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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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1 Answer 1

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

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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

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