In 5 card draw you can discard up to 4.
Say you have a pair of kings and an ace. I would like the math behind keep the ace or draw 3.
If you draw 3 you have a better chance of trips.
But two pair with a pair of aces is only 1 hand down from trips.
What are the chances of improving with discard 2 versus 3?
Drawing 3 you have more chances to hit a king (or 2).
Drawing 2 you have 5 cards to hit - 2 K and 3 A.
On
I am going to ignore drawing a pair.
Say AKKT7
52-5 = 47 cards left
I am going to ignore drawing another pair in both cases.
Discard AT7 what is the chance of trip K or quad K.
$$\dfrac{\binom{2}{2}\cdot\binom{45}{1}+\binom{2}{1}\cdot\binom{45}{2}}{\binom{47}{3}} = 0.1248843663$$
Discard T7 what is the chance of trip K or quad K or an A.
$$\dfrac{\binom{5}{2}+\binom{5}{1}\cdot\binom{42}{2}}{\binom{47}{2}} = 0.2035152636$$
You are 63% more likely to improve drawing 3 but you also are more likely to draw to a weaker hand. Intuitively that number seem reasonable. How often is 2 pair going to lose to trips? Not often but when it does you are likely to lose a big pot. So I think I would stay with draw 3.
Intuitively, the chance of improving has to be better drawing three. When you draw the first of three, you might get a match and have improved. Otherwise, you are drawing two to a pair plus one card so have the same chance for improvement as if you kept an ace.
To compute the chance of improvement, it is easier to compute the chance of no improvement and subtract from $1$. If you keep a pair plus and ace, you do not improve on the first card with probability $\frac {42}{47}$ because there are five cards of $47$ that can help you. If you didn't improve on the first the chance you fail $\frac {38}{46}$ of the time as there are now eight cards that can help. The chance of improvement is then $$1-\frac {42}{47}\frac {38}{46}\approx 0.2618$$
For drawing three to a pair, the same logic gives the chance of improvement as $$1-\frac{45}{47}\frac {41}{46}\frac {37}{45}\approx 0.2983$$