What is the probability of getting tail on every even toss when we keep tossing a coin?

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?

• 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. – JMoravitz Sep 26 '18 at 18:35
• It is the same as the probability of getting $n/2$ tails in $n/2$ tosses, assuming the remaining $n/2$ tosses are unconstrained. – Bungo Sep 26 '18 at 18:36
• can i just ignore the odd tosses when I calculate this question? – Jason Sep 26 '18 at 18:50

No, not the same. $$n/2$$ tails in $$n$$ tosses can be done in multiple ways, e.g. $$TTTHHH$$ in $$6$$ tries.
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 $$2^{-(n+1)/2}$$.