I have the next question:
In front of a player is a pitcher with N balls: 1 black and the rest white. The player will receive a prize when he takes out the black ball and for that, he can choose between two methods:
1. Randomly taking out balls without return. The price of each ball is 12 dollars.
2. Randomly taking out with return. The price of each ball is 8 dollars.
Which method has a smaller expected value?
For the first method, I have have calculated the probability function: $P(X=n)=\left[\prod_{m=0}^{n-1}\frac{N-1-m}{N-m}\right]\cdot \frac{1}{N-1-n}
$
So the expected value is: $E(x)=\sum_{n} n\cdot 12\cdot \left(\left[\prod_{m=0}^{n-1}\frac{N-1-m}{N-m}\right]\cdot \frac{1}{N-1-n}\right)
$
The second method has a geometric distribution when the success is taking out a black ball, therefore the expected value is: $E(x)=\sum_{n} n\cdot 8\cdot N
$
The problem is that I don't know how to compare the expected values. Can someone help, please?
Thanks so much for all the help in advance.
HINT
Suppose the $K_1$-th (respectively $K_2$-th) ball taken out is the black ball, in the $1$st (respectively $2$nd) method. What you are comparing is $12 \times E[K_1]$ vs $8 \times E[K_2]$.
$K_1$ is actually the easier one to calculate. Imagine the player always takes out all $N$ balls and line them up left to right, then $K_1$ is simply the position of the black ball in the line. What kind of random variable is $K_1$? What is $E[K_1]$?
$K_2$ as you correctly pointed out is a geometric random variable. You can do the math or you can "cheat" by looking up $E[K_2]$ in the article.
Answer: if you do the math above correctly, you should find that method $1$ is better.
Lemme know if you need more help!