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My simple easy homework question. Just needed some double check :D

Deal 4 cards from a deck of 52 cards. What is the probability that we get one card from each suit?

My answer

First Draw: We can get any card, and the card's suit will be done. $Chance:1$

Second Draw: Now we need to get 1 of the 3 remaining suits. There are 51 cards left. $Chance:\frac{13+13+13}{51}$

Third Draw: Now we need to get 1 of the 2 remaining suits. There are 50 cards left. $Chance:\frac{13+13}{50}$

Fourth Draw: Now we need to get the last remaining suit. There are 49 cards left. $Chance:\frac{13}{49}$

$P($One card from each suit$)=1*\frac{13+13+13}{51}*\frac{13+13}{50}*\frac{13}{49}=0.1055$

My tutor is known for giving not-so straightforward questions, so I'm wondering if I need to consider another way, or I could be wrong. Any alternatives welcome too!

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You have it just right. –  Ross Millikan Oct 12 '12 at 13:37
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Looks fine to me, except that you should write $\approx0.1055$, not $=0.1055$: reserve the equals sign for things that are genuinely equal. –  Brian M. Scott Oct 12 '12 at 13:38
    
Yes. Alternatively, you can say there are $\binom {52}{4}$ ways of picking four cards from a deck, and $13^4$ ways to pick one card from each suit, so the probability is $$\frac{13^4}{\binom{52}{4}}$$ This is the exact same value you got, just arrived at differently. –  Thomas Andrews Oct 12 '12 at 13:42
    
@ThomasAndrews How did you arrive at $13^4$? I don –  Yellow Skies Oct 12 '12 at 14:09
    
You pick one spade (13 ways) one heart (13 ways) one diamond (13 ways) and one club (13 ways). @SingaporeanDude. –  Thomas Andrews Oct 12 '12 at 14:57
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2 Answers

up vote 1 down vote accepted

first draw: Pick any card, probabilty 1 you are still OK

second draw: you must pick from 39 cards that won't wreck your hand out of 51 cards

third draw: you must pick from 26 of the remaining 50

fourth draw: you mustpick from 13 of the remaining 49.

Altogether, you get a probability of

$$1\cdot {39\over 51}\cdot{26\over 50}\cdot{13\over 49}. $$

You have it.

Here is a second solution. There are ${52\choose 4}$ hands of size 4. Now pick the four cards of different suits; there are $13^4$ ways to do this.

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The following is an (inferior) alternative. There are $\dbinom{52}{4}$ ways to choose $4$ cards, all equally likely.

There are $\dbinom{13}{1}^4$ ways to choose $1$ card from each suit. Divide.

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It's inferior from a "beginning probability" approach, but it is nicer for seeing how to approach other probiems. Example: What is the probability when selecting 8 cards that there will be two cards from each suit? Trying to answer that with the basic approach above would be painful, but your approach makes it trivial. –  Thomas Andrews Oct 12 '12 at 13:50
    
Awesome approach. Thats refreshing, Thanks for the help everyone. –  Yellow Skies Oct 12 '12 at 13:52
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