Sente, gote and endgame plays

[RobertJasiek]

I thought to have understood tally. Now that I think about it more I am not sure any more.

We have Black's and White's sequences. For Black's sequence, we use Black's perspective of counting plays: b := the number of its black plays minus the number of its white plays.

For White's sequence, we can use either Black's or White's perspective. If we use Black's perspective, we define w := the number of its black plays minus the number of its white plays and might define the tally X := b - w. If we use White's perspective, we define w := the number of its white plays minus the number of its black plays and might define the tally X := b + w.

However, b and/or w can be negative. This raises the following questions:

1) If we use Black's perspective for White's sequence, is X := b - w well-defined or should it be X := |b - w|? Why is which well-defined?

2) If we use White's perspective for White's sequence, is X := b + w well-defined or should it be X := |b + w|? Why is which well-defined?

3) If tally is well-defined without taking the absolute value, we can have a negative tally. What does this mean for counts and move values? Are they negated indeed and, in particular, can we get a negative move value in this manner? How to interpret both?

Example 1:

Black's sequence has 1 black play so b = 1. White's sequence has 1 white play followed by 4 black plays. Using Black's perspective for White's sequence, we have w = 4 - 1 = 3 and X = b - w = 1 - 3 = -2. Using White's perspective for White's sequence, we have w = 1 - 4 = -3 and X = b + w = 1 + (-3) = -2. Taking the absolute value negates this to become X = 2.

Example 2:

Black's sequence has 1 black play so b = 1. White's sequence has 3 white plays. Using Black's perspective for White's sequence, we have w = 0 - 3 = -3 and X = b - w = 1 - (-3) = 4. Using White's perspective for White's sequence, we have w = 3 - 0 = 3 and X = b + w = 1 + 3 = 4.

[Bill Spight]

I thought to have understood tally. Now that I think about it more I am not sure any more.

We have Black's and White's sequences. For Black's sequence, we use Black's perspective of counting plays: b := the number of its black plays minus the number of its white plays.

For White's sequence, we can use either Black's or White's perspective. If we use Black's perspective, we define w := the number of its black plays minus the number of its white plays and might define the tally X := b - w. If we use White's perspective, we define w := the number of its white plays minus the number of its black plays and might define the tally X := b + w.

However, b and/or w can be negative. This raises the following questions:

1) If we use Black's perspective for White's sequence, is X := b - w well-defined or should it be X := |b - w|? Why is which well-defined?

2) If we use White's perspective for White's sequence, is X := b + w well-defined or should it be X := |b + w|? Why is which well-defined?

3) If tally is well-defined without taking the absolute value, we can have a negative tally. What does this mean for counts and move values? Are they negated indeed and, in particular, can we get a negative move value in this manner? How to interpret both?

Example 1:

Black's sequence has 1 black play so b = 1. White's sequence has 1 white play followed by 4 black plays. Using Black's perspective for White's sequence, we have w = 4 - 1 = 3 and X = b - w = 1 - 3 = -2. Using White's perspective for White's sequence, we have w = 1 - 4 = -3 and X = b + w = 1 + (-3) = -2. Taking the absolute value negates this to become X = 2.

Example 2:

Black's sequence has 1 black play so b = 1. White's sequence has 3 white plays. Using Black's perspective for White's sequence, we have w = 0 - 3 = -3 and X = b - w = 1 - (-3) = 4. Using White's perspective for White's sequence, we have w = 3 - 0 = 3 and X = b + w = 1 + 3 = 4.

A negative tally indicates a ko or superko. Which means that the ko threat situation matters.

The position can be a "number". (I put number in quotes because kos are not combinatorial games. But the ko threat situation is such that the position acts like a number.) For instance, suppose that b = 0 and w = 1, so that the tally is -1. Black to play is sente, and White to play lets Black take and win a ko. Let's also suppose that the result when Black plays first is 1 and the result when White plays first is 2. Then White should not play first, but let Black do so, for a score of 1. We evaluate the position as 1.

