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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?

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Required for what? –  Qiaochu Yuan Jan 29 '12 at 1:51
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1 Answer 1

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: $(\frac {39} {52}) (\frac {38} {51}) (\frac {37} {50})$

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