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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 sequences.

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.

Kodiologist
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beta
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    I see $12$ cases of HH and $12$ cases of HT in that list, as I would expect since $16 \times 3 \times \frac14=12$. So in that sense they are equally likely. But HH is more likely to happen multiple times in a single sequence of four (the extreme case being $3$ cases of HH in HHHH) so to balance this there must be fewer sequences of four which have at least $1$ case of HH. – Henry Sep 30 '17 at 16:49
  • I am talking about counting *only* if a sequences contains either HT or HH *at least once*. So for the HH-count the sequences HHHH counts 1 because HH occurs at least *once* in that sequence. – beta Sep 30 '17 at 17:05
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    I know. My comments explains why this difference happens: it balances the different multiple appearances. One shorthand way would be to say that HH can overlap itself as in HHH, while HT cannot – Henry Sep 30 '17 at 17:18
  • ah okay. sorry, I first did not read your comment properly. I now get what you mean. Can this be somehow converted to an answer? – beta Sep 30 '17 at 18:59
  • Related: https://stats.stackexchange.com/questions/12174/time-taken-to-hit-a-pattern-of-heads-and-tails-in-a-series-of-coin-tosses – whuber Oct 01 '17 at 17:50

1 Answers1

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Your list has a total of $3 \times 16=48$ consecutive pairs of flips and you would expect a quarter of these (i.e. $12$) to be HH and a quarter to be HT. That is indeed what you get if you count your list, so in that sense HH and HT are equally likely to appear in any pair of flips.

But HH is more likely to happen multiple times in a single sequence of four flips, because it can overlap itself. In the extreme example of HHHH you get three pairs of HH. So overall you get

  • HH appearing in $8$ sequences of four flips ($1$ case three times, $2$ cases two times and $5$ cases once, with $1\times 3+2\times 2 + 5 \times 1= 12$)

  • HT appearing in $11$ sequences of four flips ($1$ case two times and $10$ cases once, with $1\times 2 + 10 \times 1= 12$)

So in a sense the fewer sequences of four slips that HH appears in at least once simply balances its greater possibility of appearing multiple times.

This self-overlap effect has other consequences. If you start flipping and stop when you first see HH, the expected number of flips is $6$; similarly if you stop when first seeing TT. But if you start flipping and stop when you first see HT, the expected number of flips is $4$; similarly if you stop when first seeing TH.

It can get even more complicated in a race:

  • If you start flipping and stop when you first see HH or HT, then each is equally likely to be what made you stop.
  • Similarly if you start flipping and stop when you first see HT or TH, then each is equally likely to be what made you stop.
  • But if you start flipping and stop when you first see HH or TH, then TH is three times more likely as HH to be what made you stop. So races are not transitive
Henry
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