# Number of cards until a diamond appears

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

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.
With replacement: $(\frac {39} {52})^3$
Without replacement: $(\frac {39} {52}) (\frac {38} {51}) (\frac {37} {50})$