What is the probability of a two pair in poker?

\(C^1_{48}\) is wrong because if you choose the value you already have in pairs you'd have full house.

What is the probability of a two pair in poker?

Cards = 52.
Suits = 4.
Values = 13.

\(P(two pair)=\frac{C^2_{13}*C^2_4*C^2_4*C^1_{44}}{C^5_{52}}\)

\(C^2_{13}\) is the number of ways to choose which 2 values will be the pairs (for example, 2 kings, and 2 aces);
\(C^2_4\) is the number of ways to choose 2 suits for the first pair;
\(C^2_4\) is the number of ways to choose 2 suits for the second pair;
\(C^1_{44}\) is the number of ways to choose the remaining fifth card from 52-8=44 (exclude 4 cards we used for the first pair and 4 cards we used for the second pair). This can also be written as \(C^1_{11}*C^1_4\): choosing value for the fifth card and choosing suit for the fifth card.

For more check: let-s-play-poker-84956.html

Hope it helps.

P.S. Please name topics properly.