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 Post subject: Value of a reverse sente move?
Post #1 Posted: Mon Sep 21, 2020 7:15 am 
Dies with sente

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Why do we consider that, on average, for comparing the value of a reverse sente to gote move, we have to double the value of the reverse sente move?

Consider an environmment made of very simple gote areas with the values, Gn, Gn-1, Gn-2 ...
In order to calculate this average value for a reverse sente move, we first of all have to define what is an ideal environment.
Obviouly (?) an ideal environment is an environment in which the sequence of Gn, Gn-1, Gn-2 ... values does not show potential tedomari considerations (I mean for example a group of large values and a goup of small values).
With this in mind suppose an environment made of the following gote (miai) values:
0.5, 1.0, 1.5, 2.0, 2.5, ......, Gn
When a player plays first in the environment it gains:
y = Gn - Gn-1 + Gn-2 - Gn-3 ...
if n is even then y = Gn/2 but if n is odd y = Gn/2 + 0.5
Because we calculate only an average value, ignoring the fine details of the environment, we cannot tell if n is even or odd. As a consequence we have to say that the player who plays first in the environmemnt will gain
y = Gn/2 + 0,25.

What implication for a reverse sente move with value x?
On average we have to compare x to y that means that you have to compare x to Gn/2 + 0,25 which is equivalent to compare 2x - 0.5 to Gn.
The gote value, equivalent to the reverse sente value x, is thus 2x - 0.5.

I know that instead of building the environment
0.5, 1.0, 1.5, 2.0, 2.5, ......, Gn
and could have chosen
0.25, 0.50, 0.75, 1.0, 1.25, ......, Gn

As consequence the reverse sente value x seems just equivalent to a value slighty less than 2x, I can note 2x-

In other words, on average, if 2x = Gn, prefer the environment rather than the reverse sente move!

What is your feeling?

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Post #2 Posted: Mon Sep 21, 2020 9:48 am 
Lives in gote

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reducing go to math hurts both my head and heart.

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 Post subject: Re: Value of a reverse sente move?
Post #3 Posted: Mon Sep 21, 2020 10:04 am 
Honinbo

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Suppose that there is an environment composed of simple gote of the form, {ti | -ti}; i.e. a gote such that Black to play can gain ti points and White to play can gain the same, i being an integer ≥ 0, and ti being a number ≥ 0. Suppose further that if j is an integer > i, then ti ≥ tj. i starts at 0. The average value of each such gote position is 0, as is the average value of the environment.

Playing first in the environment gains at least 0 and at most t0. We may estimate the gain from doing so as t0/2 with a maximum error of t0/2.

Aside from the environment let there be a simple gote, {g | -g} such that g ≥ t and a White global sente, {r || 0 | -2s}, such that s > g and r ≥ 0; i.e., Black to play can move to a position worth r points and White to play can move to the simple gote, {0 | -2s}.

Suppose that Black is to play.

1) If Black plays the reverse global sente, then White takes the gote and then Black plays first in the environment. She gains

r - g + t0/2 ± t0/2

2) If Black plays the gote first, White takes the global sente and then plays first in the environment. Black gains

g - t0/2 ± t0/2

Black's relative gain (or loss if negative) from playing the reverse global sente is

r - 2g + t0 ± t0

Note that this is also the case if g ≥ s ≥ t0, so that r is not yet reverse global sente.

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 Post subject: Re: Value of a reverse sente move?
Post #4 Posted: Mon Sep 21, 2020 10:56 am 
Dies with sente

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Bill Spight wrote:
Playing first in the environment gains at least 0 and at most t0.

1O0% OK

Bill Spight wrote:
We may estimate the gain from doing so as t0/2 ...

Here is my point.
How can you justify your estimation of the gain to t0/2.
If the number of gote points is even I quite understand this figure t0/2 but if the number of gote point is odd I think the correct figure is slightly more than this t0/2 say t0/2 + e with e very low but > 0.
On average, whitout knowing if the number of gote move is even or odd, it appears to me that the correct value is t0/2 + e/2

Let's take the obvious following examples

1) the environment is made of four gote points with the value 4, 3, 2, 1 => when plyaing the environment you gain 2 points = t0/2
2) the environment is made of five gote points with the value 5, 4, 3, 2, 1 => when plyaing the environment you gain 3 points = t0/2 + 0.5

Because we only consider an average value, this value seems to be slightly higher than t0/2.

