Average length of streaks in a coinflip sequence

[Larimani]

what is the average length of streaks in an independent identically-distributed binary infinite sequence?
ie. if you were to flip a coin to infinity what would be the average length of 'streaks' (similar continuous outcomes)

[KenProspero]

I may be missing something obvious in your question -- however, with an honest coin, wouldn't it be:

Streak of 1 -- 50% (1/2 or 1/(2^1))
Streak of 2 -- 25% (1/4 or 1/(2^2))
Streak of 3 -- 12.5% (1/8 or 1/(2^3))
Streak of 4 -- 6.25% (1/16 or 1/(2^4))
etc. etc. etc.

How are you calculating that? For example, do you consider TT to be 1 lentgh 2, and 2 length 1 streak, or just a length 2 streak?

In the later case, it's going to be:
\sum_{i=1}^{\infty}\frac{i}{2^i}
which is absolutely convergent, so we can rearange terms
\sum_{i=1}^{\infty} \sum{j=i}{\infty} \frac{1}{2^j}
which is equal to
\sum_{i=1}^{\infty} \frac{1}{2^{i-1}}
which is equal to
2

[Larimani]

[ QUOTE ]
How are you calculating that? For example, do you consider TT to be 1 lentgh 2, and 2 length 1 streak, or just a length 2 streak?

In the later case, it's going to be:
\sum_{i=1}^{\infty}\frac{i}{2^i}
which is absolutely convergent, so we can rearange terms
\sum_{i=1}^{\infty} \sum{j=i}{\infty} \frac{1}{2^j}
which is equal to
\sum_{i=1}^{\infty} \frac{1}{2^{i-1}}
which is equal to
2

[/ QUOTE ]


TTHTT. is length 2. length 1. length 2. so the average would be 5/3.

but I think you answered it right. What is the SD of the distribution of streaks?

[AaronBrown]

Using rufus' interpretation, the expected squared streak length is 6, so the standard deviation of streak length is SQRT(6 - 2^2) = SQRT(2). You can interpret that as uncertainty, in the sense of the standard deviation of the length of the next streak, or as a population parameter, if you flip a coin forever, the population standard deviation of streak length will be exactly SQRT(2).