# Two opposite events fill whole probability event space, a process selects them [duplicate]

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Probability of two opposite events

Suppose there is string of eight bits, e.g.:

00100110

Bits are randomly chosen from the string. Location of bit (in the string) does not influence its selection probability.

Probability of choosing $0$: $p_0 = \frac{5}{8} = 0.625$

Prob. of choosing $1$: $p_1 = \frac{3}{8} = 0.375$

Suppose there is an ongoing process of selecting $0$ and 1. So at each moment, $0$ or 1 is selected, and represents current state C of the process.

Probability of choosing opposite state (to current state C), and then again opposite state – such complex event is called cycle – is given with:

$$p_\text{cycle} = p_0 \cdot p_1 \space\space\space (1)$$

Question: define $p_a = p_{cycle}$. Then, opposite event is $p_b = 1 - p_{cycle}$. We have again two opposite events. How will the $p_b$ look like? I.e. what sequences of $0$ and $1$ will belong to events of kind A and events of kind B. I have problem defining B-set.

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## marked as duplicate by Dilip Sarwate, Gerry Myerson, William, Thomas, Noah SnyderOct 4 '12 at 22:53

Looks okay to me, assuming that the choices are independent. –  Brian M. Scott Apr 30 '12 at 2:10
If you're not happy with the answers you got to the earlier version of this question, don't go posting a new one - just edit the old one. –  Gerry Myerson Apr 30 '12 at 2:41
@BrianM.Scott, Gerry: I have modified the question. It is now clearly different. –  Mooncer Apr 30 '12 at 4:35

This is an answer to the updated question. Let $C$ be the current state, either $0$ or $1$. Let $\overline C$ be the opposite state, either $1$ or $0$, respectively. Let $E$ be the event not-cycle. Then $E$ occurs when the next choice is $C$, or when the next choice is $\overline C$ and the choice after that is also $\overline C$.

If $C=0$, $E$ occurs if the next choice is $0$, or if the next two choices are both $1$; the probability of this is $p_0+p_1^2$.

If $C=1$, $E$ occurs if the next choice is $1$, or if the next two choices are both $0$; the probability of this is $p_1+p_0^2$.

Now the probability of being in state $0$ at any given time is $p_0$, and the probability of being in state $1$ is $p_1$. Thus, the probability of being in state $0$ and having $E$ occur is $p_0(p_0+p_1^2)$, and the probability of being in state $1$ and having $E$ occur is $p_1(p_0+p_1^2)$. Combining the two, we find that the probability of $E$ is \begin{align*}p_0(p_0+p_1^2)+p_1(p_1+p_0^2)&=p_0^2+p_0p_1^2+p_1^2+p_1p_0^2\\ &=p_0^2+p_0p_1(p_0+p_1)+p_1^2\\ &=p_0^2+p_0p_1+p_1^2\;, \end{align*}

since $p_0+p_1=1$. And this agrees with your calculation of $p_0p_1$ as the probability of a cycle, since $$(p_0^2+p_0p_1+p_1^2)+p_0p_1=p_0^2+2p_0p_1+p_1^2=(p_0+p_1)^2=1\;:$$ the probabilities of cycle and not-cycle must add up to $1$.

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I will use regex-like symbol "+" to denote "one or more". I think that $p_1p_0$ means (for $C=0$) "wait for 1 (i.e. $\overline C$), then wait for 0", and matches sequence of randomly choosen bits $0^+1^+0$. For $\overline C$ it is $1^+0^+1$. Basing on your answer, the not-cycle is: $0^+$, $0^+1^+$, $1^+$, $1^+0^+$? –  Mooncer Apr 30 '12 at 16:11
@Steffen: No, $p_1p_0$ is the probability that the next two choices are $1$ and $0$ in that order; no waiting is involved. –  Brian M. Scott Apr 30 '12 at 16:13
If a process would select $0$ and 1 with $p_0, p_1$, and we would observe its output string, then the event: "1 observed" would mean that we wait – possibly observing $0$? And 1 would be observed ("inside" the zeroes) with prob. \$p_1? –  Mooncer Apr 30 '12 at 17:06
@Steffen: All of the calculations above, both yours and mine, refer to observing the very next output or the next two outputs; no waiting is involved. –  Brian M. Scott Apr 30 '12 at 17:09