Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $(1-3/51)\cdot1/2=8/17$.

share|cite|improve this answer
I did indeed forget that factor of 4, and that ended up fixing everything right up. When I do it my method, I get the same result as you. Of course your's is far more straightforward, but it did not come to initially. Thanks a lot! – MZimmerman6 Oct 28 '12 at 1:31
@MZimmerman6: You're welcome! – joriki Oct 28 '12 at 1:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.