Clarification: A Euchre deck consists of just A-K-Q-J-10-9 of each suit, thus four cards in each of six ranks.
I'm confused on how to approach this problem, at first I thought it was fairly straightforward and used to combinations to determine the probability but realized that it is not the correct result. This is what I did
4C2 * 20C3 = 6,840, for getting two cards of the same rank in a hand of five
4C3 * 20C2 = 760, for getting three cards of the same rank in a hand of five
4C4 x 20C1 = 20, for getting four cards of he same rank in a hand of five
I then added them together which is 7,620 and divided by the combinations of Euchre hands 42,504 which is ≈ 17.93%
To test my math I created a Monte Carlo simulation with 5,000,000 iterations and the result was, any hand with at least two or more cards of the same rank occurred ≈ 85.54086% of the time which is nowhere near my previous result.