Comparing two expected values

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.

• I would calculate the expected values for a number of values of N. Enough to compare the trends of the two methods with each other. Commented Dec 25, 2019 at 17:26
• Can you explain to me how to do this, please? Commented Dec 25, 2019 at 17:44
• I think you got a good explanation in the answer below. What you want is a domain of N large enough so you can see any changes in which methods yields the highest or lowest expected value. Merry Xmas BTW. Commented Dec 25, 2019 at 20:47
• Thank you for the explanation. Merry Christmas to you too:) Commented Dec 25, 2019 at 20:52

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!

• Thank you very much for the detailed solution. So I just what to be sure I got it right. $K_{1}$ discrete uniform value, therefore, it's expected value is:$\frac{1+N}{2}\cdot 12=6\cdot (N+1)$ and $K_{2}$ is: $8\cdot N$. So we have domains: for $0<N<3$ method 2 is better. And for $N\geq 4$ method 1 is better. Commented Dec 25, 2019 at 20:15
• yep. :) when i said "method 1 is better" i had assumed the large $N$ case. you are right that for $N=1, 2$ method 2 is better. Commented Dec 25, 2019 at 20:41
• I understand, thank you for all your help and patience :) Commented Dec 25, 2019 at 20:53