Assuming I have a bag with all $26$ letters in it, how many times would I need to draw out of the bag to have a $90\%$ chance of getting at least one of each letter?
I think drawing $26$ times with replacement would be $\frac{26!}{26^{26}}$ to get exactly one of each letter, but I'm have trouble calculating with a bigger number of draws such as drawing a letter (with replacement) $52$ times and trying to calculate the probability that I got at least one of each letter (but not caring about which letters are duplicated or if they are duplicated multiple times). I keep getting numbers greater than $1$.
Background: I bought a bag a choclates with letters on the chocolate, but I didn't get all the letters so I'm wondering how many I would have to buy to get all the letters. I'm assuming they made equal numbers of all the letters and that they made a sufficiently large amount of chocolates so we can ignore that they are not replacing them by the law of large numbers since our sample will be less than $5\%$ of the population.
Best Answer
This is a special case of the so-called "coupon collector's problem"
https://en.wikipedia.org/wiki/Coupon_collector
Calculating the expected number of draws required along with the variance is fairly easy using the linearity of expectation. From here, one can obtain probability bounds using Chebyshev's inequality. I don't know of a way to obtain the pmf directly, although you could try simulation.
Update: I played around with this in R. The Chebyshev bound says that 206 draws will be sufficient to have at least a 90% probability of drawing all 26. The exact tail probability from Ross, pointed out in @awkward's answer, gives $P(T \geq 206) \approx 0.0008$ and $P(T \geq 141) \approx 0.099$ for your example. Here's my R code if you're interested: