Expected flips for biased coin to match or differ from initial flip

[ephedyn]

Homework Statement
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 $p$.

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

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

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

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

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!

 

[Ray Vickson]

Homework Statement
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 $p$.

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

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

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

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

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!

L'Hospital's rule does not apply, because you do not have something like 0/0 or ∞/∞.

 

[ephedyn]

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 $P[H] \times E[N_H] + P[T] \times E[N_T]$ because a property that allows me to do this seems to be missing from my memory.

 

[haruspex]

Fwiw, your answer can be written as p/(1-p) + (1-p)/p.

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

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.