Probability of drawing 5 cards from a deck of 52 that will have the same suit? A standard deck of cards has 52 members consisting of 4 suits each with 13 members.  Five cards are dealt from the randomly mixed deck.  What is the probability that all cards are the same suit?
EDIT: How I went about it before posting this question was doing (1/4) as the first card probability because my thought process was that we'll draw 1 suit out of the 4 for the first probability. Then I proceeded to account of the 2nd dealt card with the probability of (12/51) since 1 card has been dealt already out of the 13 cards for that suit, also subtracting 1 from the total amount of cards able to be dealt. 
So for the 3rd card: (11/50)
4th card: (10/49)
5th card: (9/48)
Giving us the total overall probability for drawing 5 cards of the same suit: $$ (1/4) * (12/51) * (11/50) * (10/49) * (9/48) = 33/66640 $$
EDIT2: My practice quiz given by TA's is still saying I have the incorrect answer. Given how the answer should be: $ 33/16660 $ (explained in numerous ways in the thread), I contacted the TA's to see if maybe the have setup the question incorrectly. Will update when I get an answer back.
EDIT3: Got an answer back from my TA's who tested the test. They did have the answer wrong on their end. Everyone who helped me was correct!
 A: The first card has probability $\frac{52}{52}$ of having the same suit as any previously drawn cards (because there are none).  This means there is a 100% chance of the first card meeting our criteria.
The second card has probability $\frac{12}{51}$ because there are twelve left out of 51 total that match the suit of the first card.
The third, fourth and fifth cards have probabilities $\frac{11}{50}$, $\frac{10}{49}$, and $\frac{9}{48}$ because there are less and less of the suit of the first card as well as less cards to choose from.
Because I need the first thing to happen AND the second thing to happen AND, ..., I need to multiply the probabilities: $\big(\frac{52}{52}\big)\big(\frac{12}{51}\big)\big(\frac{11}{50}\big)\big(\frac{10}{49}\big)\big(\frac{9}{48}\big)$
You obviously don't need that first fraction, but I think it adds clarity.  Five cards, each with their own probability.
A: We know that the first card determines the suit, the second must be one of the $12$ out of $51$ remaining cards with that suit, the third must be one of the $11$ out of $50$ remaining cards with the same suit, etc. The reason the the numerator and denominator keep decrementing is because each time we choose a card, it takes one card away from the pool of the suit and the pool of total cards. So we should get the probability to be $$\prod_{n=1}^4 \frac{13-n}{52-n}$$ which equates to $\frac{33}{16660}$.
A: Let's use binomial coefficients to determine the number of ways that this is possible. Then I'll leave it to you to [use binomial coefficients to] divide this by the total number of outcomes to arrive at a probability. 
For a 5-card poker hand, this is how we can draw n cards of the same suit. 
$${13\choose n} \cdot{39 \choose 5-n}  $$
The left factor is the chosen/given suit, and the right factor is all three of the other suits. 
If we set $n=5$, that will give us the number of ways to draw all 5 in one of the suits. As there are 4 suits, we multiply the result by 4. 
A: Hint -
Probability of getting 5 cards of same suit -
$$\frac{\binom {13}5}{\binom{52}{5}}$$
Now multiply with 4 as we have 4 suits.
$$4 \times \frac{\binom {13}5}{\binom{52}{5}}$$
