Uberdude wrote:
Why did you multiply by 4?
If you model the game as each player taking a gote move of some clear number of points in a non interacting way (like the cards of environmental Go) then there's many ways you can end up with black having 7 more at the end (I'm using a short game to illustrate the point):
28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15
100, 100, 100, 100, 100, 93
7, 7, 7, 7, 7, 7, 7
24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11
12, 12, 12, 12, 11, 10, 45, 43, 30, 28, 2, 2, 2, 1, 1, 1, 1
28, 21, 12, 12, 10, 10, 5, 5, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1
Does it make sense to talk about the value of the first move as the gote number of points it grabs (which could be e.g 7 or 100 in the examples), or is "the value of sente" more useful, which is something closer to the difference between the first 2 numbers. Of course Go isn't so simple as grabbing a bunch of non interacting gote moves in decreasing order, but it'll do for starters.
Suppose that the stack of cards model is appropriate for go in general when you don't know the specifics of the position, but you do know how much the most valuable card (the
top card) gains. Also suppose that at each turn each player takes the top card. Also, as in your example, there could be any number of cards of the same size, and suppose that there is a 50:50 chance that the number of cards of a given size is odd or even.
Then the expected result of the stack of cards with a top card that gains T pts. is T/2 pts.
Let S(v) be the expected value of having the move when the top card gains
v pts.
Let the gain of the top card be T and the gain of the second to top card be R.
Half the time the number of cards that gain T pts. will be odd and the player with the move (the first player) will take the last one for a gain of T pts. Then the second player will take one of the second to top cards, for an expected gain of S(R). The expected result will be T - S(R).
Half the time the number of top cards will be even and the players will split them, for net 0 pts. Then the first player will have an expected gain of S(R).
The expected result will then be ((T - S(R)) + S(R))/2 = T/2. QED.
Edit: T/2 will also be the expected komi because for N cards you will have 2
^{N-1} pairs of equiprobable results that add to T. Half of the addends will be less than or equal to T/2 and half of them will be greater than or equal to T/2.
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