# Expected Value

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• Sep 20th 2008, 12:35 PM
timk19
Expected Value
QUESTION : With an infinite number of stores, half charging $1 and half charging$2. You decide to check three stores randomly and buy from a store which has the minimum of the three prices checked. What is the expected value of the price you will pay?
• Sep 20th 2008, 02:01 PM
mr fantastic
Quote:

Originally Posted by timk19
QUESTION : With an infinite number of stores, half charging $1 and half charging$2. You decide to check three stores randomly and buy from a store which has the minimum of the three prices checked. What is the expected value of the price you will pay?

$2,$2, $2 => you pay$2. Pr($2,$2, $2) = (1/2)^3 = 1/8. At least one$1 store => you pay $1. Pr(at least one$1 store) = 7/8.

Therefore ......
• Sep 20th 2008, 04:27 PM
timk19
RE:
Isn't the expected value the mean? Shouldn't the expected value be represented as a dollar amount?
• Sep 20th 2008, 04:34 PM
icemanfan
Quote:

Originally Posted by timk19
Isn't the expected value the mean? Shouldn't the expected value be represented as a dollar amount?

Expected Value could be considered a weighted average, but it's usually not the mean of a set of numbers. It is calculated as the sum of $\displaystyle n \cdot P(n)$ over all possible values of n. In the case of your problem, there is a 1/8 probability you will have all $2 stores and have to pay$2, and a 7/8 probability that at least one of the three stores will be a $1 store and you have to pay$1. No other values besides $1 and$2 are possible, so the Expected Value is (1/8)*2 + (7/8)*1. In this problem, it makes sense to give this number a dollar sign.
• Sep 20th 2008, 04:42 PM
mr fantastic
Quote:

Originally Posted by timk19
Isn't the expected value the mean? Shouldn't the expected value be represented as a dollar amount?

If you understand how to calculate an expected value I would have thought my post gave everything necessary for doing the calculation.
• Sep 21st 2008, 03:17 PM
timk19
re:
Perfect, thank you.