Combinatorics in card games With a regular set of playing cards (52, 4 of each number). Out of x draws, how can I find the number of draw combinations in which two or more cards have the same number.
 A: We interpret the problem as follows. We draw, without replacement, a given number $x$ of cards. We want to count the number of "hands" in which two or more cards have the same number. Call such a hand good.
Note: For another interpretation that gives a pretty result, see the Remark at the end.
There are $\dbinom{52}{x}$ $x$-card. We will now count the number of bad $x$-card hands, hands in which no two cards have the same number.
We must choose $x$ "numbers" from the $13$ available. This cannot be done if $x\gt 13$. If $x\le 13$, it can be done in $\dbinom{13}{x}$ ways.
For every one of these $x$ numbers, there are $4$ ways to pick the actual card with that number.  So once we have chosen the numbers, there are $4^x$ ways to make a bad hand that contains these numbers. Thus the number of bad hands is $\dbinom{13}{x}4^x$, and therefore the number of good hands is 
$$\binom{52}{x}-\binom{13}{x}4^x$$
(if $x\le 13$).
Remark: If we allow $x$ to be variable, we can obtain the total by summing over all $x$. However, there is a simpler way to find the answer.
The number of subsets of the cards is $2^{52}$. The number of "bad" hands (subsets) is 
$$\sum_{x=0}^{13}\binom{13}{x}4^x.$$
We recognize this as the binomial expansion of $(1+4)^{13}$. 
There is an easier way of seeing this. Write down the $13$ numbers, as in Ace, $2$, $3$, and so on up to King. We want to make a "bad" hand, that is, a hand with no repeated number. So at each of Ace, $2$, $3$, and so on, we have $5$ choices: Spade, Heart, Diamond, Club, and Skip, for a total of $5^{13}$. 
