Last week, we played around with probability and a deck of cards. If you haven’t had a chance to try the puzzle yet, click here.

It’s more likely that the sum of two cards is even. We can easily check this. Remember that the sum of two even values is even, the sum of two odd values is even, and the sum of an even and an odd value is odd. You have an 8/13 chance of choosing an even card (suits, of course, don’t matter) and a 5/13 chance of choosing an odd card.

The probability that that the sum of your two cards is even = (8/13)(8/13) + (5/13)(5/13).

We can write this as (8/13)^2 + (5/13)^2.

The probability that the sum of your two cards is odd = (8/13)(5/13) + (5/13)(8/13). We can write this as 2(8/13)(5/13).

We can simply see that (8/13)^2 + (5/13)^2 > 2(8/13)(5/13)

If we take the difference between the the probability that the sum is even and the probability that the sum is odd, we can also see which is more likely. If the difference is positive, than an even value is more likely; if it’s negative, then odd is more likely.

(8/13 – 5/13)^2 > 0

We could approach the second question this way and simply calculate it. But we’re going to group the second and third question together and show that we can figure out which is more likely for any arbitrary case.

We know the probability that the sum is even or odd = 1. There are only two cases, so there is a 100% chance that we’ll come up with one of them. (We can check this too, by looking at the first case, in which we’re pulling only one card: 8/13 + 5/13 = 1.)

1 = (8/13 + 5/13)^k, where k is the number of samples.

We can use the same approach to see the difference between the probability that the sum is even and the probability that the sum is odd.

(8/13 – 5/13)^k.

The key is to note that in the expansion of the above expression, all the terms that represent even sums have a positive sign and all the terms that represent odd sums have a negative sign.

We can use those results together to quickly calculate the probability that the sum is even.

(1+ (8/13-5/13)^k)/2