In a 52 deck card, you draw 4 cards without replacement, what are the odds of picking exactly 2 different suits and a 5?

Question

In a standard 52 card deck, what are the odds of picking exactly:

1. 2 Different suit
2. a 5

Picking two different suits: (52/52) * (39/51)

Picking a 5: (1/52)

0.76 * 0.019 = 0.014

What to do with the remaining card?

Answer 1

I gather from your own attempts that by “odds” you don’t actually mean the odds but the probability.

There are $\binom42$ ways to choose two different suits, and $\binom{26}4$ ways to choose the $4$ cards from those two suits, so the value to start out with for this probability is

$$ \frac{\binom42\binom{26}4}{\binom{52}4}\;. $$

But that also counts selections from a single suit $3$ times, so we have to subtract those $3$ times. And it counts selections without a $5$, so we have to subtract those. But that subtracts each selection from a single suit without a $5$ three times, and we’d already subtracted those, so we have to add those back in three times. If we put that all together, the desired probability comes out as

$$ \frac{\binom42\binom{26}4-3\cdot\binom41\binom{13}4-\binom42\binom{24}4+3\cdot\binom41\binom{12}4}{\binom{52}4}=\frac{23304}{270725}\approx0.086\;. $$

Answer 2

As an alternative to joriki's approach, let's count the number of $4$-card hands with exactly two suits and at least one $5$ as

$${4\choose2}{24\choose2}+{4\choose1}{3\choose1}\left[{24\choose3}-{12\choose3}\right]=23{,}304$$

The first part, ${4\choose2}{24\choose2}$, counts the number of hands with two $5$s: First pick the two $5$s, then pick two more cards from the remaining $24$ cards in the chosen two $5$s' suits. The second part counts the number of hands with just one $5$: First pick a $5$, then pick a second suit, then choose $3$ cards from the $24$ non-$5$s in those two suits, but don't count the choices where all $3$ cards come from the remaining $12$ cards in the suit with the chosen $5$.