Boundary plays - O Meien's method (Part 3)

[Magicwand]

Bill, so "count" is the better term than "expected value"?

expected value deals with random var.
count is not.
so you are wrong.

[Bill Spight]

Bill, so "count" is the better term than "expected value"?

Count is more precise.

[Li Kao]

Bill, so "count" is the better term than "expected value"?

expected value deals with random var.
count is not.
so you are wrong.

IMO expected value can be used on local situations because you need to average over all possible other board situations, which makes it basically a random variable. But I wouldn't talk about expectancy values in whole board situations.

Of course a deterministic result is the degenerate special case of a random variable

[RobertJasiek]

Bill, so "count" is the better term than "expected value"?

expected value deals with random var.
count is not.
so you are wrong.

Since we do not have random variables, Bill and I are right:)

[RobertJasiek]

Li Kao, looking at B or W moving first locally with 50% probability is a maybe convenient but also naive model. In reality we have deterministic game trees.

[Bill Spight]

Actually, in this example, I would say the OM count is wrong, in the sense that it does not accurately evaluate the game position. The OM count should at least include the availability of the W tesuji, weighted by 50% if playing there is double gote. However if there is one move left worth much more than anything else, the OM count will not give the correct evaluation of the position. A better count in this particular situation would be obtained by weighting the W tesuji at 100%, since it is definitely going to be the next move. The standard OM count then might or might not be applicable to the rest of the game.

One problem with the probabilistic semantics, which I am afraid that I have endorsed in earlier notes, is that it only works if you ignore who has the move. Now, traditional go evaluation, as well as CGT evaluation, ignores who has the move, so it is not much of a problem.

However, there is another approach to game tree evaluation, where who has the move is important. In that approach, the value of a tree node (whole board position) is the same as the value of the position after the optimal play by the player who has the move. Both approaches are valid, but they are inconsistent with each other.

[mitsun]

Looking at B or W moving first locally with 50% probability is a maybe convenient but also naive model. In reality we have deterministic game trees.

Good point, and game trees are a nice way of visualizing the OM method. The OM method in fact consists of writing down the game tree for a given position, down to leaves which have an unambiguous final count, including only branches where the best move is played by either side, then summing up all the final counts of each leaf, weighted by a factor of 1/2 for each branch along the way.

[mitsun]

However, there is another approach to game tree evaluation, where who has the move is important. In that approach, the value of a tree node (whole board position) is the same as the value of the position after the optimal play by the player who has the move.

But to determine the value of that position, you again must find the optimal next play and move another branch down the tree. If you carry out this recursion to its logical limit, you have to play out the entire game optimally for both sides. Of course this does give the correct solution, but it does not help a real go player evaluate the position or determine his next move.

The OM method so far has taught us how to count the value of a position, independent of who moves next. The additional ingredient not discussed so far is the value of the next move in each position. With this additional piece of information, a go player can determine his next play, without reading the entire game tree.

[mitsun]

It strikes me that the English language already has a yose proverb:
"A bird in the hand is worth two in the bush"

[daal]

It strikes me that the English language already has a yose proverb:
"A bird in the hand is worth two in the bush"

I think it's more like solomonic justice: we have one baby, so each of you gets half (not).

[flOvermind]

... But it is also correct when there are no other same-sized moves available. If you have only a single copy of the position, and everything else is already settled, then black too has exactly 1 point. And there is still a move left that gains 1 point. Whoever makes that move will change the score by 1 point. But that hasn't happened yet.

This is carrying the OM method too far. The method gives the expected value of a position, under the assumption that either side has an equal probability of playing there first, which in turn more-or-less assumes there are lots of similar sized moves left to play. If there is only one move left (or if one move is much larger than any other remaining moves), then there is no uncertainty about who will get to play there first, and the OM method is not applicable. The expected value of a position is most definitely not the same as the value of the next move in that position. But I suspect that is the topic for an upcoming chapter.

