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

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?

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!

• You have it just right. Oct 12, 2012 at 13:37
• Looks fine to me, except that you should write $\approx0.1055$, not $=0.1055$: reserve the equals sign for things that are genuinely equal. Oct 12, 2012 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. Oct 12, 2012 at 13:42
• @ThomasAndrews How did you arrive at $13^4$? I don Oct 12, 2012 at 14:09
• You pick one spade (13 ways) one heart (13 ways) one diamond (13 ways) and one club (13 ways). @SingaporeanDude. Oct 12, 2012 at 14:57

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.

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.

• 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. Oct 12, 2012 at 13:50
• Awesome approach. Thats refreshing, Thanks for the help everyone. Oct 12, 2012 at 13:52