# Playing card combination probabilities

What is the probability that when drawing 2 playing cards from a standard deck, that the second card is going to be of a lower rank than the first?

the way that I attacked this problem was to number each rank [1-13] for [2-10,J,Q,K,A] and then basically ran through each possible rank of the first card and finding the number of cards in the deck that have a lower rank using the following equation

$\sum_{n=1}^{13}4(n-1) = 312$

Becuase if you draw a 10, mapped to a 9 there are $8\cdot 4 = 32$ cards in a deck that are lower than a 10 that could be drawn that would be lower than the first card. This can then be divided by the total number of 2-card combinations, $52\cdot51 = 2652$, to get a probability of 11.76%.

However, just intuitively, this number just seems way too low. Am I doing something wrong or am I right in my assumption.

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You forgot a factor of $4$ for the first card. But there's also an easier way to do this: The probability of getting the same rank is $3/51$, and if you don't, then the probability of getting a lower rank is $1/2$, so the total probability is $(1-3/51)\cdot1/2=8/17$.