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Say I have two fair coins $a$ and $b$, and I toss them both $N$ times independently, so for $N = 1$, my event space is $\left\{ \left(H,H\right),\left(H,T\right),\left(T,H\right),\left(T,T\right)\right\} $. So for $N$ tosses I have two events $A$ and $B$, they are:

$$A="x_{1}\dots x_{N}:\ x_{i}\in\left\{ H,T\right\}" $$

$$B="y_{1},\dots,y_{N}:\ y_{i}\in\left\{ H,T\right\} "$$

What is the probability that I see a head in $a$ before I see a head in $b$? I would like to say $\frac{1}{2}$ but it's clearly not that even for $N = 1$. For $N = 1$ it's $\frac{1}{4}$.

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    It is 1/2(1-P(they both first have heads on same toss)-P(neither has heads after N tosses)). – 2017-01-07

2 Answers 2

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One way to work this out is to "list" all the ways coin $A$ might come up heads before coin $B$ does. They are: \begin{align} &(T,H)\\ &(T,T), (T,H) \\ &(T,T), (T,T), (T,H) \\ &(T,T), (T,T), (T,T), (T,H) \\ &\qquad \vdots \\ &(T,T), (T,T), \ldots, (T,T), (T,H) \\ \end{align}

where the last line is a sequence of $N$ pairs. These are disjoint events, so one need merely take the sum of their probabilities.

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For $N=1$
P(Head in A in First Throw and Tail in B in First Throw)
$=\dfrac{1}{2^2}$

For $N=2$
P[(Head in A in First Throw and Tail in B in First Throw) OR
(Tail in A in First Throw and Head in A in Second Throw and Tail in A in First Throw and Tail in B in second row)
$\left[\dfrac{1}{2^2}+\dfrac{1}{2^4}\right]$

Similarly For $N=3$
$\left[\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}\right]$

So, in general, $\sum_{p=1}^{N}\dfrac{1}{2^{2p}}$