I never said that the count of the position has anything to do with the size of future moves. It's just a coincidence that in my example both the count and the move size are 1

... This method always determines what the score is right now. It doesn't necessarily tell you who's winning, for that you have to consider the remaining moves, too. A trivial example: Let's say the count on the board is B+5. But white plays a tesuji, killing a 20 stone black group. Black resigns. Was the count of B+5 wrong before? No, the count was correct, but there was still a 20-point move left for white, with white to move.

Actually, in this example, I would say the OM count is wrong, in the sense that it does not accurately evaluate the game position. The OM count should at least include the availability of the W tesuji, weighted by 50% if playing there is double gote. However if there is one move left worth much more than anything else, the OM count will not give the correct evaluation of the position. A better count in this particular situation would be obtained by weighting the W tesuji at 100%, since it is definitely going to be the next move. The standard OM count then might or might not be applicable to the rest of the game.

Actually, OM's method (when applied correctly) will factor in the tesuji. That's why a move killing a 20 stone group is worth around 20 points (+- a few for territory), not 40

And yes, as I have already said, OM's count shows you the count right now. This is different from the expected score of the game. But for getting the expected score of the game, you would have to finish the game (i.e. do a tree search), and that's impractical.

This was my view before studying CGT. But the count is not arbitrary. If two of some position has a definite value of 3 points, it makes no sense to say that one of them is worth 2 points but the other one is worth only 1 point.

Ok, it's not entirely arbitrary. Whatever you choose, you have to be self-consistent. So if you arbitrarily choose another count, you have also have to adjust the count of all other positions and the value of moves, so in the end everything is consistent again. But that will lead to some very weird things, like I tried to show with my example where suddenly a move that clearly is worth something seems to gain no points at all

[Numsgil]

IRT the 50/50 "probability" thing, maybe we can think of it instead as the ratio of what the move gains us this "round" vs. all the points that were gained this "round" (where a "round" is the points I gained during my sente plus the points my opponent gained during his sente):

(value of my move)
-------------------------------------------------------------------
(value of opponent's next/prev sente + value of my move)

In the limit, on a board mid-endgame, we would expect this to be very close to 0.5, because there are miai endgame plays available. If there aren't any plays left, value-of-opponent's-next-sente is 0 and the ratio reaches 1.0, which agrees with the saying that tedomari is worth double. Likewise if our opponent just responds to our move and we retain sente, the ratio is 1.0, which agrees with the idea that sente is worth double. If the opponent has larger moves available, the ratio drops below 0.5, indicating either that we made a mistake or that there was some sort of tedomari play involved.

Note also that there are two ratios for any given move: the ratio considering your opponents prior move and the ratio considering your opponents future move. I'm not exactly sure how you resolve them...

I'm not sure if this would help in determining last-play tedomari problems. And it's unwieldy for actual board play. But for those that don't like probabilistic language for a deterministic game tree, I like this way of looking at it better.

[Time]

Li Kao, looking at B or W moving first locally with 50% probability is a maybe convenient but also naive model. In reality we have deterministic game trees.

Obviously if you can evaluate the whole board position (e.g., the endgame problems in Train Like a Pro), then you would just do that.

When the situation is too difficult to evaluate correctly, using EV in the way described seems pretty useful to me. You guys can get out of here with your "well there aren't really any random variables so you can't use EV." There aren't any random variables when I shuffle a deck of cards and then draw a card, but I can't reasonably evaluate the situation, so we treat it as random.

[cyclops]

Li Kao, looking at B or W moving first locally with 50% probability is a maybe convenient but also naive model. In reality we have deterministic game trees.

Obviously if you can evaluate the whole board position (e.g., the endgame problems in Train Like a Pro), then you would just do that.

When the situation is too difficult to evaluate correctly, using EV in the way described seems pretty useful to me. You guys can get out of here with your "well there aren't really any random variables so you can't use EV." There aren't any random variables when I shuffle a deck of cards and then draw a card, but I can't reasonably evaluate the situation, so we treat it as random.

Picking a random card out of a deck is a stochastic proces that gives rise to a stochastic variable: the value of that card. If previously you attached numerical values to the cards you can calculate the EV. In a game of go there are no stochastic variables. That does not mean that it cannot be useful to treat some variable as stochastic in some theoretical model about go.