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For Bernoulli process, what is the probability of getting tail on every even toss when we keep tossing a coin? It is not same as getting n/2 tails in n tosses, right?

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  • $\begingroup$ Correct, it is not the same as getting $n/2$ tails in $n$ tosses. Let $n=5$. Possible acceptable sequences of coinflips would be $H\color{red}{T}H\color{red}{T}H, T\color{red}{T}T\color{red}{T}T, H\color{red}{T}T\color{red}{T}H,\dots$ and so on. We are merely wanting every even position (colored in red in the examples) to all contain tails. $\endgroup$ – JMoravitz Sep 26 '18 at 18:35
  • $\begingroup$ It is the same as the probability of getting $n/2$ tails in $n/2$ tosses, assuming the remaining $n/2$ tosses are unconstrained. $\endgroup$ – Bungo Sep 26 '18 at 18:36
  • $\begingroup$ can i just ignore the odd tosses when I calculate this question? $\endgroup$ – Jason Sep 26 '18 at 18:50
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No, not the same. $n/2$ tails in $n$ tosses can be done in multiple ways, e.g. $TTTHHH$ in $6$ tries.

Compute the number of choices for the coin to fall on even tosses, and then on odd tosses, and multiply them.

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Let $n>1$ be even. The number of total possibilities is $2^n$.

Number of desired ways (x T x T x T ...) is $2\times1\times2\times1\times2\times1 \cdots=2^{n/2}$, so the probability of desired outcome is $2^{-n/2}$.

For odd $n$, there are $2^{(n-1)/2}$ desired ways and the probability becomes $2^{-(n+1)/2}$.

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