Probability in the game of draw poker

Question

Suppose you are playing a game of draw poker and you have 3 aces and two different small cards. If you discard two small cards ,what are your chances of improving your hand on the draw? Improvement can be done by drawing another ace and any other card or by drawing a pair. Because it's a fair game so other players cards are not known so theirs can be treated as if they were part of the deck. I tried it as Total no of possible outcomes 47C2 as 5 cards are already out and we had to choose 2 out of remaining. To improve the chances Favourable cases by drawing ace and any card are 1C1 * 46C1 as one one ace is remain in the deck and any other card can be chose. Favorable cases for drawing a pair are 46C2*1C0 as 47 does not make pair. So probability should be (1C1*46C1 + 46C2 * 1C0 )/47C2 But my answerr is not matching.can someone please point out where is the miatake

Answer 1

Your count of outcomes with the fourth ace is correct.

You can’t just choose $2$ out of $46$ to get a pair; they actually have to be a pair.

There are $10$ ranks with $4$ cards left and $2$ ranks with $3$ cards left. Thus there are $10\binom42+2\binom32=66$ pairs left, and that’s the number of favourable outcomes to get a pair.

So in total the probability to improve your hand is

$$ \frac{\binom11\binom{46}1+10\binom42+2\binom32}{\binom{47}2}=\frac{112}{1081}\;. $$