I understand how to do this for a single hand.
12/51 which is about 24%
But what if I am playing three separate hands. What is my chance of one of those hands having two suited cards?
I'm not sure how to calculate the probability of 3 chances to hit a 24% chance.
Before beginning, here is a much easier solution (which is unfortunately incorrect) that gives us some insight as to how probabilities work.
Since there is (as calculated correctly by you) a $\frac{12}{51}$ chance of a single 2-card being monosuit, we can model this via a binomial distribution with $n=3, p=\frac{12}{51}, q = 1-p = \frac{39}{51}$. The probability of at least one mono-suit hand is then $Pr(X\geq 1) = 1 - Pr(X<1) = 1 - Pr(X=0) = 1 - \binom{3}{0}(\frac{12}{51})^0(\frac{39}{51})^3 = \frac{2716}{4913}\approx 0.55282$.
However, it is only valid to use a binomial distribution when each chance of success/failure is independent of one another (which would be the case had we shuffled each hand back into the deck before dealing the next hand). Unfortunately, we need to take a much more tedious approach to get the exact value.
(Read more at Binomial-Distribution and Inclusion-Exclusion)
We consider a deal of three 2-card hands from a standard 52 card playing card deck and ask the question of the probability that at least one of the three hands are monosuit. An equivalent question is what is the opposite probability of none of the hands being monosuit.
I will approach this with the sample space being All ways of drawing six cards from a deck where order of cards matters. There are $52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47 = 14~658~134~400$ different ways to deal six cards from a deck (where order matters).
As all hands are not monosuit, we have the following cases:
Let us inspect how many possible outcomes of each such scenario exist:
This could only occur if the cards are in groups of two with each pair having one of each suit. E.g.$(\spadesuit \heartsuit)(\spadesuit\heartsuit)(\heartsuit\spadesuit)$ Pick the two suits that appear ($\binom{4}{2}=6$ choices), pick which order the suits appear ($2\cdot 2\cdot 2=8$ choices), pick the cards to go in each slot ($13\cdot 13\cdot 12\cdot 12\cdot 11\cdot 11$). For a total of $6\cdot 8\cdot 13^2\cdot 12^2\cdot 11^2 = 141~343~488$ possibilities in case 1.
Since the suit with 3 cards cannot be paired together, it must look something like $(\spadesuit\heartsuit)(\spadesuit\heartsuit)(\spadesuit\clubsuit)$. Pick the suit with 3 cards ($4$ choices), pick the suit with 2 cards ($3$ choices), pick the suit with one card ($2$ choices), pick which hand receives the suit with one card ($3$ choices), pick the order of the suits in each hand ($8$ choices), pick the cards for each suit ($13\cdot 13\cdot 12\cdot 12\cdot 11\cdot 13$ choices) for a total of $4\cdot 3\cdot 2\cdot 3\cdot 8\cdot 13^3\cdot 12^2\cdot 11 = 2~004~507~648$ different deals in the second case.
Pick the suit with the 3 cards ($4$ choices), pick which hand each of the cards from individual suits go ($3! = 6$ choices), pick the order of the suits in each ($8$ choices), pick the cards for each suit ($13^4\cdot 12\cdot 11$ choices) for a total of $4\cdot 6\cdot 8\cdot 13^4\cdot 12\cdot 11 = 723~849~984$ possibilities in the third case.
Each suit must be paired with one of each of the other suits such as $(\spadesuit\heartsuit)(\spadesuit\clubsuit)(\heartsuit\clubsuit)$. Else if you tried to do the same pairing twice you run into problems such as $(\spadesuit\heartsuit)(\spadesuit\heartsuit)(\clubsuit\clubsuit)$. Pick which three suits appear $\binom{4}{3} = 4$, (we already know in what way they are paired), Pick the specific arrangement of hands ($3!=6$), Pick the order of suits in each hand ($8$ choices), Pick the specific cards for each suit ($13^3\cdot 12^3$ choices) for a total of $4\cdot 6\cdot 8\cdot 13^3\cdot 12^3 = 728~911~872$ different deals in the fourth case.
This case is the most difficult, as there are multiple ways they could be paired. Break this into sub cases:
(note, it is impossible for them to be paired in none of the three hands)
Sub case 1: looks something like $(\spadesuit \heartsuit)(\spadesuit\heartsuit)(\clubsuit\diamondsuit)$. Glossing over the details (same method as before) there are $\binom{4}{2}\cdot 3\cdot 8\cdot 13^4\cdot 12^2 = 592~240~896$ total possibilities
Sub case 2: looks something like $(\spadesuit \heartsuit)(\spadesuit \clubsuit)(\heartsuit \diamondsuit)$. Pick the two suits that have two cards each ($\binom{4}{2}=6$ choices), pick the way in which the suits with one card get matched with the suits with 2 cards ($2$ choices), pick the order of the three hands ($3!=6$ choices), pick the order of suits in each hand ($8$ choices), pick the specific cards for each suit ($13^4\cdot 12^2$ choices) for a total of $6\cdot 2\cdot 6\cdot 8\cdot 13^4\cdot 12^2 = 2~368~963~584$ possibilities.
Thus, there are:
$$141~343~488 + 2~004~507~648 + 723~849~984 + 728~911~872 + 592~240~896 + 2~368~963~584 = 6~559~817~472$$
possible deals of six cards into three hands which do not have any monosuit hands.
The probability of no monosuit hands is then:
$$\frac{6~559~817~472}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47} = \frac{25~766}{57~575}\approx 0.44752$$
Finally the probability of getting at least one monosuit hand is:
$$1 - \frac{25766}{57575} = \frac{31809}{57575} \approx 0.55248$$