# Reverse sente is worth half of the second biggest gote

"Reverse Sente Is Worth Half Of The Second Biggest Gote" is a reformulation of the well known "__Reverse sente is worth double__?" proverb. Both of them are quite inaccurate, but in different ways. In some cases their application, somewhat surprisingly, produces identical results.

## Origins of the old proverb

The old proverb is based on the idea that taking gote to prevent a sente "saves" one stone compared to taking gote that only prevents a gote move. This is because if the situations were reversed so that the opponent plays first, then only in the latter case would the opponent need to play one more stone than the player himself.

Therefore, preventing sente "gains a stone" by playing a stone, and is "worth double".

## Origins of the new proverb

On the page "when to play a reverse sente", some clever endgame simplifications are used to arrive at the conclusion that the value of a reverse sente is the point value of the move itself, plus the value of sente after the biggest gote is taken.

Using even more simplifications, one can arrive at a very simple (and accordingly very unaccurate) estimate for the value of said sente.

## Application

When the biggest gote move is 6 points, and there is a reverse sente of 3 points available, the old proverb says that the moves are approximately equal.

Because the new proverb is using BOSE simplification, one has to assume (because of simplification 2) that the second biggest gote move is very likely also 6 points. In that case, the value of the reverse sente move would be 3 points + 3 points for being reverse sente. Therefore, the border case would behave exactly like the old proverb. Also, if one were to change either the size of the reverse sente or the size of the big gote, both the proverbs behave exactly alike.

So, if you only know the sizes of the reverse sente move and the biggest gote move, the two proverbs always agree as to which one you should play.

However, if one happened to know the size of the second biggest gote play also, then the results of the proverbs might differ even dramatically. In this case, the new proverb would get a reasonable estimate (by luck, the accurate result) out of the situation where those 3 point reverse sente and the 6 point gote plays were the only ones remaining.

## Comments

Bill: I think I know what is intended. To simplify, suppose that we have a reverse sente in an environment of gote, and that we know how much the reverse sente, the largest gote, and the second largest gote gain. Since we do not know the sizes of smaller plays, we estimate the residual gain from the second largest gote as one half its actual gain. We also suppose that if we do not take the reverse sente now, the opponent will take the sente.

Bill: That gives us two lines of play:

Bill: First, let us take the reverse sente, gaining its value, R, then the opponent takes the largest gote, gaining its value, G, and then we take the second largest gote, gaining approximately half its value, H/2. Result: R - G + H/2.

Bill: Second let us take the largest gote, then the opponent takes the sente, and then the opponent takes takes the second largest gote. Result: G - H/2

Bill: Comparing these results, we should take the reverse sente when R + H > 2*G and take the gote when R + H < 2*G.

Bill: With deiri counting the size of the largest gote is 2*G and the size of the second largest gote is 2*H. In the example given, R = 3 and 2*G = 6. Since G >= H, we should normally play the gote instead of the reverse sente. :)

tapir: Something very common in our proverb section, but I do not like calling any phrase somebody came up with a proverb.

Bass, 2011-03-16: @Bill, Well spoken, you make it sound very simple in comparison to my ramblings :-)

Bass: I used plain point swing sizes for the moves the whole time, so my math did look a bit different. (I'll use lower case letters to denote swing values). So, taking the reverse sente, we would get the result "r - g + h/2", taking the gote would result in "g - r - h/2", and the conclusion comes out as "we should take the reverse sente if r + h/2 > g".

Bass: @tapir, you are correct, of course. It seems that my use of the word is lacking in many aspects: "proverb, noun: a condensed but memorable saying embodying some important fact of experience that is taken as true by many people". While certainly condensed and embodying an important fact, the memorability could use some work, and certainly there are not (yet) many people taking it as true. Hopefully that may change. :-) In the meantime, please feel free to improve the wording as you see fit. Maybe "a rule of thumb" would work better?

Andre Engels: In fact, looking at the value of the gote plays one by one, one gets a sequence of 'rules'. Let *g1*, *g2*, *g3*... be the )size of) the various gote plays, ordered by decreasing size (so *g1* is the largest etcetera), and *s* be the size of the reverse sente play. In that case, one should play the sente move if:

*s > g1-g2+g3-g4+g5-g6+...*

Under reasonable assumptions (we assume either that the size of moves decreases linearly, or that for every point value equal to or lower than *g1* there is a 50% chance that there is an odd number of moves that is that size or larger), *g1-g2+g3-g4+g5... ~= 0.5*g1* (where *~=* is used for lack of the official 'is approximately' sign).

This leads to the old rule:

*s > 0.5*g1*

However, if we also have the value of *g2*, we could apply the same approximation to *g2* instead of *g1*, and get

*s > g1 - 0.5*g2*

instead, which is equivalent to the rule proposed here.

Of course, we can also go on further, and then get a sequence of rules:

s > 0.5*g1s > g1 - 0.5*g2s > g1 - g2 + 0.5*g3s > g1 - g2 + g3 - 0.5*g4s > g1 - g2 + g3 - g4 + 0.5*g5...

If there are only small moves left, and more than a few of them, either the old or the new rule will generally suffice. If there are just a few moves left, you can calculate the exact rule. If there are a few large moves, take the *n+1*th move, where *n* is the number of large moves. If there are too many, use an earlier rule, adjusting the right-hand term by *+0.5*G* if there is an odd number of large moves, *-0.5*G* if there is an even number of them, with *G* the difference in size between the smallest large move and the largest small move. - Andre Engels

Bass: Ah, very clever! So in an actual game one can count the actual values of the biggest moves, preferably until a tedomari if an obvious one exists, and then approximate the rest by the biggest remaining move. And in addition, one can use the same approximation to the part **before** the tedomari by treating the tedomari move as size zero.

Dieter: adding a diagram to try and understand ...

A is a 6 point gote, miai value 3 B is a 3 point reverse sente, miai value 3 C is a 1 point reverse sente, miai value 1

According to the heuristic posted here, the deiri value of B is 3 plus 0,5