I want to solve this problem.
Consider a deck of 52 cards, ordered such that Ace > King > Queen > ... > 2. I pick a card, then you pick one. What is the probability that my card is larger than yours?
I thought of doing it like this, but it is wrong and I don't see why.
Denote the cards by values 1 to 13, so the card "2" is denoted by 1 and the card "Ace" is denoted by 13. This is just a transformation.
Let $p$ be the desired probability. Then $$p = \sum_{i=1} P(\text{my card} > \text{your card } | \text{ I pick card of value i})P(\text{I pick card of value i})$$ by law of total probability.
We have $$P(\text{I pick card of value i}) = \frac{4}{52} = \frac{1}{13}$$ since there are 4 cards of value $i$ in a deck (eg. there are 4 aces in a deck).
We also have $$P(\text{my card} > \text{your card} | \text{I pick card of value i}) = \frac{i-1}{12} + \frac{i-1}{13}\times 3$$ because if I pick a card of value $i$ then the cards $1, ..., i-1$ are smaller than my card, and there are 12 cards from the suite my card is from and 13 cards in the 3 remaining suites.
So we get
$$p= \frac{1}{13}\times (\frac{1}{12} + \frac{3}{13})\times \sum_{i=1}^{13} i-1 = \frac{1}{13}\times (\frac{1}{12} + \frac{3}{13})\times 66$$ and this is bigger than $1$. What's the error in my method and how do I fix it?
The correct (and simpler) expression is $$P(\text{my card} > \text{your card} | \text{I pick card of value i}) = \frac{(i-1)\times 4}{51}$$
Simply: the numerator is the possible cards that are lower than $i$, and $51$ are the available cards.
AS noted in the comments, this can be expressed as
$$ \dfrac{(i-1) + (i-1)\times 3}{12+13\times 3}$$
where the first terms in each sum correspond to the selected rank. But this does not really help. And of course, the mistake amounts to assuming $\frac{a}{b}+\frac{c}{d}\neq \frac{a+c}{b+d}$ This is already noted by
JMoravitz' comment.