A fair coin is tossed 6 times. All 64 sequences of heads and tails are equally likely…

Step 1. Count. A block starting at toss $1$: HHH and any $3$ tosses after it, $8$ sequences. A first block starting at toss $2$: T at toss $1$, then HHH, then any $2$ tosses, $4$ sequences. A first block starting at toss $3$: toss $2$ is T, toss $1$ is free, then HHH, then any last toss, $4$ sequences. A first block starting at toss $4$: toss $3$ is T, tosses $1$ and $2$ are free, then HHH, $4$ sequences. The total is $8 + 4 + 4 + 4 = 20$.

Step 2. The probability. $\dfrac{20}{64} = 0.3125$.

Step 3. The expected number. $\left(\dfrac{1}{2}\right)^3 + 3 \times \dfrac{1}{2} \times \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8} + \dfrac{3}{16} = \dfrac{5}{16} = 0.3125$.

Why they are equal. Two separate blocks of $3$ heads need at least $7$ tosses: HHH, a T, and HHH. So in $6$ tosses a sequence has either no block or exactly one. The average number of blocks is then the share of sequences with a block.