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 Post subject: Re: Value of a reverse sente move?
Post #5 Posted: Mon Sep 21, 2020 11:38 am 
Honinbo

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Gérard TAILLE wrote:
Bill Spight wrote:
Playing first in the environment gains at least 0 and at most t0.

1O0% OK

Bill Spight wrote:
We may estimate the gain from doing so as t0/2 ...

Here is my point.
How can you justify your estimation of the gain to t0/2.


By minimizing the maximum error. :)

Your environment contains only one gote of each size. Mine may include any number of gote of the same size. The difference between consecutive gote in your environment is fixed. This means that as that difference approaches zero, the error of the estimate for playing first in the environment also approaches zero. This fact is useful for working with limits.

Quote:
If the number of gote points is even I quite understand this figure t0/2 but if the number of gote point is odd I think the correct figure is slightly more than this t0/2 say t0/2 + e with e very low but > 0.
On average, whitout knowing if the number of gote move is even or odd, it appears to me that the correct value is t0/2 + e/2


For evaluating any given go position or finite combinatorial game it is possible to construct an ideal environment such that for each relevant ti the value of playing first in the environment is ti. The ideal environment is an analytical device. Real environments on the go board are usually such that the error of such evaluations is small. :)

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 Post subject: Re: Value of a reverse sente move?
Post #6 Posted: Mon Sep 21, 2020 11:58 am 
Honinbo

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Bill Spight wrote:
Aside from the environment let there be a simple gote, {g | -g} such that g ≥ t and a White global sente, {r || 0 | -2s}, such that s > g and r ≥ 0; i.e., Black to play can move to a position worth r points and White to play can move to the simple gote, {0 | -2s}.

Suppose that Black is to play.

1) If Black plays the reverse global sente, then White takes the gote and then Black plays first in the environment. She gains

r - g + t0/2 ± t0/2

2) If Black plays the gote first, White takes the global sente and then plays first in the environment. Black gains

g - t0/2 ± t0/2

Black's relative gain (or loss if negative) from playing the reverse global sente is

r - 2g + t0 ± t0

Note that this is also the case if g ≥ s ≥ t0, so that r is not yet reverse global sente.


Note 2:

Suppose that r = g, i.e., that the theoretical size of the gote and reverse sente are the same. Then the estimated gain from playing the reverse sente is

g - 2g + t0 ± t0
=
-g + t0 ± t0

Since g ≥ t0, let h = g - t0 ≥ 0

Then the estimated loss from playing the reverse sente is

h ± t0

So in general Black should play the simple gote rather than the reverse sente of the same size.

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 Post subject: Re: Value of a reverse sente move?
Post #7 Posted: Mon Sep 21, 2020 1:51 pm 
Dies with sente

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Bill Spight wrote:
Suppose that there is an environment composed of simple gote of the form, {ti | -ti}; i.e. a gote such that Black to play can gain ti points and White to play can gain the same, i being an integer ≥ 0, and ti being a number ≥ 0. Suppose further that if j is an integer > i, then ti ≥ tj. i starts at 0. The average value of each such gote position is 0, as is the average value of the environment.


Because I am not quite familiar with your notation it seems rather difficult to find where could existe some misundertanding.
Here is maybe a difference I identified:

You suppose {ti | -ti gote with t ≥ 0 though I excluded {0|0} gote point which looks like a dame point.
If I am right I can always suppose t > 0. Is it true?

If that is the case then call n the number of gote points with the value t0, t1, ... tn-1
with t0 ≥ t1 ≥ ... ≥ tn-1 > 0

Now we can calculate the gain env for the player who plays first in this environmement.

if n is even the value env is a value such that t0 ≥ env ≥ 0 and we may estimate env to
env = t0/2
with an error +-t0/2

if n is odd the value env is a value such that t0 ≥ env ≥ tn > 0 and we may estimate env to
env = t0/2 + tn-1/2
with a error +-(t0/2 + tn-1/2)

Supposing your are not able to know if n is even or odd then the average value of env is
env = t0/2 + tn-1/4
with tn-1being the smallest non zero gote value of the environmement.

