4-p)*(3-p)/2162]*[n*(3-d)*(4-d)/1980]*[1-(n-1)*(2-d)*(1-d)/3612]+[(4-d)*(3-d)/2162]*[n*(4-p)*(3-p)/1980]*[1-(n-1)*2-p)*(1-p)/3612]+[(4-p)*(4-d)/1081]*[n*(3-p)*(3-d)/990]*[1-(n-1)*(2-p)*(2-d)/1806+(n-1)*(n-2)*(2-p)*(2-d)*(1-p)*(1-d)/4442760].

Source: https://books.google.de/books?id=4gkrMlwj_hIC&lpg=PP1&hl=de&pg=PA60#v=onepage&q&f=false

So, i don't have any clue how the author gets to this formula. I hope you can explain it to me :D Greetings

  • 1
    It's hard to tell without seeing the whole book. It could be some of the math is explained elsewhere. – Herb Aug 22 '17 at 21:09
  • 1
    What book else then? And do you have another mathemical approach, for how to calculate the odds for an opponent? -It does not get explained, you can read ~90% of the book online via Google Books. – Neo_unleashed Aug 22 '17 at 22:05
  • 1
    Thanks for the link. Ok I get the main idea of binomial distribution, but how I understand it, it only calculates the probability of drawing exact these two asses (as an example). So when I play against one enemy - I have no ass in hand and there is no on the community board. I use binomial distribution to calculate the chance for drawing two, and then I know what's the probability for my opponent having it in hand. But my problem: When I play against more than one enemy, how do I know that exact one player got 2 asses and not that two players just got one?? – Neo_unleashed Aug 24 '17 at 15:00
  • 1
    So @Paparazzi? What's your answer for that? – Neo_unleashed Aug 29 '17 at 12:12
2
+50

I will try to give some insight. Won't get into much mathematical detail (no room here), I assume there is some basic probability & algebraic understanding.


Introductory

(1) Binomial Coefficient : C(x,y), calculates how many ways there are to choose y-pieces from x total pieces. For example (from the book p.35) if you draw two random cards from a total of 45 hole cards, you have C(45,2)=990 possible combinations.

(2) A way to do event calculation, for example to get AK as starting hand in poker is :

a) Calculate AK combinations in your two cards (4 Aces for first card,4 Kings for the second, total=16)

b) Calculate possible two hand combinations: 52 cards in the deck, you want to check 2-hand combinations, so C(52,2)=1326

c) Event [Hold an AK as starting hand]=16/1326

*Note: If we were to calculate the event of holding AA preflop, the possible combinations would be 4/1326 (4 Aces total for first card, 3 for the second, minus duplicates)


Explanation of the Equation

In the snippet you provided, the book calculates the event of an opponent holding two pair (not in general, but a specified pair denoted as PPDD) when we have only seen the flop and there are no P's or D's in the first three cards (community cards),since p"=0 and d"=0

The variables p and d work like this. If you hold a P or a D, you "block" the possible combinations of your opponents to have two pairs PPDD. So they are entered in the equation to capture this effect. Ignore them for the time being.

After that, the author does pretty much the following. For each case he identifies all possible scenarios to get to the result and then for each scenario he calculates the possibility of it (exactly as we did in the introductory above).

Event A:
One Opp only
Will achieve 2 pair PPDD 
no P's or D's in the flop

Three possible scenarios to do that:
1) He holds PP and last 2 cards come as DD 
2) He holds DD and last 2 cards come as PP 
3) He holds PD and last 2 cards come as PD (or DP, as order does not matter) 

These are encapsulated in this equation you provided: 

P(A) =
[(4-p)*(3-p)/2162]*[(3-d)*(4-d)/1980]    (scenario 1)
+ [(4-d)*(3-d)/2162]*[(4-p)*(3-p)/1980]  (scenario 2)
+ [(4-p)*(4-d)/1081]*[(3-p)*(3-d)/990]    (scenario 3)


For Event B:
Exactly the same scenarios as before but for multiple players.

P(B') = [(4-p)*(3-p)/2162]*[n*(3-d)*(4-d)/1980]*[1-(n-1)*(2-d)*(1-d)/3612]   (scenario 1)
+[(4-d)*(3-d)/2162]*[n*(4-p)*(3-p)/1980]*[1-(n-1)*2-p)*(1-p)/3612]            (scenario 2)
+[(4-p)*(4-d)/1081]*[n*(3-p)*(3-d)/990]*[1-(n-1)*(2-p)*(2-d)/1806+... (scenario 3)

To generalize for multiple opponents (event B) we do the following:
(i will explain for scenario B1 only)

a) Get the possibility of single event (was done in Event A above)   
this is done here: [(4-p)*(3-p)/2162]*[(3-d)*(4-d)/1980]

b) Apply it for each player, assuming that the rest don't achieve it 
this is done here:[1-(n-1)*(2-d)*(1-d)/3612]

c) Merge it (multiplication by n)
| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.