# How many draws to have a 90% chance of getting the ace of spades?

You have a standard 52-card deck, and you want to take the minimum number of draws from a random/shuffled deck such that you have a 90% chance of drawing the ace of spades. How would you find the minimum number of draws to achieve this 90% probability of succeeding in drawing the ace of spades at least once, for both the case of replacement and non-replacement?

• do you replace the cards when you draw them? – clark Apr 29 '16 at 2:47
• – Frenzy Li Apr 17 '17 at 9:34

Without replacement is (for once) easier. The Ace of Spades is equally likely to be in any of the $52$ positions, so we need to draw $n$ cards, where $n$ is the smallest integer $\ge (0.90)(52)$.
For with replacement, the probability we don't see the Ace of Spades in $n$ draws is $\left(\frac{51}{52}\right)^n$. We want this to be $\le 0.1$. Thus we want $n\ln(51/52)\le \ln(0.1)$, or equivalently $n\ge \ln(10)/(\ln(52/51)$.