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!
Best Answer
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.