Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Calculate the probability that at least 4 draws are required (until a diamond appears) both with and without replacement.

With Replacement:

$$1- ((1-13/52)^0(13/52) + (1-13/52)^1(13/52) +(1-13/52)^2(13/52))$$

Without Replacement: $$1-((13/52)+ (39/51)(13/51)+(39/52)(38/51)(13/50))$$

Is this correct?

share|cite|improve this question
Required for what? – Qiaochu Yuan Jan 29 '12 at 1:51

There is a small error in your "with replacement" answer: The second term of the big parenthetical should be $\frac {39} {52}$ and not $\frac {39} {51}$. Otherwise both are correct.

To calculate it more directly we could reframe the question as, "What is the probability that the first three cards are not diamonds?" Then the answers become:

With replacement: $(\frac {39} {52})^3$

Without replacement: $(\frac {39} {52}) (\frac {38} {51}) (\frac {37} {50})$

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.