Probability of getting the same 3 cards from a 60 card deck?

Question

What is the probability of getting the same 3 cards from a 60 card deck?

Order doesn't matter and the cards are placed back into the deck and shuffled between deals. Not the best with probability, so I felt a little uneasy trying to solve this.

My initial thought was it was just two independent cases of [60 choose 3] = 34220² Making the outcome 1,171,008,400 and the probability 1 / 1,171,008,400.

Not sure if I made a mistake somewhere and don't feel confident in that answer

Answer 1

From a close reading of your problem statement, and attempted solution,
I get that $3$ cards are dealt, replaced and shuffled, and the probability you seek is that the same three cards are drawn in the second deal.

Now any three cards might have been drawn on the first deal, it doesn't matter which,
only the second deal has to match, so the probability is just $\frac 1{\binom{60}3} = \frac1{34,220}$

Answer 2

You have to make three independent choices.

This means that you choose the first card with a probability of $\frac{60}{60}$.

Then you put it into the deck and draw the second that as a probability of $\frac{1}{60}$ to be equal to the first.

Then you put it into the deck and draw the third that as a probability of $\frac{1}{60}$ to be equal to the first and the second.

At the end you get that the probability of drawing for three times the same card is: $$1\cdot\frac{1}{60}\cdot\frac{1}{60}=\color{blue}{\frac 1{3600}}$$