A negative tally can also indicate a mistake. Suppose that Black's play is actually gote, as is White's. The final scores are the same. Then we may find that at a high ambient temperature White should play to the Black ko, but at a low temperature White should win the ko instead, so that the final result is 1 instead of 2.

[RobertJasiek]

So you are saying that we do not take the absolute so that we can appreciate negative tallies?

In your ko example with the black sente follower's count S = 1, white follower's count W = 2 and tally X = -1 , the initial count is given due to White's decision to avoid the mistake of local play to the larger, less favourable white follower's count W = 2 and choose the more favourable, smaller black sente follower's count S = 1. Hence the correct calculation of the initial count C is to inherit it from the sente follower's count S, that is, C = S = 1. If we calculate the move value as (S - W) / X = (1 - 2) / (-1) = 1 by dividing by the tally, it does, however, not express White's gain, as it would for a reverse sente. The move value is meaningless here (or expresses the contrary of the usual because of White's mistake)! Instead, we better calculate White's gain (a loss, to be honest) as G(W) := C - W = 1 - 2 = -1 (a negative gain so it is a loss, as promised). Have I understood this right?

Depending on study purposes, a negative tally need not indicate a ko or mistake but can also indicate a certain non-ko local playing which might be correct in a global context of ordinary tenukis.

[Bill Spight]

Depending on study purposes, a negative tally need not indicate a ko or mistake but can also indicate a certain non-ko local playing which might be correct in a global context of ordinary tenukis.

In that case you need to include the other positions as part of the analysis.

[RobertJasiek]

Bill, you have shown the following two multiples:





You rely on the CGT definitions of local temperatures and mean. Can you explain the relation between the definitions and their application, please?

You show application for a local endgame with one player's simple follow-up and without kos. For such, application is straightforward. How about local endgames with more complicated follow-ups and without kos? Is the method exactly the same, except for needing more multiples?

Can you write down your general method as a procedure applicable to all examples of a class?

Do we know a priori how many multiples we need at least? How do we find out the minimal necessary number of multiples? (Why) is it always 4 for a local endgame with one player's simple follow-up and without ko?

How would you define as a procedure the gote strategy and the sente strategy a) for a local endgame with one player's simple follow-up and without ko and b) for an arbitrary local endgame possibly with follow-ups and without ko?

For a local endgame with both players' follow-ups, how to determine for whom to test a sente strategy? Must a procedure be more complicated by possibly having to test either player's gote strategy and either player's sente strategy?

Suppose we test one player's gote versus sente strategies with the resulting counts G versus S. Does the more favourable count for the starting player determine the correct strategy? Is the count of one local endgame the resulting count of the more favourable strategy divided by the number of multiples? Why is this so explained by the CGT definitions? If we determine a local sente, is the reverse sente gain S - G (with counts calculated in favour of the starting player)? Why is this so explained by the CGT definitions?

Suppose we have a local endgame with follow-ups and without ko and have to test both players' gote versus sente strategies with the resulting counts Gb, Sb, Gw and Sw. How to determine the correct strategy, calculate the correct count, explain by the CGT definitions, possibly calculate a reverse sente gain and explain by CGT definitions?

For all cases, how to calculate the move value and explain it by CGT definitions?

In summary, your examples look convincing but how and why can they be generalised as methods?

[Bill Spight]

Many questions, which I will gloss over, if you don't mind too much.

In college, when I was trying to come up with a convincing argument for the equivalent of the Monty Hall problem, I realized that, instead of asking what the correct probabilities were, I could ask what the winning strategy was. People could argue about probabilities, but they could not argue about the results of the different strategies. OC, the winning strategy was consistent with a certain probability. Much later I used this strategy with various puzzles, and found, for instance, that the two envelope puzzle cannot be turned into a finite game.