It is not easy to explain but that is my point : the gain by playing in the environment is slighly larger than t0/2, simply because the number of gote points is finite => it is relevant to take tn-1 into consideration.

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 Post subject: Re: Value of a reverse sente move?
Post #8 Posted: Mon Sep 21, 2020 2:13 pm 
Honinbo

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On the average value of a sente position

Reference: https://senseis.xmp.net/?MethodOfMultiples

First, let's derive the average value of a simple gote

Given a gote, {r | -r}, the average value is 0.

Suppose there are two of them. Then Black to play gets one and White to play gets the other, for a score of r - r = 0, no matter who plays first. The same is true for any multiple of two of them.

Next, let's derive the average value of a simple White sente.

Given a sente, {r || 0 | -2s}, the average value is 0.

Given any number, N, of them, when White plays first with sente the result is 0. If Black plays first and then White plays with sente the result is r. Thus the average value lies between 0 and r/N. The limit as N approaches infinity is thus 0. Any numerical value greater than 0 will be violated by sufficiently large N.

Still, the value of this sente is greater than 0, by the definition in combinatorial game theory (CGT). Black wins this game regardless of who plays first, because the second player wins a 0 game in CGT; it is not a tie. This value is not a number, OC.

Thus, a move by Black to r gains less than r, even though the average value of that gain is r. Strange as that sounds.

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 Post subject: Re: Value of a reverse sente move?
Post #9 Posted: Mon Sep 21, 2020 3:36 pm 
Honinbo

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Gérard TAILLE wrote:
Bill Spight wrote:
Suppose that there is an environment composed of simple gote of the form, {ti | -ti}; i.e. a gote such that Black to play can gain ti points and White to play can gain the same, i being an integer ≥ 0, and ti being a number ≥ 0. Suppose further that if j is an integer > i, then ti ≥ tj. i starts at 0. The average value of each such gote position is 0, as is the average value of the environment.


Because I am not quite familiar with your notation it seems rather difficult to find where could existe some misundertanding.
Here is maybe a difference I identified:

You suppose {ti | -ti} gote with t ≥ 0 though I excluded {0|0} gote point which looks like a dame point.
If I am right I can always suppose t > 0. Is it true?


No. In an actual game you typically have lots of dame. In fact, you can have an environment with t = -1. Most Japanese rules beasts have such an environment, as there is usually no problem if there are enough dame. :)

Quote:
If that is the case then call n the number of gote points with the value t0, t1, ... tn-1
with t0 ≥ t1 ≥ ... ≥ tn-1 > 0

Now we can calculate the gain env for the player who plays first in this environmement.

if n is even the value env is a value such that t0 ≥ env ≥ 0 and we may estimate env to
env = t0/2
with an error +-t0/2

if n is odd the value env is a value such that t0 ≥ env ≥ tn > 0 and we may estimate env to
env = t0/2 + tn-1/2
with a error +-(t0/2 + tn-1/2)

Supposing your are not able to know if n is even or odd then the average value of env is
env = t0/2 + tn-1/4
with tn-1being the smallest non zero gote value of the environmement.

It is not easy to explain but that is my point : the gain by playing in the environment is slighly larger than t0/2, simply because the number of gote points is finite => it is relevant to take tn-1 into consideration.


IIUC, you are not assuming that the difference between successive temperatures is fixed, but you are assuming that all temperatures are above 0, and you are not assuming only one gote at each temperature.

Well, for convenience let's look at temperatures that are integers and an environment where the maximum temperature is 4. Also for convenience, let's write the gote {t | -t} as ±t.

Here are some of those environments.

1) ±4. Playing first in this environment gains 4. Max error =2.
2) ±4, ±4. Playing first gains 0. Max error = 2.
3) ±4, ±4, ±4. Playing first gains 4. Max error = 2.
Etc.

Regardless of the parity of the number of gote in the environment, the maximum error will always be the same.

----

I am not trying to prove you wrong. Your approach is valid. :)

But an ideal environment is an analytical device. Given a non-ko position to analyze, you can always find an ideal environment such that, when the time comes for the players to play in the environment, the gain from playing in the environment will be ½ the ambient temperature. That simplifies the task. :)

(Analyzing kos may require environments with negative temperatures, and the rules are different.)

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 Post subject: Re: Value of a reverse sente move?
Post #10 Posted: Mon Sep 21, 2020 9:36 pm 
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For the late endgame, we can use exact analysis.

