Probability of unfair coin not giving 2 heads Problem: An unfair coin with probability $p$ to flip Tails is flipped 4 times independently. What is the probability of no Heads twice in a row?
My answer: I calculated the probability that of the event
$$A = \text{Heads twice in a row}$$
by splitting it into the three events
$$A_i = \text{i Heads, Heads twice in a row}$$
, took the complement and got:
$\mathbb{P}(A^c) = 1- 3p^2(1-p)^2 - 4p(1-p)^3 - (1-p)^4$
Is there a quicker way to derive the answer? 
 A: Flipping 4x with no heads twice in a row means either you had 2 heads separated by at least one tail, or 1 head, or no heads. (This is because if you had 3 or 4 heads with 4 coins, you must have had at least 2 in a row.)
So break this down:


*

*2 heads (but no consecutive heads). There are 3 configurations (HTHT, HTTH, THTH). Each has probability $p^2(1-p)^2$.

*1 head: There are four configurations, each has probability $p^3(1-p)$.

*No heads: There is one configuration, with probability $p^4$.


Add these up and you get
$$ 3p^2(1-p)^2 + 4p^3(1-p) + p^4 $$
Edit- Your approach is also correct, though you could stand to show your work :) You can verify this by expanding these polynomials in $p$ and showing that they are equal.
A: Your solution appears to be correct, but ill suggest my own interpretation and you'll be the judge.
The probability to have $HHTT$ is obviously $(1-p)(1-p)(p)(p)$ and there is 3 ways to permute $HHTT$ (those being $HHTT$, $THHT$, and $TTHH$) so the probability of $P(two-heads- in-a-row)$ is $3*(1-p)^2p^2$ the compliment of this being $1-3(1-p)^2p^2$
