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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?

Ice Tea
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Kyle
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  • Well, since there must be at least one bid before you accept any, so the expectation can't be less than one. – saulspatz Feb 25 '18 at 05:12
  • @saulspatz You're right, I know that. But I thought the problem is asking the expectation value, isn't it? – Kyle Feb 25 '18 at 05:16
  • It's asking for the expected number of bids, not the average bid. – saulspatz Feb 25 '18 at 05:17
  • Are the $\Delta_k$ supposed to be the bids, or the differences between successive bids? The way the question is written, it seems to be the former, but then the bids would not necessarily increase, as in most auctions, and it's hard to envision the process. – saulspatz Feb 25 '18 at 05:20
  • @saulspatz It's the bids, not the differences between them. I just use a random notation. – Kyle Feb 25 '18 at 05:22

1 Answers1

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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."

saulspatz
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