For the early endgame, the even / odd cases amounting to an epsilon can be simplified by ignoring (rounding down) the epsilon for practical purposes. For perfect analysis, we would keep the epsilon but must also consider many other aspects, such as non-ideal environments including kos. As long as we do not perfectly understand all those aspects, the epsilon is the second last thing we care about because this number is tiny. (The last thing is the CGT infinitesemals analysis of getting the last move in the game.)

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Post #11 Posted: Tue Sep 22, 2020 2:13 am 
Dies with sente

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Bill Spight wrote:
Well, for convenience let's look at temperatures that are integers and an environment where the maximum temperature is 4. Also for convenience, let's write the gote {t | -t} as ±t.

Here are some of those environments.

1) ±4. Playing first in this environment gains 4. Max error =2.
2) ±4, ±4. Playing first gains 0. Max error = 2.
3) ±4, ±4, ±4. Playing first gains 4. Max error = 2.
Etc.

Regardless of the parity of the number of gote in the environment, the maximum error will always be the same.

----

I am not trying to prove you wrong. Your approach is valid. :)

But an ideal environment is an analytical device. Given a non-ko position to analyze, you can always find an ideal environment such that, when the time comes for the players to play in the environment, the gain from playing in the environment will be ½ the ambient temperature. That simplifies the task. :)

(Analyzing kos may require environments with negative temperatures, and the rules are different.)


Though you think my approach may be valid it seems to me there remains some misunderstanding.
I looked at your reference: https://senseis.xmp.net/?MethodOfMultiples and I feel it is really an very interesting approach.
In an other hand when in you post above you take an environmemnt this only ±4 gote points I think you miss my point.

When I wrote
env = g0/2 + gn-1/4
you have to understand (of course you may not agree on that !) that you cannot forget that, at the very end of the yose, a player will play the last yose point which is the smallest one.

Let's take an area counting which is far easier to understand:
Putting aside ko complications gn-1 is typically equal to 1 which the value of a move on a dame point. Thus
env = g0/2 + 0.25

The point is that the number of gote moves in the environment is finite (say equal to n) and, as a consequence one of the player will take the last dame point (it looks the CGT infinitesemals analysis of getting the last move in the game Robert Jasiek is refering to).

Another to way to see the point is the following calcultation:
Let's call ri the result for playing the environment of gote moves gi, gi+1, ..., gn-1.
In other terms
ri = gi - gi+1 + gi+2 - gi+3 ... gn-1

What about an environment such that ri = gi / 2 for all i.
Of course this is impossible because it is not true when i = n-1 but you can also, more generally, analysing ri as follows:
ri = gi - ri+1 = gi - gi+1/2
=> gi / 2 = gi - gi+1/2
=> gi = gi+1

As a conclusion you cannot ignore the very last gote point (typically a dame point) when calculating the gain from the environment.
That why my estimation of the gain when playing first the environment is

env = g0/2 + 0.25

Now, can I build an ideal environment for which ri = gi / 2 + 0.25 for all i.
The answer is yes and you find easily the environment:
gn, gn - 0.5, gn - 1.0, ..., 1.5, 1.0, 0.5

You note that the last value of this ideal environment is 0.5 instead of 1.0 for a dame point. This is due to the averaging approach in which you do not know which player will take the last dame. By building an ideal point with this last 0.5 point you achieve to build a really ideal environment.

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Post #12 Posted: Tue Sep 22, 2020 5:39 am 
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Gérard TAILLE wrote:
Bill Spight wrote:
Well, for convenience let's look at temperatures that are integers and an environment where the maximum temperature is 4. Also for convenience, let's write the gote {t | -t} as ±t.

Here are some of those environments.

1) ±4. Playing first in this environment gains 4. Max error =2.
2) ±4, ±4. Playing first gains 0. Max error = 2.
3) ±4, ±4, ±4. Playing first gains 4. Max error = 2.
Etc.

Regardless of the parity of the number of gote in the environment, the maximum error will always be the same.

----

I am not trying to prove you wrong. Your approach is valid. :)

But an ideal environment is an analytical device. Given a non-ko position to analyze, you can always find an ideal environment such that, when the time comes for the players to play in the environment, the gain from playing in the environment will be ½ the ambient temperature. That simplifies the task. :)

(Analyzing kos may require environments with negative temperatures, and the rules are different.)


