A chest contains 4 rubies, 8 emeralds, 3 diamonds, and 3 sapphires. Nate will draw jewels one at a time. After each draw, he puts the jewels back into the chest. Nate will go free if he picks some diamonds. How many draws does Nate need until the expected number of diamonds he picks is 1?
The expected value is the number of diamonds. I assume there are $n$ draws. Since he puts the jewels back, I think $P(diamond)=\frac{3}{18}=\frac{1}{6}$. Therefore, $P(not diamond)=\frac{5}{6}$.
I assume there are $n$ draws. There are n possible outcomes. I use $A$ to represent diamonds and $B$ to represent rubies & emeralds & sapphires.
- $1$ draw: $P(A)=\frac{1}{6}$
- $2$ draws: $P(BA)=\frac{5}{6}*\frac{1}{6}$
- $3$ draws: $P(BBA)=\frac{5}{6}^2*\frac{1}{6}$ ...
- $n$ draws: $P(BB...A)=\frac{5}{6}^{n-1}*\frac{1}{6}$.
$E[X]=\frac{1}{6}+2*\frac{5}{6}*\frac{1}{6}+3*\frac{5}{6}^2*\frac{1}{6}*\frac{1}{6}+...+n*\frac{5}{6}^{n-1}*\frac{1}{6}$.
Can anyone tell me if I'm correct?
