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