Probability of picking 3 playing cards that belong to different suits

Question

If you have 52 playing cards with 4 different suits (13 spades, 13 clubs, 13 hearts, and 13 diamonds), what is the probability of picking 3 random cards that are all from a different suit?

I tried going about this by writing down all 52 cards and I found out that, if you were to pick a second card, the probability of the second card being from a different suit is 39/51, since there are 51 cards left, with 12 of them being from the same suit as the first card you picked (51-12=39).

However, when I tried to do the same thing with a third card, I got 26/50, since there are 50 cards left, and, assuming the first 2 cards you picked are from a different suit, the probability of the third card being from a different suit would be 50 cards - 12 cards of the same suit * 2 different suits =26 out of 50.

Unfortunately, I realized that it's also possible for the second card you pick to belong to the same suit as the first one, so, in that case, there would also be 50 cards left, and the probability is not the same since there would also be 50 cards left, but 39 cards would belong to a different suit, so I got 39/50.

After that, I realized that I had just managed to confuse myself more than I should have and couldn't really figure out where to go from there. Where did I go wrong and what is the proper way to solve this problem?

Answer 1

You are correct that your chance to get an acceptable second card is $\frac {39}{51}$ and that assuming the second is acceptable the chance for the third to be acceptable is $\frac {26}{50}$. You don't need to worry about the chance the third is acceptable if the second is not because you have already failed. Now you can use the multiplication principle to get the probability of success $$\frac {39}{51} \cdot \frac {26}{50}=\frac {169}{425}$$

Answer 2

There are 4 suits from which you choose 3, C(4,3)=4. There are 4 ways of choosing which suits to represent. For each of the 3 chosen suits, there are 13 options from which to choose one representative, C(13,1)=13. So there are $4*13^3$ ways of selecting these cards.

Overall, the number of ways of choosing 3 cards from a group of 52 is :

$C(52,3)=\frac{52!}{3!49!}=\frac{52*51*50}{6}=26*17*50$

So the probability of choosing 3 cards each of different suits is : number of ways of selecting 3 cards of different suits/ total number of ways of picking 3 cards out of 52.

ans = $\frac{4*13^3}{26*17*50}=\frac{4*13^2}{2*17*50}=\frac{169}{425}$