Boundary plays - O Meien's method (Part 2)

[Bill Spight]

Bill used a phrase saying that we "estimate Black's territory as 2 points". Not quite right either, for some of us, I suggest. It's quite possibly the correct usage in mathematics, but in ordinary speech "estimate" suggests vagueness (cf. the well known issue of estimate over quotation when dealing with tradesmen).

Would it be better to say that our estimates are fuzzy, rather than vague? Vagueness implies something like the sorites problem (if I estimate +1 in this area it could just as easily be 0, and if I could estimate 0 it could just as well be -1...), whereas fuzziness implies that my expectations bear a somewhat loose relation to reality.

Yes, the values of positions are technically fuzzy. You can also give them a probabilistic semantics.

Similarly, as Bill says, if my estimates of the score in all the remaining contested areas are all correct (that is, if they are good estimates), then even if I estimated too high for Black here and too high for White there, I estimated the total score perfectly.

I do not recognize what I said in that.

For people like me, it is easy enough to understand that the second figure (1, relating to 'b') has to be treated differently from the 'a' figure, to cover the fact that it is lower down the food chain and less likely to be used. But the reason for treating it in this particular way is far from obvious - I'm still making a leap of faith. OM admits it is fiddly but claims you get used to it.

Chaining together the probabilities in this way is very standard. What are the odds of flipping two coins and having them both come up heads? What are the odds of just missing your train at Harvard Square, and then again at Park Street? What are the odds of rolling ten or higher on two dice?

To figure out the probability of any event that's composed of two independent events, you just multiply the probability of the first event by the probability of the second. (So you get 25%, 100%, 11%.) In this case, OM is suggesting that the probability of Black getting a move that's worth a specific number of points is 50%.

At my level, that's clearly a fiddly assumption But multiplying the probability of the follow-up by the probability of the initial move is the only thing to do.

There is the practical value of the probabilistic semantics.

[ChradH]

I think both methods for calculation presented here are equivalent. I would call them 'recursive':

The value V of a position is always the average of black's gain minus white's: V = (B - W)/2.
Let's assume point "a" gets played first and call the black/white gains Ba and Wa.

Black playing "a" gains 7 points immediately, so B(a) is 7.
White playing "a" leads to a board position which has to be further evaluated, but there is nothing special about it. It's exactly the same procedure: Wa = (Bb - Wb)/2

With white "a" in place, black playing "b" gains 2 points, so Bb = 2.
White playing "b" gains 0 points, so Wb = 0, net result Wa = (2 - 0)/2 = 1.

Having values for both Ba and Wa, we get final value V = (7 - 1)/2 = 3 (no surprise here).

What we did was recursively substituting the basic formula (B - W)/2 into itself because one of the gains (W in our example) depended on playing another move.

In principle calculation of further 'dingleberries' works the same, but this might take some practice if you want to do it in your head.
Consider V = (Ba - (Bb - (Bc - Wc)/2)/2)/2. Recursion is much easier for computers

Edit: typo corrected, thanks Robert
Edit2: some more errors detected, see below

[RobertJasiek]

ChradH, do you use "gain" in a specialized meaning? If yes, which?

To mention a typo, Wa = (2 - 0) = 1 should be Wa = (2 - 0) / 2 = 1.

[cyclops]

ChradH, do you use "gain" in a specialized meaning? If yes, which?

Allow me to try this one. JF taught us how to get a net result of +3 ( 83 - 80 ) for black in his initial diagram in his first thread from his crosses and triangles and the numbers of prisoners, taken and to be taken. How he got the crosses and triangles he is still going to explain, I hope. For the time being we neglect fractions.
Without further definition I call this number 3 the "smell" of his diagram ( that includes the number of prisoners taken ). Why smell: I dont want to reserve a more sensible name that we/JF might need later for better purposes.
I think the "gain" of a sequence Chradh is talking about is the increase in smell if the diagram ( including the prisoners and ( by using JF's still undisclosed method) the crosses and triangles ) is updated with that sequence.
Hope this answers your question, Robert.

[ChradH]

ChradH, do you use "gain" in a specialized meaning? If yes, which?

I think the "gain" of a sequence Chradh is talking about is the increase in smell if the diagram ( including the prisoners and ( by using JF's still undisclosed method) the crosses and triangles ) is updated with that sequence.
Hope this answers your question, Robert.

Yes and no. The "smell" value of this position is indeed 3. Only if black really plays "a", he gains 7 solid, countable points, that's how "gain" was meant.
Of course, this is gote, so white will get compensation elsewhere. If all positions get played out, I've got an inkling that because then white will get the bigger "gain" Wx of playing the next biggest point "x", black gets By of the next biggest point "y" and so on, we might end up with nearly the same value as if we simply counted the averaged "smell" points in the first place.
But that's only speculating, let's see what JF has yet to disclose.

[topazg]

Black at a: 7 points

There's no need to process anything more in this case.

White at a: Relies on b.

Now, to apply the method to b for the "White at a" case ...

Black at b: 0
White at b: -2

Average at b: (0-2)/2=-1

Now the whole thing:

Black at a: 7
White at a: -1

Average (for a): (7-1)/2=3

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...

[ChradH]

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).

[jts]

Similarly, as Bill says, if my estimates of the score in all the remaining contested areas are all correct (that is, if they are good estimates), then even if I estimated too high for Black here and too high for White there, I estimated the total score perfectly.

I do not recognize what I said in that.

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

As my demonstration in the previous thread shows, if you add four such positions together, their combined score is a constant.

[Bill Spight]

Black at a: 7 points

There's no need to process anything more in this case.

White at a: Relies on b.

Now, to apply the method to b for the "White at a" case ...

Black at b: 0
White at b: -2

Average at b: (0-2)/2=-1

Now the whole thing:

Black at a: 7
White at a: -1

Average (for a): (7-1)/2=3

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...

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.

[Bill Spight]

Similarly, as Bill says, if my estimates of the score in all the remaining contested areas are all correct (that is, if they are good estimates), then even if I estimated too high for Black here and too high for White there, I estimated the total score perfectly.

I do not recognize what I said in that.

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

As my demonstration in the previous thread shows, if you add four such positions together, their combined score is a constant.

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.)

[jts]

Yes, fair enough fair enough.

Normally, if you add four estimates together, you get a larger error than any of the individual errors, not exactness.)

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