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Expected number of coin tosses before landing heads

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Given a fair coin, how many times will you flip it (on average) before you get heads?

    2. Relevant equations
    E(x) = x1p1 + x2p2 + x3p3 + ... + xnpn

    3. The attempt at a solution
    I know this is easy, but I'm stumped with the probability part.

    Probability of heads is 1/2 and probability of tails is 1/2. So, after 1 flip, the probability of having landed heads right away is 1/2. So x1p1 = (1)(1/2). Otherwise, flip again: x2 = 2. My question: does p2 = (1/2)2? I got this by arguing that the probability of the sequence TH is P(T) times P(H). Similarly, for p3, the sequence is TTH, so does p3 = (1/2)3?

    I'm also guessing that arbitrarily long sequences TTT...TH (of length n) are possible. Assuming I'm right so far, would I take the limit of E(x) as n approaches infinity?

    In symbols,
    [tex] \lim_{n \to \infty} E(x) := E = \sum_{n=1}^\infty \frac{n}{2^n} = 2 [/tex]
    So is the expected value 2 tosses? Will I usually have gotten heads on my second toss?

    Thanks!
     
  2. jcsd
  3. Jun 19, 2011 #2

    lanedance

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    not quite if you average things you will get an expected value of 2 (mean)
    [tex] E(x) = \sum_{n=1}^{\infty} n p_n = \sum_{n=1}^{\infty} n \frac{1}{2^n} [/tex]

    "usually have gotten" implies most likely value (mode) which in this case will be 1. ie if you repeat this experiment over and aver you will get heads on the first toss more than any otehr number.
     
  4. Jun 20, 2011 #3
    Thanks lanedance! So what is the real-life explanation of "expected value" in this case?
     
  5. Jun 20, 2011 #4

    lanedance

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    The expected value is the average or mean.

    For example if someone said they would run the experiment 1,000,000 times, and each time pay you a dollar or each tosses, then you would make:
    $2*1,000,000 = $2,000,000
     
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