Probability Average amount of rolls

Question

I have a question regarding probability. I'll start by saying I've never taken a statistics or other similar course and was trying to work out this for a game. On average how many attempts will it take to get all 24 outcomes if there is a 1/16 chance of getting an outcome and there can be duplicates of the outcome. My thought was $\sum_{i=1}^n (256/i)$ where n = 24 This would make this answer ~ 966.6 attempts. Wondering if I did this right. If not, could someone please explain the correct answer to me.

Answer 1

Trying to interpret your question:

Suppose you have $24$ very biased coins so that there is a $\frac{1}{16}$ probability that each shows heads when flipped. Each round you flip all the coins. What is the expected number of rounds until you have seen each coin show heads at least once?

• The probability that a particular coin has never shown heads in $n$ flips is $\left(\frac{15}{16}\right)^n$
• The probability that a particular coin has shown heads at least once in $n$ flips is $1-\left(\frac{15}{16}\right)^n$
• The probability that all $24$ coins have shown heads at least once in $n$ rounds is $\left(1-\left(\frac{15}{16}\right)^n\right)^{24}$
• The probability that all $24$ coins have shown heads at least once at the $n$th round is $\left(1-\left(\frac{15}{16}\right)^n\right)^{24}-\left(1-\left(\frac{15}{16}\right)^{n-1}\right)^{24}$
• The expected number of rounds needed so that all $24$ coins have shown heads at least once $\displaystyle \sum_{n=1}^{\infty} n\left(\left(1-\left(\frac{15}{16}\right)^n\right)^{24}-\left(1-\left(\frac{15}{16}\right)^{n-1}\right)^{24}\right) $
• This seems to be about $59.00705$