The problem is as follows. Suppose that a baseball trading card for sale receives successive bids $$\Delta_1,\,\Delta_2,\cdots,$$ which are independent random variables with geometric distribution $$\mathbb P(\Delta=k)=0.01\cdot(0.99)^k$$ for $k=0,\,1,\cdots$. If you decide to accept any bid over \$100, how many bids on average will you receive before an acceptable bid appears.
I am not quite sure if I understand the problem or not, but that's what I think:
I should calculate the expectation value of $0$ to $100$, $$0.01\times\left[0.99+2(0.99)^2+3(0.99)^3+....+100(0.99)^{100}\right]= 0.6276.$$ But the answer is $2.73$, which is totally different. Can anyone help?
We have that the probability that a bid is $k$ is $\frac{.01}{.99^k},$ so the probability that a bid is over $100$ is $$p=.01\sum_{k=101}^{\infty}{.99^k}=.01\frac{.99^{101}}{1-.99}=.99^{101}$$ So now we have a succession of Bernoulli trials: we walk through the trading fair, getting independent bids on the card. The bids are acceptable with probability $p$. The expected waiting time until the first success in a sequence of such trials is $1/p,$ so the answer I get is $.99^{-101} \approx 2.7595949761909377$
EDIT $.99^{-100} \approx 2.7319990264290284,$ so perhaps the statement of the problem should be that you will accept any bid of $100$ or more, not "over $100."