If I flip four coins, what is the probability that the first coin is heads AND there are more heads than tails? I don't know why this answer is eluding me, but I had to figure it out somehow. I know that the probability of flipping heads first is 1/2. 8/16 outcomes have H first. The probability of having more heads than tails is 5/16: HHHH, HHHT, HHTH, HTHH, THHH. I realize that I can count it and that four of these results have H first, meaning that the answer is 4/16 or 1/4. However, how is this calculated mathematically? You can not simply multiply the two probabilities as they are inclusive. I know that they are independent as well. The first coin flip doesn't matter to having more heads than tails as it is still possible regardless. What probability formula would I need to solve this?
 A: There is nothing "un-mathematical" about your approach; it is perfectly fine.
Another approach is to realize that after the first flip is heads, you need at least 2 heads in the last 3 flips, which has probability $4/8=1/2$. So,
\begin{align}
&P(\text{first flip is heads, more heads than tails in 4 flips})
\\
&= P(\text{first flip is heads}) P(\text{more heads than tails in 4 flips} \mid \text{first flip is heads})
\\
&= P(\text{first flip is heads}) P(\text{at least two heads in the last 3 flips})
\\
&= \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
\end{align}
A: You are correct, you cannot multiply the probabilities. However this is because the events are not independent.
The outcomes for obtaining heads first and more heads are: $\sf\{HHHH, HHHT,HHTH,HTHH\}$, which are four from the sixteen total outcomes. (Thus the answer is $1/4$).
The outcomes for obtaining tails first and more heads are: $\sf\{THHH\}$, which is one from the sixteen total outcomes.
Thus there is a dependency between the event of "the first is a head" and "there are more heads than tails."

You seek the probability that you have obtained a head first, and two or three heads among the remaining three. That is expressing the favored event as an intersection of independent events, so...$$\dfrac 12\cdot\left(\binom 32+1\right)\dfrac 1{2^3}=\dfrac 14$$
