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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!

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    $\begingroup$ How many ways are there of drawing five cards from the same suit? Note that there are four suits, so the number of ways of drawing five cards from the same suit is four times, say, the number of ways of drawing five clubs. And how many ways are there of drawing five cards in general? $\endgroup$ – joeb Apr 5 '17 at 1:05
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    $\begingroup$ When you pose a computational question such as this, you should show what you have attempted and indicate where you are stuck so that you receive responses appropriate to your skill level. $\endgroup$ – N. F. Taussig Apr 5 '17 at 1:12
  • $\begingroup$ @N.F.Taussig, you're right, I apologize. I was just in a bit of a hurry, but I'll edit in on what I did initially. $\endgroup$ – ProjectDefy Apr 5 '17 at 3:26
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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.

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  • $\begingroup$ I went ahead and implemented the equation and for some reason my practice quiz is still saying it's wrong. I definitely see your reasoning, and along with others giving the same solution, I have a feeling my TA's setup the practice quiz incorrectly. *ps, I also added how I initially tried to solve it to give some background on my process. $\endgroup$ – ProjectDefy Apr 5 '17 at 3:34
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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 $\frac{33}{16660}$.

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  • $\begingroup$ Along with the others who also gave the same answer, for some reason my practice quiz given by my TA's is saying it's wrong. I do see the reasoning behind what everyone is saying, but no luck. I'm wondering at this point if they setup the question wrong. *ps, I also added how I initially tried to solve it to give some background on my process. $\endgroup$ – ProjectDefy Apr 5 '17 at 3:33
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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.

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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.

$$4 \times \frac{\binom {13}5}{\binom{52}{5}}$$

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  • $\begingroup$ I went ahead and tried your solution (answer: 33/16660), and for some reason it's still saying it's wrong. I do believe this solution (along with others) are correct, but maybe the TA's setup the question wrong online... Will update the main post once I get a word back. *ps, I also added how I initially tried to solve it to give some background on my process. $\endgroup$ – ProjectDefy Apr 5 '17 at 3:36

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