Probability of getting at least two cards in the same suit Three cards are drawn from one standard deck of 52 cards. What is the probability of at least two of the cards being in the same suit?
There are a total of $52C3$ sets of 3 cards which could be chosen. You are looking for two positive outcomes: all 3 from the same suit; or two from the same suit and one from another. There are $4 \times (13C3)$ sets of three cards which are all from the same suit. And there are $4 \times (13C2) \times (3\times13)$ sets of cards where two are from the same suit and one is from another. Note here: the $4 \times (13C2)$ comes from choosing a suit (4 options) and then choosing 2 cards from that suit $(13C2)$ options, and then choosing any other card from the other 3 suits ($13\times3$ choices). Together, this gives
$$
\frac{4 \times(13C3) \quad + \quad 4 \times(13C2) \times 39}{(52C3)} = 0.60235...
$$
Alternatively, look at the negation. I pick 3 cards. In how many ways can they all be from different suits? The first card can be anything. The second card can be from any of the other 3 suits (so 39 choices). The third can be from any of the other two suits (so 26 choices). Together, this gives
$$
1 - \frac{52 \times 39 \times 26}{52\times51\times50} = 0.60235...
$$
also giving the same result.
Note, here $(nCr)$ denotes the choice function.
There are in total $C_{52}^{3}$ options if you are going to take random 3 cards from a standard deck.
To have at least 2 cards being in the same suit, you either need to withdraw 3 cards of a same suit, or 2 cards of one suit and one of the other. So there are $$4 \cdot C_{13}^{3} + 4 \cdot C_{13}^{2} \cdot 3 \cdot C_{13}^{1}$$ available options.
Therefore, the probability is $(4 \cdot C_{13}^{3} + 4 \cdot C_{13}^{2} \cdot 3 \cdot C_{13}^{1} )/ C_{52}^{3}$.
Check the combination calculator out if you need the precise result.