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Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only once are among the fairest sequences of all. Their score differences are 0.

Let's use free-monoid notations and write $1^40^2$ for $111100$. Then

$v_0(1^40^2) = v_1(1^40^2) = 5$

$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$

$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$

$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$

$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$

$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$

More generally,

$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$

for any integer $k\ge 0$.

EDIT: As RaphaelB4 commented, his insight about the simple multiplicative form of his recursion, $$v_1(0b_1\dots b_n)+1 = \frac{n+3}{n+2}\left(v_1(b_1\dots b_n)+1\right)$$ iteratively yields $$v_1(0^{i}b_1\dots b_n)+1 = \frac{n+2+i}{n+2}(v_1(b_1\dots b_n)+1)$$ so that

$$v_0(1^{2k(k+1)}0^{2k}) + 1 = v_1(0^{2k(k+1)}1^{2k}) +1$$ $$ = \frac{2k+2+2k(k+1)}{2k+2} (v_1(1^{2k})+1)$$ $$ = (k+1) (2k+1) = k(2k+3) + 1$$

and

$$v_1(1^{2k(k+1)}0^{2k}) + 1 = 2k(k+1) + v_1(0^{2k})+1$$ $$ = 2k(k+1) + \frac{0+2+2k}{0+2} = k(2k+3) + 1$$

which proves the equality.

Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only once are among the fairest sequences of all. Their score differences are 0.

Let's use free-monoid notations and write $1^40^2$ for $111100$. Then

$v_0(1^40^2) = v_1(1^40^2) = 5$

$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$

$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$

$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$

$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$

$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$

More generally,

$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$

for any integer $k\ge 0$.

EDIT: for exactly the same lengths $n = 2k(k+2)$ as above, there are perfectly fair sequences with as many $0$'s as $1$'s, provided you allow two alternations instead of only 1.

$v_0(0^2 1^3 0) = v_1(0^2 1^3 0) = 5$

$v_0(0^6 1^8 0^2) = v_1(0^6 1^8 0^2) = 14$

$v_0(0^{12} 1^{15} 0^3) = v_1(0^{12} 1^{15} 0^3) = 27$

$v_0(0^{20} 1^{24} 0^4) = v_1(0^{20} 1^{24} 0^4) = 44$

$v_0(0^{30} 1^{35} 0^5) = v_1(0^{30} 1^{35} 0^5) = 65$

$v_0(0^{42} 1^{48} 0^6) = v_1(0^{42} 1^{48} 0^6) = 90$

and more generally

$$v_0(0^{k(k+1)}1^{k(k+2)}0^k) = v_1(0^{k(k+1)}1^{k(k+2)}0^k) = k(2k+3)$$

for any integer $k\ge 0$.

$Proof$: As RaphaelB4 commented, his insight about the simple multiplicative form of his recursion, $$v_1(0b_1\dots b_n)+1 = \frac{n+3}{n+2}\left(v_1(b_1\dots b_n)+1\right)$$ iteratively yields $$v_1(0^{i}b_1\dots b_n)+1 = \frac{n+2+i}{n+2}(v_1(b_1\dots b_n)+1)$$ so that

$$v_0(1^{2k(k+1)}0^{2k}) + 1 = v_1(0^{2k(k+1)}1^{2k}) +1$$ $$ = \frac{2k+2+2k(k+1)}{2k+2} (v_1(1^{2k})+1)$$ $$ = (k+1) (2k+1) = k(2k+3) + 1$$

and

$$v_1(1^{2k(k+1)}0^{2k}) + 1 = 2k(k+1) + v_1(0^{2k})+1$$ $$ = 2k(k+1) + \frac{0+2+2k}{0+2} = k(2k+3) + 1$$

which proves the first equality. The second equality with two alternations is similar. It is also easy to prove there isn't any other solutions with one or two alternations.

Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only are among the fairest sequences of all. Their score differences are 0.

Let's use free-monoid notations and write $1^40^2$ for $111100$. Then

$v_0(1^40^2) = v_1(1^40^2) = 5$

$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$

$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$

$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$

$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$

$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$

More generally,

$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$

for any integer $k\ge 0$.

