P versus other plays
First let us simplify things a bit. Let Black have a move to a position worth G, a positive number, with gote, but let White have a move to a position worth -R, a negative number, and let Black also have a sente move to a position worth 0. Now we know that if G > R, the position is gote, if R > G, it is sente, and if G = R, it is ambiguous.
Next, let us add a play, a simple gote where Black can move to a position worth T and White can move to a position worth -T. (All variables are greater than 0.) T is smaller than BIG, so Black still has a sente play.
Black to play.
1) Black plays sente, then plays to T. Result: T.
2) Black plays to G, then White plays to -T. Result: G - T.
3) Black plays to T, then White plays to -R. Result: T - R.
T > T - R, so option 3) is out. Comparing 1) to 2) we find that if G > 2T then Black should play to G (gote) and if 2T > G Black should play sente first.
(This was my approach in the 1970s.
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Now let us add another simple gote, where Black can move to T1 and White can move to -T1, with T >= T1 > 0.
1) Black plays sente, then plays to T, then White plays to -T1. Result: T - T1.
2) Black plays to G, then White plays to -T and Black plays to T1. Result: G - T + T1.
3) Black plays to T, then White plays to -R, then Black plays to T1. Result: -R + T + T1.
4) Black plays to T, then White plays to -T1, then Black plays sente. Result: T - T1, the same result as 1).
5) Black plays to T, then White plays to -T1, then Black plays to G. Result: G + T - T1, which is at least as good as the result for 2).
4) and 5) show that if a White reply to -T1 is correct, a Black play to T dominates the other plays. However, if 3) is inferior to 1) or 2), then Black should not play to T. So we can ignore White’s response to -T1.
Comparing 2) to 3), we find that 2) is better if G + R > 2T. I. e., if playing P as a gote is better that playing T. T1 does not enter the picture.
Comparing 1) to 2), we find that playing to G is better than the sente if G > 2(T - T1).
Comparing 1) to 3), we find that playing sente is better than playing to T if R > 2T1. T does not enter the picture.
Now let us add other simple gote similar to T and T1 to make an environment, such that T >= T1>= T2 >= . . . > 0
We still have three comparisons.
1) Black plays sente. Result: T - T1 + T2 - . . . .
2) Black plays to G. Result: G - T + T1 - T2 + . . . .
3) Black plays to T, then White plays to -R. Result: -R + T + T1 - T2 - . . . .
Comparing 2) to 3) we get that Black should not play to T if G + R > 2T. Same as before.
If 2T > G + R we compare 1) to 3) and get that Black should play sente if R > 2(T1 - T2 + . . .).
If G + R > 2T we compare 1) to 2) and get that Black should play gote if G > 2T - 2(T1 - T2 + . . .). I have separated T from T1, T2, etc., because T does not appear in the comparison between 1) and 3). It really should not be considered part of the environment.
We may estimate T1 - T2 + . . . as T1/2. The estimate does not affect the comparison between 2) and 3), but it does affect the others. Using the estimate we get
1) vs 3): Compare R to T1.
1) vs 2): Compare G to 2T - T1.
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We may also consider a generalized environment with a temperature of T. In that case thermography will tell us that the comparison between 1) and 2) is between G and T. It applies when P is classified as sente or ambiguous.
My method is more powerful than thermography, as it suggests that the environment should start with T1, and it applies to the case with a losing sente, i. e., when G > R and P is classified as gote.
The sente option does not show up in the thermograph in that case.