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A coin is tossed repeatedly forever.

Let $A$ be the event that $r$ heads in a run occurs earlier than $s$ tails in a run;

Let $B$ be the event that the first toss is a head;

Let $C$ be the event that the first $s$ tosses are all tails.

My textbook said that $P(A|B)=P(A|B^c\cap C^c)$. And I have no clue why it's true.

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I misread the question, earlier comment withdrawn. –  André Nicolas Sep 29 '12 at 5:02

1 Answer 1

up vote 2 down vote accepted

Let $X=\{H,T\}^{\Bbb N}$ be the set of all infinite sequences of coin tosses. For $n\in\Bbb Z^+$ let $$E_n=\{x\in X:x_n=H\text{ and }x_k=T\text{ for }k=1,\dots,n-1\}\;;$$ $B=E_1$, and $B^c\cap C^c=\bigcup_{n=2}^sE_n$. Fix $n\in\{2,\dots,s\}$, and let

$$\varphi_n:E_n\to E_1:x\mapsto\langle x_{k+n-1}:k\in\Bbb Z^+\rangle\;,$$

in other words, $\varphi_n$ simply shifts $x$ to the left $n-1$ places, stripping off the initial string of $n-1$ tails. Clearly $\varphi_n$ is a bijection, and $\varphi_n[A\cap E_n]=A\cap E_1=A\cap B$. Moreover, if $p$ is the probability of tossing a tail, then $P(E_n)=p^{n-1}P(E_1)$ and $P(A\cap E_n)=p^{n-1}P(A\cap E_1)$, so

$$P(A|E_n)=\frac{P(A\cap E_n)}{P(E_n)}=\frac{P(A\cap E_1)}{P(E_1)}=P(A|E_1)=P(A|B)\;.$$

Finally,

$$\begin{align*} P(A|B^c\cap C^c)&=P\left(A\bigg|\bigcup_{n-2}^sE_n\right)\\\\ &=\frac{P\left(A\cap\bigcup_{n=2}^sE_n\right)}{P\left(\bigcup_{n=2}^sE_n\right)}\\\\ &=\frac{P\left(\bigcup_{n=2}^s(A\cap E_n)\right)}{P\left(\bigcup_{n=2}^sE_n\right)}\\\\ &=\frac{\sum_{n=2}^sp^{n-1}P(A\cap E_1)}{\sum_{n=2}^sp^{n-1}P(E_1)}\\\\ &=\frac{P(A\cap E_1)}{P(E_1)}\\\\ &=P(A|B)\;. \end{align*}$$

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