Though you think my approach may be valid it seems to me there remains some misunderstanding.
I looked at your reference: https://senseis.xmp.net/?MethodOfMultiples and I feel it is really an very interesting approach.


It is not mine, it is the way that mathematicians proved what is called the mean value theorem for games such as independent non-ko go positions. I call it the method of multiples. With it you can derive the average value of positions and moves without appealing to an environment. If you evaluate them by means of an environment but get different answers, you are talking about something else. To get the same answers by your approach you have to take the difference between successive gote and the value of the last gote to 0 in the limit.

Quote:
In an other hand when in you post above you take an environmemnt this only ±4 gote points I think you miss my point.


I assure you I have not missed your point. Your model of the environment is quite similar to the one I started off with many years ago. :) But in that particular note I thought that you were allowing a more general model of the environment than your own, for the purpose of discussion.

Quote:
When I wrote
env = g0/2 + gn-1/4
you have to understand (of course you may not agree on that !) that you cannot forget that, at the very end of the yose, a player will play the last yose point which is the smallest one.


Consider this environment:

±4, ±3, ±3 ±2, ±2, ±1, ±1

Here the smallest gote gains 1 pt. for the player who takes it. However, the gain from playing first in this environment is 4 pts. the same as it is from playing first in this environment.

±4

Consider this environment:

±4, ±4, ±3, ±3 ±2, ±2, ±1, ±1

The gain from playing first in it is 0.

If we allow both environments with a maximum gain from the largest play of 4 pts., then we can estimate the gain from playing first in that environment as 2 pts. with a maximum error of 2 pts. Any other estimate will have a larger maximum error. Now, such a large error may be undesirable for certain purposes, but the estimate yields the same values as the method of multiples. :)

Quote:
Let's take an area counting which is far easier to understand:
Putting aside ko complications gn-1 is typically equal to 1 which the value of a move on a dame point. Thus
env = g0/2 + 0.25


Putting aside ko considerations, the method of multiples indicates that the average value of a dame is 0 and taking a dame gains 1 pt. at area scoring. Playing first in an environment of an unknown number of dame points gains ½ ±½. But your equation gives a different estimate.

Quote:
The point is that the number of gote moves in the environment is finite (say equal to n) and, as a consequence one of the player will take the last dame point (it looks the CGT infinitesemals analysis of getting the last move in the game Robert Jasiek is refering to).


On SL and here I have written by far more than anyone else on go infinitesimals. :) It seems to me that Robert's post addresses your concerns.

All models of actual environments on the go board are idealizations. When applied to evaluation of positions and moves, different models may give different answers. That does not mean that one model is right and the other one is wrong. No model pretends to do more than approximate actual boards. The method of multiples yields the same values for non-ko positions, and for many ko positions, as the traditional values. There are good reasons that these values are useful. Your model yields the same values in the limit. That's fine. The method of multiples yields the value of a sente position only in the limit, as well. :)

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Post #13 Posted: Tue Sep 22, 2020 6:33 am 
Dies with sente

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Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:
Well, for convenience let's look at temperatures that are integers and an environment where the maximum temperature is 4. Also for convenience, let's write the gote {t | -t} as ±t.

Here are some of those environments.

1) ±4. Playing first in this environment gains 4. Max error =2.
2) ±4, ±4. Playing first gains 0. Max error = 2.
3) ±4, ±4, ±4. Playing first gains 4. Max error = 2.
Etc.

Regardless of the parity of the number of gote in the environment, the maximum error will always be the same.

----

I am not trying to prove you wrong. Your approach is valid. :)

But an ideal environment is an analytical device. Given a non-ko position to analyze, you can always find an ideal environment such that, when the time comes for the players to play in the environment, the gain from playing in the environment will be ½ the ambient temperature. That simplifies the task. :)

(Analyzing kos may require environments with negative temperatures, and the rules are different.)


Though you think my approach may be valid it seems to me there remains some misunderstanding.
I looked at your reference: https://senseis.xmp.net/?MethodOfMultiples and I feel it is really an very interesting approach.


