I'm currently reading "Winning Ways for Your Mathematical Plays" at http://annarchive.com/files/Winning%20Ways%20for%20Your%20Mathematical%20Plays%20V1.pdf. I'm on page 5, and I don't really understand the first paragraph on that page. How does figure 6(a) come into 6(b)? And what exactly does a "1/2 positive advantage" means?
Thanks!
On
Figures $6$(a), $6$(b), and $6$(c) are independent of one another: they are three different Hackenbush positions. The discussion on pages $4$ and $5$ shows that
Thus, $6$(c) is a zero position as defined at the bottom of page $2$: neither Left nor Right has an intrinsic advantage. Left clearly does have an advantage in position $6$(a), however, and we want to know just how big this advantage is. Let’s say that it’s an advantage of $x$ moves.
The position with one red edge and nothing else is clearly a win for Right no matter who starts, and it makes sense to say that it represents a one-move advantage for Right: Right has one move that can be taken no matter what Left does, and Left has zero moves. Position $6$(b) adds this position to the one in $6$(a); thus, $6$(b) is better for Right than $6$(a) by exactly one move — which amounts to saying that Left’s advantage in $6$(a) has been reduced by one full move. And since $6$(b) is a win for Right, $6$(a) must have been worth less than one full move to Left: Left has an advantage in $6$(a), but reducing it by one move goes past neutral to give Right the advantage. In other words, $x>0$, but $x-1<0$.
Now consider the position consisting of two copies of $6$(a); Left’s advantage here is $2x$ moves. To get $6$(c) we add a free-standing red edge to these two copies, decreasing Left’s advantage by $1$ just as before, so in $6$(c) Left’s advantage must be $2x-1$ moves. But $6$(c) is a zero position, so Left’s advantage is $0$. Thus, $2x-1=0$, and $x=\frac12$ a move.
The statement that Left has a half-move advantage in some position $P$ is really just shorthand for the following statement:
The Hackenbush position that consists of two copies of $P$ and one free-standing red edge is a zero position, one in which the second player can always win.
In other words, Left has an advantage in $P$, and that advantage is exactly enough so that doubling it (by adding a second copy of $P$ to the position) gives Left exactly a one-move advantage, one that is exactly balanced by giving Right a free move (by adding a free-standing red edge).
A game $H$ with a value of $\frac12$ is a game such that if you play a combined game consisting of $2$ copies of $H$ and one copy of a game of value $-1$, then the value of that combined game is zero.
Figure 61 does not "come into" 6b, they are two different examples of games. figure 6b proves that the stick in figure 6a (I will call that game $H$) has value less than $1$. Figure 6c shows that twice $H$ plus the single red line is a loss for whoever has to move first; that is, it has value $0$.