Flipping Fair Faces

Let us denote the event of obtaining four consecutive heads before six consecutive tails by $E$ . Now, the following equations must be satisfied $\mathbb{P}(E| \textrm{ 1st toss is H })= \frac{1}{2^3}+ (1-\frac{1}{2^3})\mathbb{P}(E| \textrm{ 1st toss is T})\\ \mathbb{P}(E| \textrm{ 1st toss is T })= (1-\frac{1}{2^5})\mathbb{P}(E| \textrm{ 1st toss is H })$ The first equation may be derived by considering the fact that, when the result of the first toss is $H$ , the event $E$ occurs when the next $3$ consecutive tosses are $H$ (with probability $\frac{1}{2^3}$ ), or there is a $T$ before three consecutive $H$ occurs (with probability $(1-\frac{1}{2^3})\mathbb{P}(E| \textrm{ 1st toss is T})$ ). The second equation may also be argued similarly.

Solving the above equations, we have $\mathbb{P}(E| \textrm{ 1st toss is H})= \frac{32}{39}, \mathbb{P}(E| \textrm{ 1st toss is T}) =\frac{31}{39}.$ Hence, $\mathbb{P}(E)= \frac{1}{2}\mathbb{P}(E| \textrm{ 1st toss is H})+ \frac{1}{2}\mathbb{P}(E| \textrm{ 1st toss is T})=\frac{21}{26}. \blacksquare$

Let $p_j$ , for $j=1,2,3,-1,-2,-3,-4,-5$ be the probability that $HHHH$ occurs before $TTTTTTT$ , given that your current streak is of $j$ Heads. Note that a negative streak of Heads represents a streak of Tails, so that $p_{-3}$ is the probability that $HHHH$ occurs before $TTTTTT$ given that your current streak is $TTT$ .

Conditional probability considerations tell us that $\begin{aligned} p_3 & = \tfrac12 + \tfrac12p_{-1} \\ p_2 & = \tfrac12p_3 + \tfrac12p_{-1} \\ p_1 & = \tfrac12p_2 + \tfrac12p_{-1} \\ p_{-1} & = \tfrac12p_{-2} + \tfrac12p_1 \\ p_{-2} & = \tfrac12p_{-3} + \tfrac12p_1 \\ p_{-3} & = \tfrac12p_{-4} + \tfrac12p_1 \\ p_{-4} & = \tfrac12p_{-5} + \tfrac12p_1 \\ p_{-5} & = \tfrac12p_1 \end{aligned}$ so that $p_1 \; = \; \tfrac18 + \tfrac78p_{-1} \hspace{1cm} p_2 \; = \; \tfrac14 + \tfrac34p_{-1} \hspace{1cm} p_3 \; = \; \tfrac12 + \tfrac12p_{-1}$ and $p_{-1} \; = \; \tfrac{31}{32}p_1 \hspace{1cm} p_{-2} \; = \; \tfrac{15}{16}p_1 \hspace{1cm} p_{-3} \; = \; \tfrac78p_1 \hspace{1cm} p_{-4} \; =\; \tfrac34p_1 \hspace{1cm} p_{-5} \; =\; \tfrac12p_1$ Solving the simultaneous equations for $p_1$ and $p_{-1}$ given here, we deduce that $p_1 \; =\; \tfrac{32}{39} \hspace{2cm} p_{_1} \; = \; \tfrac{31}{39}$ and so the overall probability of $HHHH$ occurring before $TTTTTT$ is $\tfrac12(p_1 + p_{-1}) \; = \; \boxed{\tfrac{21}{26}}$