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 $|i-j|>1$?
Tossing 10 coins, what's the easiest way to see that these two events are independent?
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probability
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0@copper.hat: $I=E_i$, $J=E_j$ – 2012-10-23
1 Answers
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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.