Combinatorics and probability for cards With a standard 52 deck (13 types and 4 suites), what is the probability that a poker hand contains cards of five different types? A poker hand is a set of 5 cards.
The obvious answer is A / B where A is the number of poker hands that have 5 different types of cards and B is the total number of poker hands which is C(52,5).
Right now I think A = C(13,5) since there are 13 different types of cards I choose 5 from them but I am unsure if A = 4 * C(13,5) instead, where 4 is the different suites.
 A: An alternate way to compute the probability that the types are all different is to imagine that the cards are dealt one at a time. We keep on smiling as long as the cards are all of different types.
After the first card, we are smiling. The probability we are still smiling after the second card is $\frac{48}{51}$, for there are $3$ cards that will stop the smile. Given that we were smiling after the second card, the probability that we are still smiling after the third card is $\frac{44}{50}$, since we need to avoid the first two types. And so on. The required probability is therefore
$$\frac{48}{51}\cdot\frac{44}{50}\cdot\frac{40}{49}\cdot\frac{36}{48}.$$
A: According to the classical definition of probability you should count the number of favourable hands and divide it with the number of possible hands. That is 


*

*Number of possible hands (denominator). There are as you say $$\dbinom{52}{5}$$ different poker hands.

*Number of favourable hands (numerator). There are now $$\dbinom{13}{5}\cdot 4^5$$ ways to choose a hand that has $5$ different hands. That is choose first $5$ types of cards with $C(13;5)$ ways and then from each of these types choose $1$ out of it's $4$ cards with $$4\times 4\times 4\times 4\times 4 =4^5$$ ways. 


Combining the above you have that the requested probability is equal to $$\frac{\dbinom{13}{5}\cdot 4^5}{\dbinom{52}{5}}$$

For example, choose $3, 5, 7, 10, J$. Then out of the $4$ existing $3$'s choose $1$, out of the $4$ existing $5$'s choose $1$ and so on.
A: The correct solution is
$$P = 4^5\frac{\binom{13}{5}}{\binom{52}{5}} = 0.50708$$
For a flush, your approach is correct and gives 
$$P = 4\frac{\binom{13}{5}}{\binom{52}{5}}=0.00198$$
A: There are $52 \choose 5$ total ways of drawing $5$ cards, where order doesn't matter, as you say.  Then there are $13 \choose 5$ ways to choose the five ranks (types) of your cards.  But you now have to choose for each of the five cards which suit it is, and there are not one, but five such choices you have to make.  So the answer is
$$
\frac{{13 \choose 5} \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4}{52 \choose 5}
$$
