Probability of Flipping A Coin Ten Times and Tails Occur Only in Pairs

Question

Suppose a coin is flipped 10 times and the outcomes are recorded. Find the probability that any tails occur only in consecutive pairs. For example, with 4 flips, TTHH and HHHH are both valid, but HTHH and TTTH are not valid. I know that I'm supposed to use inclusion-exclusion principle, but I'm not sure how to go about it.

Answer 1

Why not just add up sequences of "doubles" and "singles",
eg (in order doubles-singles), $5-0,4-2,\,...\,0-10$, thus

$$\dfrac{\frac{5!}{5!0!} + \frac{6!}{4!2!} + \frac{7!}{3!4!} + \frac{8!}{2!6!}+\frac{9!}{1!8!} + \frac{10!}{0!10!}}{2^{10}} = \frac{89}{1024}$$

Answer 2

Markov Chain Solution:

• Define $A$ the event: "The current sequence has an even (including 0) number of T".
• $B$ is the event: "The current sequence has an ood number of T"

We have:

$$ P_k(A) = \frac{1}{2}(P_{k-1}(B)+P_{k-1}(A)) \\ P_{k}(B) = \frac{1}{2} P_{k-1}(A) $$ Stacking the probabilities in a vector $A_k=(P_k(A),P_k(B))$, we have:

$$ A_k= \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} A_{k-1} $$ We have $A_1=(.5,0.5)$ and using the recursion: $$ A_k= \frac{1}{2^9} \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}^9 A_{0} $$ We find $P_{10}(A) = \frac{89}{2^{10}}$