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We toss 10 identical biased coins. Let $I$ and $J$ be the event that the $i$-th and $j$-th toss ($i,j\in\mathbb{N}_{[2,10]}$) are different from their predecessors, respectively. What's the easiest way to see that $I$ and $J$ are independent whenever $|i-j|>1$?

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Well, one way would be to verify that $P(I \cap J) = P(I)P(J)$ by direct computation. – Alan Guo Oct 23 '12 at 4:00
@AlanGuo: You are right. But I'm expecting an explanation on how to understand it more intuitively or why exactly this rule fails when $|i-j|=1$ – Voldemort Oct 23 '12 at 4:03
Is $i \in \{0, 1\}$? Can you be more specific with your notation? – Alex Oct 23 '12 at 4:09
@AlexOlssen: $i,j\in\{2,3,...,10\}$ – Voldemort Oct 23 '12 at 4:14
@AlexOlssen: $I$ and $J$ are events that those tosses are "different from their predecessors," instead of whatever their outcomes. – Voldemort Oct 23 '12 at 4:20
up vote 2 down vote accepted

When $|i - j| = 1$ then one of the tosses is the predecessor of the second toss. Since the coins are biased ($p \in (0,1)$, and $p \neq \frac{1}{2})$, knowing that a coin agrees with its predecessor changes our probability of it being heads or tails.

That is $P(Heads |\text{ agrees with predecessor}) = \frac{P(Heads \text{ and agrees with predecessor})}{P(\text{agrees with predecessor})} = \frac{p^2}{p^2 + (1-p)^2} = \frac{p^2}{1 - 2p(1-p)} \neq p$ if $p \neq 0,\frac{1}{2},1$.

So, intuitively, if $p > \frac{1}{2}$, then if $I$ is true, then $i$ is most likely heads, and $i-1$ is most likely heads, and that means that $i-1$ is also more likely to agree with its predecessor (since $i-2$ is more likely heads), etc.

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