EDIT: As RaphaelB4 commented, his insight about the simple multiplicative form of his recursion, $$v_1(0b_1\dots b_n)+1 = \frac{n+3}{n+2}\left(v_1(b_1\dots b_n)+1\right)$$ iteratively yields $$v_1(0^{i}b_1\dots b_n)+1 = \frac{n+2+i}{n+2}(v_1(b_1\dots b_n)+1)$$ so that

$$v_0(1^{2k(k+1)}0^{2k}) + 1 = v_1(0^{2k(k+1)}1^{2k}) +1$$ $$ = \frac{2k+2+2k(k+1)}{2k+2} (v_1(1^{2k})+1)$$ $$ = (k+1) (2k+1) = k(2k+3) + 1$$

and

$$v_1(1^{2k(k+1)}0^{2k}) + 1 = 2k(k+1) + v_1(0^{2k})+1$$ $$ = 2k(k+1) + \frac{0+2+2k}{0+2} = k(2k+3) + 1$$

which proves the equality.

Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only once are among the fairest sequences of all. Their score differences are 0.

Let's use free-monoid notations and write $1^40^2$ for $111100$. Then

$v_0(1^40^2) = v_1(1^40^2) = 5$

$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$

$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$

$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$

$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$

$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$

More generally,

$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$

for any integer $k\ge 0$.

EDIT: As RaphaelB4 commented, his insight about the simple multiplicative form of his recursion, $$v_1(0b_1\dots b_n)+1 = \frac{n+3}{n+2}\left(v_1(b_1\dots b_n)+1\right)$$ iteratively yields $$v_1(0^{i}b_1\dots b_n)+1 = \frac{n+2+i}{n+2}(v_1(b_1\dots b_n)+1)$$ so that

$$v_0(1^{2k(k+1)}0^{2k}) + 1 = v_1(0^{2k(k+1)}1^{2k}) +1$$ $$ = \frac{2k+2+2k(k+1)}{2k+2} (v_1(1^{2k})+1)$$ $$ = (k+1) (2k+1) = k(2k+3) + 1$$

and

$$v_1(1^{2k(k+1)}0^{2k}) + 1 = 2k(k+1) + v_1(0^{2k})+1$$ $$ = 2k(k+1) + \frac{0+2+2k}{0+2} = k(2k+3) + 1$$

which proves the equality.

Get ready for a counter-intuitive conjecture that get us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only once seem to be among the fairest sequences of all. Their score differences are 0.

Let's use free-monoid notations and write $1^40^2$ for $111100$. Then

$v_0(1^40^2) = v_1(1^40^2) = 5$

$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$

$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$

$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$

$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$

$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$

This is true and easily checked with small sums of binomials, or with RaphaelB4 recursion.

More generally, it looks like

$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$

for any integer $k\ge 0$. I have not been able to prove it.

Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only are among the fairest sequences of all. Their score differences are 0.

Let's use free-monoid notations and write $1^40^2$ for $111100$. Then

$v_0(1^40^2) = v_1(1^40^2) = 5$

$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$

$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$

$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$

$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$

$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$

More generally,

$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$

for any integer $k\ge 0$.

EDIT: As RaphaelB4 commented, his insight about the simple multiplicative form of his recursion, $$v_1(0b_1\dots b_n)+1 = \frac{n+3}{n+2}\left(v_1(b_1\dots b_n)+1\right)$$ iteratively yields $$v_1(0^{i}b_1\dots b_n)+1 = \frac{n+2+i}{n+2}(v_1(b_1\dots b_n)+1)$$ so that

$$v_0(1^{2k(k+1)}0^{2k}) + 1 = v_1(0^{2k(k+1)}1^{2k}) +1$$ $$ = \frac{2k+2+2k(k+1)}{2k+2} (v_1(1^{2k})+1)$$ $$ = (k+1) (2k+1) = k(2k+3) + 1$$

and

$$v_1(1^{2k(k+1)}0^{2k}) + 1 = 2k(k+1) + v_1(0^{2k})+1$$ $$ = 2k(k+1) + \frac{0+2+2k}{0+2} = k(2k+3) + 1$$

which proves the equality.