I'm sorry. I'm really bad with combinatorics, so I'm asking you. Lets say I'm holding $2$ cards $5\spadesuit 6\spadesuit$ preflop. My opponent holds $A\clubsuit5\clubsuit$. That makes it only two $5$s and three $6$s in the deck to come. Deck now consists of $48$ cards that can come. What are the odds I would hit exactly $2$ pairs by the river (in the next $5$ cards to come there would be exactly one $5$ and one $6$ from the remaining $5$s and $6$s)?
This is No Limit Holdem game where both players hold 2 hole cards and than comes the shared flop (3 cards), shared turn (1 card), shared river(1 card). So its 5 extra cards to come after both players receive their 2 hole cards. The situation starts preflop (there are no shared cards come up yet). So what are the odds that within the next 5 cards that comes on the board (shared cards) there will come exactly one of the remaining 5's(only 2 left in the deck) and one of the remaining 6's(only 3 left in the deck)
I think this might be the solution :
$$\frac{\left(_1^2\right)\left(_1^3\right)\left(_3^{43}\right)}{\left(_5^{48}\right)} = 0.043$$
what do you guys think, could this be correct?