What is worth of a stalk in red-blue Hackenbush?? I was studying about Red-Blue hackenbush from this link http://www.link.cs.cmu.edu/15859-s11/notes/Hackenbush.pdf

http://math.ucsd.edu/~wgarner/math168a/blueredhackenbush.htm this url shows a Red-Blue Hackenbush Calculator.

However I do not get the idea of worth of a stalk. What does worth of a stalk mean??
The worth (hereafter "value") of a Red-Blue Hackenbush stalk (sometimes known as a "Hackenstring", when written the RRBRB way) represents how many moves it gives to Blue (the player who can remove bLue edges is usually called "Left", but I won't follow that convention) when combined with other stalks. Since Hackenbush is a game you lose when it's your turn to move, the more moves you get the make, the safer you are. 

The rest of this post is a long explanation of where the particular values all come from, which you can skip if you're not interested in how the above idea translates into all these weird numbers with denominators that are powers of 2 (called dyadic fractions/rationals).



Integer Values

For example, BBB is worth three moves to Blue in any combination. The reason is that Red (conventionally called "Right") can't touch this stalk, and Blue could move once in it to BB, then again to B, and then again to an empty stalk. Similarly, the value of an empty stalk is 0, and the value of n Bs in a row is just n.

Now, what about something like RRR? That's worth three moves to Red, and Blue isn't allowed to make a move on it. So should its value be 0? Well, RRR doesn't act like the empty stalk in combinations: If you play a two-stalk game with the empty stalk and BBB, then it's just like BBB, so the empty stalk doesn't change the fact that BBB gives Blue 3 moves. But if you play a two-stalk game with RRR and BBB (or a bigger game with those two stalks), then any move in one of those stalks can be mirrored in the other, so Blue's "extra moves" have been nullified. 

If we use the plus sign to represent playing games with multiple stalks, then RRR+BBB is like an empty stalk (because of mirroring). If the value of BBB is 3, and the value of the empty stalk is 0, then we should have RRR+3=0, so that the value of RRR is -3. 3 moves for Red are like 3 negative moves for Blue. 

In general, if a stalk is all one color, it has an integer value. n Rs is -n, the empty stalk is 0, and n Bs is n.



What is BR?

Let’s consider the game BR. If Red goes first, they can remove the R, leaving B=1. If Blue goes first, they can remove the B (and the R to the right of it) leaving “”=0. Since Blue can win by taking the B, this is better for Blue than the empty string - we should have BR>0. How does BR compare to B=1? Well, BR-B=BR+R. In that game, Red can move to B+R=1-1=0 (so Blue loses), and Blue can move to R=-1 (which Blue still loses). Thus, BR+R=BR-B is a game that Blue loses, so it's good for Red and we should have BR-B&lt;0 so BR

We can get an even better understanding of the value of BR by comparing BR+BR with B=1. We must play BR+BR-B=BR+BR+R. If Blue goes first, they move to BR+R&lt;0 and Blue loses. If Red goes first and takes the lone “R” then BR+BR>0 remains and Red loses. If Red goes first another R, then they leave BR+B+R=BR and Red loses. In any case, the person who goes first loses (if the other player plays perfectly). But when that happens, if you were to combine BR+BR+R with any other stalks, no one would want to move first in that piece, so it's like adding on the empty stalk. Therefore BR+BR+R=0, so BR+BR=B=1! These are pretty good reasons to call the value of BR 1/2, and that’s exactly what people do. (There's also a natural multiplication on Red-Blue stalks that I won't define, but it turns out that BB*X=X+X, so BB*BR=1, which is another reason to call it 1/2.)



Other fractions?

BR=1/2 and similarly, RB=-1/2. but what happens with more complicated stalks? Consider BRR; it’s clearly positive since Blue wins even if Red goes first, and it’s less than 1 since BRR+R is a win for Red. If you play BRR+BRR-BR=BRR+BRR+RB, then if Red goes first and they leave BRR+BR+RB=BRR>0, they lose. If Red goes first and they leave BRR+BRR>0, they lose. If Blue goes first and they leave BRR+RB then Red can respond with BR+RB=0 and they (Blue) lose. If Blue goes first and they leave BRR+BRR+R then Red can move to BRR+BR+R and Blue’s only moves are to BR+R (a win for Red) or BRR+R (also a win for Red). In all cases, the second player has a winning strategy, so BRR+BRR+RB is like the empty stalk, so BRR+BRR=BR. Thus, we can/should call BRR “1/4″. This pattern continues: BRRR=1/8, BRRRR=1/16, etc.

Let’s look at another family of positions, starting with BBR. Blue can take the middle B which the R is sitting on, but the bottom B intuitively seems independent of that top BR action. For this reason, I’d guess that BBR=B+BR. To test this, play BBR+R+RB. If Red goes first, they might take the first R, leaving BB+R+RB=B+RB>0, but they’d lose. If they take the third R, they’d leave BBR+R and Blue could respond with B+R=0 and Red would still lose. BBR+RB>BBR+R, so Blue would win if Red started by taking the other R, too. If Blue goes first, they can move to BBR+R+R

Finally, let’s look at one more family before I make a general claim. Consider BRB; this is potentially better for Blue than BR, but worse than B, so it’s somewhere in between 1/2 and 1. It actually turns out to be 3/4. We can prove that BRB+BRR=B by playing BRB+BRR+R. It’s easy to see (from previously known facts) that all of Blue’s starting moves result in negative games, so this is ≤0. If Red takes the first R, they leave B+BRR+R=BRR>0. If Red takes the third R (the second is worse) then Blue moves to 0 by leaving BR+BR+R. If Red takes the last R, Blue leaves BRB>0. (There is some advanced theory of games that helps to avoid these case analyses.) BRBB is even better for Blue, and equals 7/8, etc.



General Stalk Values

Since the negative of a basic stalk switches all of the Bs and Rs, and we know “”=0, it suffices to describe the values of strings starting with B. A string consisting only of n Bs just has the integer value n. Otherwise, suppose there are n Bs before the first R. Then the value turns out to be  $(n-1)+$the binary number you get by reading Bs as 1s and Rs as 0s after the first BR, with a 1 appended. This can be seen in: 10.110010 $1_{(2)}=$ BB BR BBRRBR.


  Note: The sections after "Integer Values" were hastily adapted from a
  blog post I wrote, so there may be typos/things that are unclear.
When you cut off a stalk, you remove the connection of other stalks to the ground.  The worth of a stalk is the value of that stalk and all those above it, which are dependent on that stalk.  The reason they start adding $\textbf{after}$ a change in color is because, as long as it is the same color, the opponent can cut from the bottom and benefit by $+n$.  However, with the addition of one of the opposite color, you can cut optimally cut from that stalk, making it a different value from the rest.  This value should be lower since you are removing less of the hackenbush game.  Like many numbers in game theory, the factor of $1/2$, to my knowledge, is arbitrary.