What is the probability of drawing 5 cards of the same type from a deck of 52 cards?

Question

My textbook has the following problem:

There is 52 cards in a deck and 13 cards of each type/color. You are drawing 5 cards. Whats the probability of all these 5 cards being the same type?

My solution:

There is a $\frac{52}{52}$ probability of drawing the first card, then a $\frac{12}{51}$ chance of drawing a second card of the same type as the first one and so on... $$\frac{52}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48} = \frac{33}{16660}$$

I solved the same problem by calculating the individual probabilities for each card type and then adding all the probabilities together:

$$4\cdot(\frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48}) = \frac{33}{16660}$$

My textbook says the solution is $(\frac{13}{51} \cdot \frac{12}{50} \cdot \frac{11}{49} \cdot \frac{10}{48})$ without any explanation. Although I don't see how you arrive at this answer.

Whats wrong with my way of solving the problem and how do you arrive at the textbooks solution?

Answer 1

As has been discussed in the comments, your solution is correct and the one in the textbook is wrong. It seems they forgot to subtract the first card from its suit, which makes the numerators off by $1$.

Answer 2

Your solution $4\cdot(\dfrac{13}{52}\cdot\dfrac{12}{51}\cdot\dfrac{11}{50}\cdot\dfrac{10}{49}\cdot\dfrac{9}{48})$ is correct. This does simplify to $\dfrac{33}{16660}$. Your textbook actually just has the incorrect answer.