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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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    1731 views..? Wow...2013-02-28

2 Answers 2

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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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    Awesome approach. Thats refreshing, Thanks for the help everyone.2012-10-12