In a standard poker game (no wild cards), suppose you are dealt five cards and your hand contains exactly one pair. You trade in the three worthless cards for new ones. What is the probability that your hand improves? meaning that there is a substantive transformation from one "kind" of hand into another
[Math] Poker game probability question
pokerprobability
Best Answer
You have a pair, so these are the (mutually exclusive) ways of improving your hand:
Note that you cannot get a flush or a straight. Also observe that the three cards you discarded all had distinct numbers, otherwise you would have had two pair.
Let us assume that there are no other players, so that there are $47$ cards remaining. (If there are other players, you need to do essentially the following for each possible hand they're dealt, then sum over all of their possible hands times the probability of those hands. Probably intractable by hand.) Fix the hand you were dealt. We count the number of ways of being dealt each of the preceding possibilities.
Since there are $\binom{45}{3}$ ways of being dealt three replacement cards, we have the probability of improvement is: $$\frac{45\cdot 42 + 45 + 43\cdot 9\binom{4}{2} + 45\cdot 3\binom{3}{2} + 2\cdot 9\binom{4}{2} + 3\binom{3}{2} + \bigg(\binom{9}{1}\binom{4}{3} + \binom{3}{1}\binom{3}{3}\bigg)}{\binom{45}{3}}$$
Google's calculator says that is about $34\%.$
I think I covered everything here without error, but these counting arguments can be tricky.