I am not sure how to work this out but I split the outcomes of 3 coin flips into the 8 possible outcomes: HHH, HHT, HTH, HTT | TTT, TTH, THT, THH
And have gotten a probability mass function for the random variable out of the 3 flips(only 1 head is 1/4, only 2 heads is 3/8, only 3 heads is 1/8) -
I have calculated the entropy as $H(x) = -sum(P(xi)*logP(xi)) =-1/4*log(1/4)-3/8*log(3/8)-1/8*log(1/8)$ - is that method correct?
Your method is right, except that you did not compute the correct mass function reflecting the random variable of interest. For more on entropy, this is a lecture were i explain the intuition behind it.
Let $X$ be the random variable denoting the number of heads in your experiment \begin{align} p_0 \triangleq \Pr(X = 0) &= \Pr(TTT) = \frac{1}{8}\\ p_1 \triangleq\Pr(X = 1) &= \Pr(HTT) +\Pr(THT)+\Pr(TTH) = \frac{3}{8}\\ p_2 \triangleq\Pr(X = 2) &= \Pr(HHT) +\Pr(HTH)+\Pr(THH) = \frac{3}{8}\\ p_3 \triangleq\Pr(X = 3) &= \Pr(HHH) = \frac{1}{8} \end{align} This is a mass function because $\sum p_i = 1$, in contrast to yours which is not. Now, we can easily compute the entropy $$H(X) = - \sum p_i \log p_i$$