Prob of sum of cards being multiple of die roll Suppose you roll a die. If the roll yields a 1, you win. If you roll above a 1, you take the corresponding number of cards from a deck consisting of one suit (13 cards). 
To win, the sum of the cards pulled needs to be a multiple of the die roll (i.e., if I roll a 3, I need the sum to be 3,6,9...18, etc). The card values go from ace = 1 to jack,queen,king = 10. Say I roll a 2. How do I find the probability of the 2 cards sum being a multiple of 2?
Edit
Roll: 5
{5,10,j,q,k} {1,6} {2,7} {3,8} {4,9}
(5c5) + (2c2)(2c2)(5c1) + (5c1)(2c1)(2c1)(2c1)(2c1) / (13c5)
 A: The number of ways to choose $2$ out of $13$ cards is $\dbinom{13}{2}=78$.
To get an even sum, we need to choose $2$ out of $5$ odd cards or $2$ out of $8$ even cards.
The number of ways to choose $2$ cards with an even sum is $\dbinom{5}{2}+\dbinom{8}{2}=38$.
So the probability of choosing $2$ cards with an even sum is $\dfrac{38}{78}\approx0.487$.
A: As barak manos answered, to pick 2 cards which sum to an even number you must pick either 2 of the 8 even value cards $\{2,4,6,8,10,\mathrm J,\mathrm Q,\mathrm K\}$, or 2 of the 5 odd value cards $\{\mathrm A, 3,5,7,9\}$. This is out of all the ways to pick any 2 of 13 cards.
$$\frac{{8\choose 2}+{5\choose 2}}{13\choose 2}$$

The same principle is applied to higher values. For example, to pick three cards that sum to a multiple of three you need either three from one of the following sets, or else one from each of the sets.  The sets being: 
$$\{3,6,9\}, \{\mathrm A,4,7,10,\mathrm J,\mathrm Q,\mathrm K\}, \{2,5,8\}$$  
These sets are formed of cards whose values are, respectively, $3n, 3n+1, 3n+2$ for various $n\in\{0,1,2,\ldots\}$.  (Going by your post, the face cards all have a value of $10$, and the Ace has a value of $1$).
There is $1$ way to pick three cards from $\{3,6,9\}$.  There are ${7\choose 3}$ ways to pick three cards from $\{\mathrm A,4,7,10,\mathrm J,\mathrm Q,\mathrm K\}$.  There is $1$ way to pick three cards from $\{2,5,8\}$.  Finally there are $3\times 7\times 3$ ways to pick one card from each set.  This is out of a total of $13\choose 3$ ways to pick three cards from the deck.
$$\frac{1+{7\choose 3}+1+3\times 7\times 3}{13\choose 3}$$

To extend to four cards who sum to multiple of 4, you need to select from sets of cards whose values are $4n, 4n+1, 4n+2, 4n+3$, in appropriate combinations.
$$\{4,8\}, \{\mathrm A,5,9\}, \{2,6,10,\mathrm J,\mathrm Q,\mathrm K\}, \{3,7\}$$  
$$\frac{{2\choose 2}{3\choose 1}{2\choose 1}+{2\choose 2}{6\choose 2}+{2\choose 1}{3\choose 2}{6\choose 1}+{2\choose 1}{6\choose 1}{2\choose 2}+{3\choose 2}{2\choose 2}+{3\choose 1}{6\choose 2}{2\choose 1}+{6\choose 4}}{13\choose 4}$$
