I'm going to call your "streams" "strings" instead, because "streams" looks too much like "streaks" to me. This becomes a much easier problem if we translate it into an enumeration problem. Since each of the 2^{10} possible bit strings occur with equal probability, it suffices to count the total number of streaks of length k in all these strings (and then as Kristal says just divide by 2^{10} for the expected value per string). Let S(k,n) be the number of k-streaks in all strings of length n. For k = 1 we can make a straightforward recursion:

$S(1,n) = 2S(1,n-1) + 2^{n-1} - 2^{n-2} = 2S(1,n-1) + 2^{n-2}$

since for every n string, we have two copies of every (n-1)-string as a prefix (with their associated 1-streaks), plus half the time we add a new 1-streak at the end (if the bit we add at the end is different from the old last bit). Sometimes, though, we break a 1-streak at the end; this happens if our previous last bit was a 1-stream, so it happens once for each possible n-2 string as a prefix. (Note that this means the recursion requires n > 2!).

This recursion is easy to solve; we have S(1,2) = 4 from which it follows $S(1,n) = (n+2)2^{n-2}$.

Now there's a bijection between (k-1)-streaks in (n-1)-strings and k-streaks in n-strings (just add / remove another bit to the streak), so S(k-1,n-1) = S(k,n). We conclude

$S(k,n) = (n-k+3)2^{n-k-1}$

for k < n, and S(n,n) = 2. For n=10, we have the sequence

3072, 1408, 640, 288, 128, 56, 24, 10, 4, 2

for streaks of length 1, 2, 3, ..., 10 out of the 2^{10} = 1024 possible strings, so dividing through should give the expected number in a random string.

The sequence S(1,n) is not in Sloane, but the total number of streaks in all n-strings
$\sum_{k=1}^n S(k,n)$
is - this sequence starts 2, 6, 16, 40, 96, 224,... and has general form
$\sum_{k=1}^n (n-k+3)2^{n-k-1} = (n+1)2^{n-1}$.

This sequence A057711 apparently has quite a few combinatorial interpretations.