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

A standard deck of 52 cards. It seems obvious that the probability of getting a particular suit (say diamonds) on the first try is 13/52. After that, if we do not get diamonds, our chances are 13/51. But how should we calculate the probability of getting at least once diamonds if we have two attempts?

share|cite|improve this question
The easiest way is to calculate the probability of not getting a diamond in the first two cards and subtract that from $1$. – Brian M. Scott Sep 24 '12 at 21:22

Only read this after you've figured it out.

The number of trials that do not satisfy your condition is $\frac{39\cdot 38}{2}$: Select a non-$\diamondsuit$, then another one, and all you care about is, at the end, which suits you have, not the order in which they arrived, so divide by 2 to account for the same result occurring in two distinct ways.
Similarly, the number of total trials is $\frac{52\cdot 51}{2}$. You can use these to find the probability of failure, which easily leads to the probability of success.

share|cite|improve this answer

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.