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Expected value problem

  1. Jan 31, 2007 #1
    1. The problem statement, all variables and given/known data
    A client is trying to connect to a web server on an overloaded machine that only accepts an average of one in 20 connections.

    How many times would the client expect to try before getting a connection? Explain your answer.

    3. The attempt at a solution

    This is part 5 of a problem and I don't know where to start. I know that to find expected value you take a sum of all events times the probability of each event. I don't see how you can use that here though. Please help.
     
  2. jcsd
  3. Jan 31, 2007 #2

    Dick

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    Let's get warmed up. What is the probability of connecting on the first call? What is the probability of failing the first call and connecting on the second? Probability of failing on k calls and then connecting?
     
  4. Jan 31, 2007 #3
    1/20 for first call
    (1/20)*(19/20)^(k-1) in general to get a connection on the kth try
     
  5. Jan 31, 2007 #4

    Dick

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    Great! So now with these probabilities how do you get an expectation value for k?
     
  6. Jan 31, 2007 #5
    I honestly don't know. I'm back to where I started. I don't know how to figure out how many times the user should expect to try before getting connected.
     
  7. Jan 31, 2007 #6

    D H

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    Sometimes it helps to look at the complementary problem. For example, the probability of not getting a connection on the first attempt is 19/20. What is the probability of not getting a connection on the first and second attempts? Now generalize: What is the probability of not getting a connection on the first through nth attempts? (Hint: These are easy probabilities to calculate.) Once you generate a general expression for this latter probability the answer to the original question is fairly easy to answer.
     
  8. Feb 1, 2007 #7

    Dick

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    For expectation values, you gave the answer in your problem post. Calling your computed probability p_k for any k, then the expectation value of k is the sum k*p_k over all k. This is a 'fairly' easy infinite series.
     
  9. Feb 1, 2007 #8
    OK I talked to a friend of mine who says the answer is simply 20. The question was how many times would the client expect to try before connecting and since the probability of connection is 1/20 then he would expect to get connected 1 time if he makes 20 attempts.

    Do others here think that infinite series is the way to go?
     
  10. Feb 1, 2007 #9

    Dick

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    And 20 IS the sum of the series. Whether you HAVE to do the series depends on what you already know. If you already have a formula for this case (which was derived using the series) then no, you don't need to resum the series again.
     
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