It is not mine, it is the way that mathematicians proved what is called the mean value theorem for games such as independent non-ko go positions. I call it the method of multiples. With it you can derive the average value of positions and moves without appealing to an environment. If you evaluate them by means of an environment but get different answers, you are talking about something else. To get the same answers by your approach you have to take the difference between successive gote and the value of the last gote to 0 in the limit.

Quote:
In an other hand when in you post above you take an environmemnt this only ±4 gote points I think you miss my point.


I assure you I have not missed your point. Your model of the environment is quite similar to the one I started off with many years ago. :) But in that particular note I thought that you were allowing a more general model of the environment than your own, for the purpose of discussion.

Quote:
When I wrote
env = g0/2 + gn-1/4
you have to understand (of course you may not agree on that !) that you cannot forget that, at the very end of the yose, a player will play the last yose point which is the smallest one.


Consider this environment:

±4, ±3, ±3 ±2, ±2, ±1, ±1

Here the smallest gote gains 1 pt. for the player who takes it. However, the gain from playing first in this environment is 4 pts. the same as it is from playing first in this environment.

±4

Consider this environment:

±4, ±4, ±3, ±3 ±2, ±2, ±1, ±1

The gain from playing first in it is 0.

If we allow both environments with a maximum gain from the largest play of 4 pts., then we can estimate the gain from playing first in that environment as 2 pts. with a maximum error of 2 pts. Any other estimate will have a larger maximum error. Now, such a large error may be undesirable for certain purposes, but the estimate yields the same values as the method of multiples. :)

Quote:
Let's take an area counting which is far easier to understand:
Putting aside ko complications gn-1 is typically equal to 1 which the value of a move on a dame point. Thus
env = g0/2 + 0.25


Putting aside ko considerations, the method of multiples indicates that the average value of a dame is 0 and taking a dame gains 1 pt. at area scoring. Playing first in an environment of an unknown number of dame points gains ½ ±½. But your equation gives a different estimate.

Quote:
The point is that the number of gote moves in the environment is finite (say equal to n) and, as a consequence one of the player will take the last dame point (it looks the CGT infinitesemals analysis of getting the last move in the game Robert Jasiek is refering to).


On SL and here I have written by far more than anyone else on go infinitesimals. :) It seems to me that Robert's post addresses your concerns.

All models of actual environments on the go board are idealizations. When applied to evaluation of positions and moves, different models may give different answers. That does not mean that one model is right and the other one is wrong. No model pretends to do more than approximate actual boards. The method of multiples yields the same values for non-ko positions, and for many ko positions, as the traditional values. There are good reasons that these values are useful. Your model yields the same values in the limit. That's fine. The method of multiples yields the value of a sente position only in the limit, as well. :)


Good news Bill, with your help I can see now where I am wrong. Thank you very much for that Bill!

Let me summarize what looks like a better analysis:

Let's take again the gotes values g0 ≥ g1 ≥ ... gn-1 > 0

if n is odd then the max possible gain for the player playing first in the environment is g0 and the min value is gn-1 for on average value g0/2 + gn-1/2

But here is where I missed the point : if n is even then the max possible gain for the player playing first in the environment is g0 - gn-1 and the min value is 0 for on average value g0/2 - gn-1/2

Finally, I you cannot say if n is odd or even then the average gain for playing the environment is really g0/2.

It was not such obvious but it appears now for me that we do not care who will take the last gote point. Thank you again Bill for having challenged my reasonning. It is the best way to progress isn't it?

BTW Bill, have you seen my following post https://lifein19x19.com/viewtopic.php?p=259925#p259925 ?


This post by Gérard TAILLE was liked by: Bill Spight
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 Post subject: Re: Value of a reverse sente move?
Post #14 Posted: Tue Sep 22, 2020 7:08 am 
Honinbo

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Gérard TAILLE wrote:
It was not such obvious but it appears now for me that we do not care who will take the last gote point. Thank you again Bill for having challenged my reasonning. It is the best way to progress isn't it?


I am glad that we have reached agreement about a more general model than your original one. :) It is not that your reasoning was wrong with it. It led you to see that the value of a reverse sente is less than its theoretical value. That matters on a finite board, as its theoretical value is a limit which is approached from below.

Quote:
BTW Bill, have you seen my following post https://lifein19x19.com/viewtopic.php?p=259925#p259925 ?


Yes, thank you for reminding me. I have neglected those problems for too long. :(

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