During certain sequences of coin flips, why does HHHT have a greater chance than HHHH?

Question

I was reading an article, They were comparing sequences of coin flips. They said:

"HHHT and HHHH are equal only if flipping an unbiased coin exactly four times or infinitely many times. For values in between these two extremes, probabilities will not be the same. Imagine flipping a coin, say, 20 times and checking whether either HHHH or HHHT arise at least once in that series. Given that the wait time for HHHH is longer than that for HHHT, HHHH will also be less likely to occur at all."

Can someone help me understand why the wait time for HHHH is longer than HHHT for values in between 4 and infinite flips? Naturally, I expected that either of these sequences occurring has the same probability. Since Heads and Tails, each have a 50/50 chance with an unbiased coin. What am I missing in my understanding? 

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5215234/#!po=6.13208

Answer 1

Consider the simpler problem of HH vs HT in 3 coin tosses. \begin{array} 3Flips & HH & HT \\ \hline HHH & 1 & 0 \\ \hline HHT & 1 & 1 \\ \hline HTH & 0 & 1 \\ \hline HTT & 0 & 1 \\ \hline THH & 1 & 0 \\ \hline THT & 0 & 1 \\ \hline TTH & 0 & 0 \\ \hline TTT & 0 & 0 \\ \hline Total & 3 & 4 \\ \hline \end{array}

Why does this happen? One explanation is that in HHH we have two counts of HH, but we only count it once. This is slightly similar to the explanation given by Peter that after an HHH, you have to start counting from scratch for HHHH, in case a T occurs, but for HHHT, even if another H occurs, you can reuse 3 H's.

You can now extend the explanation to larger flip-sets (3 throws vs n throws) and larger patterns (HH vs HHHH)

Answer 2

The casual, yet intuitive answer is that whenver $\textsf{HHHH}$ occurs, you are basically guaranteed to get $\textsf{HHHT}$, as the sequence of $\textsf{H}$ must stop with a $\textsf{T}$ at some point. This is not the case vice-versa.

(Sure, there is the limit case where you only get $\textsf{H}$ from one point onwards, but this case is highly unlikely.)

Answer 3

The difference is the following : If $HHH$ has occured, in the case we want $HHHT$ , we are done as soon as $T$ occurs ($HHHT$ has then occured as well). But to get $HHHH$ , we do not just need another $H$, we have to build up $HHHH$ again from beginning unless the next result is already $H$. This is the reason why $HHHH$ requires more flips in average and hence the average waiting time is longer.