Suppose you are given 6 cards. Which is more likely, you get $3$ different value cards with value having $2$ suits. (e.g. two aces two kings and two jacks). Or $2$ different value cards with $3$ different suits. (e.g. three kings and three 10's)?
I calculated that you would be more likely to get the hand with the two different values (e.g. three kings and three 10's) but I'm not sure if this is right. Any thoughts?
We assume the cards are drawn at random, without replacement. There are $\binom{52}{6}$ equally likely hands. Now it is a matter of counting the "favourables" in each case.
For $3$ different values, and $2$ of each, the number of favourables is $\binom{13}{3}\binom{4}{2}^3$. For $2$ different values, $3$ of each, the number is $\binom{13}{2}\binom{4}{3}^2$. The first number is much larger than the second. This is the opposite of your conclusion.