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Coin flipping question

  1. Feb 18, 2013 #1
    The problem statement, all variables and given/known data
    A biased coin lands heads with probability p and tails with probability 1-p.

    (1) What is the expected number of flips, after the initial flip, to land a match to the initial flip?
    (2) What is the expected number of flips, after the initial flip, to land a different side from the initial flip? Comment on the extreme values of [itex]p[/itex].

    The attempt at a solution
    Without loss of generality, assume we land heads on the initial flip. Let [itex]N_H[/itex] be the number of flips required until we land heads again. Since [itex]N_H[/itex] is a geometric random variable, with pmf [itex]f(x) = p(1-p)^{x-1}[/itex], then [tex]E[N_H]=\sum^{\infty}_{x=1} x \cdot f(x)= \sum^{\infty}_{x=1} px(1-p)^{x-1}=\dfrac{p}{1-(1-p)^2}=\dfrac{1}{p}[/tex] and similarly we have [itex]\dfrac{1}{1-p}[/itex] for tails. Let [itex]H[/itex] and [itex]T[/itex] denote the event of landing heads and tails on the initial flip respectively.

    So (1) for matching flips, the expected number of flips is [tex]P(H) \times \dfrac{1}{p} + P(T) \times \dfrac{1}{1-p} = p \times \dfrac{1}{p} + (1-p) \times \dfrac{1}{1-p}=2[/tex].

    Similarly, (2) for different flips, the expected number of flips is [tex]P(T) \times \dfrac{1}{p} + P(H) \times \dfrac{1}{1-p} = \dfrac{1-p}{p} + \dfrac{p}{1-p}=\dfrac{p^2+(1-p)^2}{p(1-p)}=\dfrac{2p^2-2p+1}{p(1-p)}[/tex]

    For the extreme cases, by L'Hopital's rule, we have [tex]\lim_{p\rightarrow0} \dfrac{2p^2-2p+1}{p(1-p)}=\lim_{p\rightarrow0} \dfrac{4p-2}{1-2p}=-2[/tex] and similarly, [tex]\lim_{p\rightarrow1} \dfrac{4p-2}{1-2p}=-2[/tex]

    So I realize I must be doing something wrongly because I'm getting negative expectation values in the final part. Any guidance on my working?

    Thanks!
     
  2. jcsd
  3. Feb 19, 2013 #2

    Ray Vickson

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    L'Hospital's rule does not apply, because you do not have something like 0/0 or ∞/∞.
     
  4. Feb 19, 2013 #3
    Ah! You're right! So I have the expected number grow asymptotically in both cases. Makes sense intuitively, since it should become nearly impossible to land the other side on a flip. Thanks!

    Does the rest of my approach make sense? I'm not too convinced about taking [itex]P[H] \times E[N_H] + P[T] \times E[N_T][/itex] because a property that allows me to do this seems to be missing from my memory.
     
  5. Feb 19, 2013 #4

    haruspex

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    Fwiw, your answer can be written as p/(1-p) + (1-p)/p.
    Yes, that's fine. You can justify it by considering the prob that it takes N tosses given the outcome of the initial toss, P[N|H], P[N|T]. P[N] = P[N|H]P[H]+ P[N|T]P[T]. E[N|H] = ƩNP[N|H], etc.
     
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