Why isn't it 7+1 Bill? All my boundary play would be screaming another 2 point gote (/2) move following the 7 points makes the next one effectively worth 8 (swing value, right?). So I ended up with 7+1/2 = 4...

Re-reading Robert's remarks about double negation, I see I fell into a sign trap.

If I really want to calculate a position's value as (B - W)/2, I must count white's gain as positive for white. (It also implies I'm looking from black's perspective. A value of 3 means three points in favour of black.)

But this means when evaluating white gain Wa after playing point "a", I'd have to calculate it as (Wb - Bb)/2, because now it's seen from white's perspective. Yuck.
Switching signs depending on whose gain to count is quite impractical, so it seems indeed simpler always to add the terms and simply express the opponent's gain as negative numbers. Incidentally, that is what OM used, apparently...

Let's again look at diagram 5 from from black's point of view (Black's gains are positive, white's negative):

V = (Ba + Wa)/2
Ba = 7
Wa = (Bb + Wb)/2
Bb = 0
Wb = -2 (A two point loss from black's point of view)

and (again), we arrive at V = (7 + (0 - 2)/2)/2 = 3, this time even using consistent calculations (I hope).

_________________
To sig or not to sig, that is the question.

Sorry Bill, I thought my point was a different way of stating this:

Well, first, I was quoting someone who quoted someone else. That does not mean that I agree with what either one said.

That said, let's move on.

First, let's establish that the average value of the position is indeed 3 points for Black.

[sgf-full](;GM[1]ST[2]FF[4]SZ[19]CA[ISO8859-1]AP[GOWrite:2.2.21]AW[qk][rm][qn][ro][so][sq][sr][rr][br][ar][aq][ao][bo][cn][bm][ck][ci][bg][cf][be][ae][ac][ab][bb][rb][sb][sc][se][re][qf][rg][qi]PW[ ]GN[ ]FG[259:]PM[2]PB[ ]AB[ss][qs][qr][qq][rq][rp][qo][po][as][cs][cr][cq][bq][bp][co][do][de][ce][bd][bc][cc][cb][ca][aa][sa][qa][qb][qc][rc][rd][qe][pe]
(
;B[sd]
;W[sp]
;B[ap]
;W[ad]
;B[ba]
;W[rs]C[The score for the 4 positions is 14 - 2 = 12 points for Black.

The average for each position is 3.]
)
(
;W[sd]
;B[sp]
;W[ap]
;B[ad]
;W[ra]
;B[bs]C[The score for the 4 positions is 14 - 2 = 12 points for Black.

The average per position is 3.]
)

)[/sgf-full]

Es bueno?

What you have done is calculate how much each play gains, on average. Since Black can move to a position worth 7 points, and the original position is worth 3 points, the gain to Black for the gote is 4 points. Likewise, White can move from a position worth 3 points to one worth -1, for a gain of 4 points.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

I did not mean to say that any estimate was too high or too low, but that they were just right, a la Goldilocks. (Also that they were not just estimates, but more than that. Normally, if you add four estimates together, you get a larger error than any of the individual errors, not exactness.)

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Yes, fair enough fair enough.



As a fraction of the total estimate, you should expect it to be a good deal smaller. (And for forty estimates, to be a great deal smaller.) I'm just trying to gently nudge at people's intuitions about how probabilistic estimates work. The key point is that it's fine to treat your estimates as cold, hard facts (rather than something vague and unknowable) if you have many estimates to make. Talking about missing trains and throwing dice is cheating a little bit because we're talking about dependent events here (if there are 7 2-point gote plays, there's not a 0.8% chance Black will get all 7 of them...), but it seems like a good way to encourage people to overcome any reluctance to treat a set of estimates as solid and reliable. I apologize for associating our different points about aggregation