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

00100110

Bits are randomly chosen from the string. All choices are done equally likely.

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

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

Suppose you have already chosen $0$ or $1$. Probability of choosing opposite char, and then again opposite char, is given with: $p(0 \wedge 1) = p_0 p_1 = 0.234$.

Without the "you have already chosen $0$ or $1$ ...", the probability would be: $p(0\wedge 1) = 2p_0 p_1 = 0.468$

Correct?

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5  
I have no idea what's going on... –  David Mitra Apr 26 '12 at 21:09
    
Are you choosing one bit from the 8 given bits at random? –  David Mitra Apr 26 '12 at 21:10
    
I have updated the question. –  Mooncer Apr 27 '12 at 2:03

1 Answer 1

The problem is not described with complete clarity. So some assumptions are needed in order to produce an answer. We assume that you are picking one of the $8$ locations independently and at random $3$ times, with all choices equally likely. In particular, it is assumed that repetition of location is allowed.

The probability of getting the bit sequence $0$, $1$, $0$ is then $\frac{5}{8}\cdot\frac{3}{8}\cdot\frac{5}{8}$, and the probability of getting the bit sequence $1$, $0$, $1$ is $\frac{3}{8}\cdot\frac{5}{8}\cdot\frac{3}{8}$. Add. Our probability is $$ \frac{5}{8}\cdot\frac{3}{8}\cdot\frac{5}{8}+\frac{3}{8}\cdot\frac{5}{8}\cdot\frac{3}{8}.$$ This simplifies to $\dfrac{15}{64}$.

Remark: The number $\dfrac{15}{64}$ is precisely your number $p_0p_1$. There is good structural reason for that. However, I think that the detailed analysis above is more informative, since it generalizes readily to other situations.

If repetition of location is not allowed, the analysis is quite similar, but the numbers change. We get $$ \frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{3}{8}\cdot\frac{5}{7}\cdot\frac{2}{6}.$$

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I have updated the question. Can you elaborate on the structural reason? –  Mooncer Apr 27 '12 at 1:59
1  
Whatever the probabilities $p_0$ and $p_1$ are, they add up to $1$. We are adding $p_0^2p_1$ and $p_1^2p_0$. This is $p_0p_1(p_0+p_1)$, which is just $p_0p_1$. But if we were interested in the probability of ($0101$ or $1010$), we would not get the pleasant simplification. –  André Nicolas Apr 27 '12 at 5:14
    
Thanks that is interesting to me, because in general I am trying to exploit the opposite relation between $p_0$ and $p_1$. For example, suppose that there is a moveable head, like in Turing machine, pointing at a char. We are interested in "cycle" i.e. in moving to opposite char, and then to opposite-opposite char. The sum up to 1.0 that you described corresponds to fact, that head is always at some char when starting the cycle, either 0 or 1. –  Mooncer Apr 27 '12 at 6:40

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