Probabilities of the sum of two cards - the sum of two cards equaling a number

Question

I'm trying to find the probability of a card game that is summarized as the sum of two cards - the sum of two cards Ace = 1 Jack = 11 Queen = 12 King = 13 (Normal deck of 52 cards) I'm unsure how to find the probability of different outcomes How would you find the probability of getting 13 for example? - The only way that I can think of is writing every single possible outcome down and then calculating the probability for each individual outcome then adding it up. e.g. (13,13) - (12,1)

Answer 1

I am assuming that you are simply drawing two different cards at random from the deck and desire the probability that their sum is some given value. I’ll illustrate calculations for a couple of values.

Suppose that $k+\ell=13$, where $k\le\ell$. Note that $k$ and $\ell$ actually cannot be equal, since $13$ is odd, so $1\le k\le 6$. For each integer $k$ with $1\le k\le 6$ there are $4$ cards with value $k$ and $4$ cards with value $\ell=13-k$, so there are $4^2=16$ pairs of cards with values $k$ and $\ell$. There are $6$ possible choices for $k$, so there are altogether $6\cdot16=96$ pairs whose sum is $13$. There are $\binom{52}2=1326$ pairs of cards altogether, so the probability of drawing a pair whose sum is $13$ is $\frac{96}{1326}=\frac{16}{221}$. A similar approach will work for any odd total; you just have to determine which values of $k$ are possible.

It’s a little trickier if you want an even total, say $12$, because now $k$ can equal $\ell$: two of the cards can have the same value, in this case $6$. For each integer $k$ with $1\le k\le 5$ there are again $4^2=16$ possible pairs with values $k$ and $12-k$, and there are $\binom{4}2=6$ pairs of cards each with value $6$, for a total of $5\cdot16+6=86$ pairs whose sum is $12$, and the probability of drawing a pair whose sum is $12$ is $\frac{86}{1326}=\frac{43}{661}$. Again, a similar approach will work for any even total, once you determine which values of $k$ are possible.

Answer 2

Let $S$ be the sum and consider the following four cases:

1. $S\le 14$, $S$ odd: $N=4\cdot4\cdot(S-1)$

2. $S\le 14$, $S$ even: $N=4\cdot4\cdot(S-2)+4\cdot3$

3. $S> 14$, $S$ odd: $N=4\cdot4\cdot(27-S)$

4. $S> 14$, $S$ even: $N=4\cdot4\cdot(26-S)+4\cdot3$

Dividing $N$ by $52\cdot 51$ you will obtain the corresponding probability.