Occurrences of permutations in sequences of 4 coin flips

Question

I was reading this interesting article on hot hands and streaks in sports. The article revolves around the 16 possible sequences of 4 coin flips (H = heads, T = tails):

HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT

The article states the following:

[I]n the sixteen length-four sequences, there are only eight that have any occurrence of HH, but there are eleven that have an occurrence of HT. That is, the distribution of HH and HT is not uniform in the fourteen (sic!) sequences.

NB: The quote mentions fourteen sequences, because in the linked article, only 14 of the 16 sequences are considered.

And then there is a footnote:

At first, this may seem paradoxical since the two counts might be expected to be equal by “symmetry”. But, the two occurrences are not symmetric, which I leave you to ponder.

Since then I try to think of an explanation on why this is the case? Why are there more sequences in which at least one HT occurs than there are sequences where at least one HH occurs?

Of course, by simply looking at each sequence, I can see (and count) the HH and HT occurrences, but I would like to know why exactly (which property of the sequences) leads to this fact.

Answer 1

I believe this is a mistake in comprehension. If there are $x$ $\text{HH}$ sequences, then there must be $x$ $\text{TT}$ sequences. $\text{HT}$ sequences are not related.

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The number of $HT$ sequences must be same as the number of $TH$ sequences.

Answer 2

We have a pair (HH) 3 in HHHH, 2 in HHHT, 1 in HHTH, 1 in HHTT 1 in HTHH, 2 in THHH, 1 in THHT, 1 in TTHH$. In total $12$. We have a pair (HT): 1 in HHHT, 1 in HHTH, 1 in HHTT, 1 in HTHH, 2 in HTHT, 1 in HTTH, 1 in HTTT, 1 in THHT, 1 in THTH, 1 in THTT, 1 in TTHT. In total 12. You can count (TH), (TT) are also the same in total 12.

Since (HH) can appear in a way: (1st H 2nd H),(2nd H, 3rd H), (3rd H, 4th H) in a 4 coin tosses as a sequence, you can use maximum 3 (HH) pairs in a sequence. And you count it as a single time. Whereas, (HT) can appear in a way (1st H 2nd T),(3rd H 4th T) in a sequence. So, you can use maximum 2 pairs in a sequence. And you count it as a single time. Therefore you have more(HT) than (HH).

Answer 3

It is important to keep in mind that we are counting sequences in which the subsequence HH occurs at least once, not the total number of appearances of HH. For instance, HH appears three times in the sequence HHHH, but we only count that sequence once. The subsequence HT will occur in any sequence with both heads and tails provided that not all the tails appear before the first head.

Consider cases.

Four heads: There is $\binom{4}{4} = 1$ such sequence.

The subsequence HH must occur.

The sequence HT cannot occur since there are no tails.

Three heads and a tail: There are $\binom{4}{3} = 4$ such sequences.

The subsequence HH must occur in all of these sequences.

The subsequence HT must occur unless T occurs in the first position, so three of these four sequences contain the subsequence HT.

Two heads and two tails: There are $\binom{4}{2} = 6$ such sequences.

For the subsequence HH to occur, it must begin in the first, second, or third position. Hence, three of these sequences contain the subsequence HH.

The subsequence HT will occur unless both heads occur after both tails. Hence, five of these sequences contain the subsequence HT.

One head and three tails: There are $\binom{4}{1} = 4$ such sequences.

The subsequence HH cannot occur since there are not enough heads.

The subsequence HT will occur unless heads is in the final position. Hence, the subsequence HT will occur in three of these four sequences.

Four tails: There is $\binom{4}{0} = 1$ such sequence.

Since there are no heads, neither the subsequence HH nor the subsequence HT can occur.