My Probability Paradox concerning Series Imagine a random sequence of coin tosses, the length of the sequence is n.
I call the appearance of a single Head or Tail as a series of 1, the appearance of exactly 2 Heads or Tails as
a series of 2, etc...
Lets denote a series of 1 as S1, a series of 2 as S2 etc.
I believe the average number of S1=n/4, average number of S2=n/8, av.no. of S3=n/16 etc...
Am I correct so far?
So, if n=1024, the average number of S1=256, S2=128, s3=64 etc...
So, 256 of length 1,
    128 of length 2 = length of 256,
     64 of length 3 = length of 192 etc...
The sum of these lengths will approach n while n is a finite number, but not
reach n.
Am I still correct?
Now, as the sum < n,
P(S1) > (  P(S2) + P(S3) + P(S4) ... + P(Sn)  )
(Where P denotes Probability)
So I would have an advantage if I bet for singles, S1, to form!
If not, why is this not so?
This question I have explained as simply as I can, I devised it whilst reading about series and sequences
and probabilities. It seems to be a paradox to me!
Any help is appreciated...
 A: Your "I believe the average number of $S_1=n/4$, average number of $S_2=n/8$, av.no. of $S_3=n/16$, etc." is not exact.  For example with $n=2$ you have the expected number being $1$ and $\frac12$ and $0$, rather than $\frac12$ and $\frac14$ and $\frac18$
More precisely:


*

*the expected number of $S_1$ runs is $\frac n4 + \frac12$ if $n \gt 1$, and is $1$ if $n=1$

*the expected number of $S_2$ runs is $\frac n8 + \frac18$ if $n \gt 2$, and is $\frac12$ if $n=2$, and $0$ if $n \lt 2$ 

*the expected number of $S_3$ runs is $\frac n{16}$ if $n \gt 3$, and is $\frac14$ if $n=3$, and $0$ if $n \lt 3$

*the expected number of $S_m$ runs is $\frac n{2^{m+1}} -  \frac{m-3}{2^{m+1}}$ if $n \gt m$, and is $\frac1{2^{m-1}}$ if $n=m$, and $0$ if $n \lt m$
Your statement that the expected number of runs of length $1$ is greater than the expected number of all runs of longer length is indeed correct.  The expected total number of runs is $\frac n2 +\frac12$, and the expected number of $S_1$ runs is clearly very slightly more than half this.  There is no particular paradox there
What is peculiar is that this may not directly give you useful information about your bet.  If you are betting on the length of the initial run, then with $n\gt 1$ it is equally likely to be of length $1$ or be longer, as this just depends on the second toss  
If you are betting on whether the number of runs of length $1$ is greater than or less than the total number of runs of longer length then it gets more complicated:


*

*With $n=1$ you will obviously win your bet since the only run will be of length $1$, either H or T

*With $n=2$ you win your bet with probability $\frac12$ when HT or TH appear, but lose it with the same probability if HH or TT appear

*With $n=3$ you win your bet with probability $\frac14$ when HTH or THT appear, but lose it with the same probability if HHH or TTT appear, and the result is a tie if HHT, HTT, THH or TTH appear 

*With $n=4$ you win your bet with probability $\frac12$ when HTHT, THTH, HTHH, HHTH, THTT, TTHT, HTTH or THHT appear, but lose it with lower probability $\frac14$ if HHHH, TTTT, HHTT, TTHH appear, and the result is a tie if HHHT, HTTT, THHH or TTTH appear


so the benefit of your bet would depend on the particular value of $n$.   The explanation of this is that the actual number of runs of particular lengths is not independent of the number of runs of other lengths. I have not properly checked $n=5$ but a first attempt suggested probabilities of you winning of $\frac 12$, losing of $\frac38$ and tying of $\frac18$