With the method of multiples I had simply accepted the fact the mean value for sente games is a limit as the number of games goes to infinity. That's fine in terms of mathematics, but unsatisfactory for those whose eyes glaze over. Also, correct play in multiples is not always obvious, so proving that the mean value is correct is also not obvious.

However, this past winter I realized that you can compare strategies without worrying whether the result is correct. And since there are only two strategies with these simple positions, you can find out which (if either) is better and classify the position without calculating mean values or temperatures or doing any math except counting the score. Once you know the correct classification you can figure out the mean values and temperatures.

I am pretty sure that the correct number of multiples (sans ko, OC) is 2^d, where d is the depth of the tree. However, increasing depth increases the number of possible strategies, and increasing the number of options at the same depth also increases the number of possible strategies. Things can quickly become unwieldy.

Right now, I think that the main value of this method is to get the basic concepts across. Currently a lot of players resist the idea of local sente. The fact that the sente strategy wins in certain positions shows that local sente exist. A lot of players also resist the idea of sente and gote positions. I think that this illustrates that we can classify positions in that manner.

[RobertJasiek]

Why do you think that many would be sceptical about local gote versus sente positions? There is no fundamental difference between a local position (not) having a sente move / sequence and the local position (not) being one having a sente move / sequence.

[EdLee]

Hi Bill,

In college, when I was trying to come up with a convincing argument for the equivalent of the Monty Hall problem, I realized that, instead of asking what the correct probabilities were, I could ask what the winning strategy was. People could argue about probabilities, but they could not argue about the results of the different strategies. OC, the winning strategy was consistent with a certain probability. Much later I used this strategy with various puzzles, and found, for instance, that the two envelope puzzle cannot be turned into a finite game.

The MH puzzle is interesting to me (and to many others, apparently, who wrote passionately to what was it, Scientific American?) -- could you explain a bit more about "why people couldn't argue over the results of different strategies", using MH as an example. (For me, a nice way to explain MH, even to people with very little background, is the variation to change it to from 3 to a million doors, and then to open (1M-2) of them; I found this variation much easier to digest intuitively.)

I also don't know what's the 2-envelope puzzle.

[Bill Spight]

Hi, Ed.

The Monty Hall problem is usually presented in an imprecise way. In the late 70s or early 80s Monty, who had retired, was interviewed and asked about the Monty Hall problem. He had not heard of it. But his reply was that he could always manipulate the person to make the wrong choice. Marilyn vos Savant presented the problem in one of her columns without restricting Monty's actions. You have to assume that Monty always gives you the choice to switch doors. Anyway, even making that assumption she got letters from math professors telling her that her answer was wrong, that switching doors was a 50-50 proposition.

For those who have not heard of the problem, Monty Hall ended his show, Let's Make a Deal by offering a contestant the choice of three doors. Behind one door was a fabulous prize, behind each of the other two doors was a "goat", a prize you might not want to take. Typically after the contestant chose a door, Monty would open one of the other doors to reveal a goat, and then would ask the person if they wanted to switch doors and take what is behind the other unopened door. Whatever the person said, Monty would usually try to get them to change their mind, offering cash for them to do so. These extra offers are not part of the Monty Hall Problem. The question is, should you switch doors or not?

The usual answer given is that you should switch doors, because the probability is ⅔ that the fabulous prize is behind the other door. This is where the arguments arise. Many people will insist that it is a 50-50 choice, that the probability is ½.

What I realized in college was that arguing probabilities was usually futile, but you could easily show that the switching strategy paid off. You do not have to appeal to Bayesian probability or the Principle of Restricted Choice or anything except the idea that you have chosen the prize ⅓ of the time. Monty can always show you a goat behind one of the other doors, by assumption he must always do so. In two of the three cases switching gets the prize, in one it loses it, so switching is the superior strategy. Once you have found the superior strategy you can work out the probabilities.

In the Two Envelope Problem you are informed, correctly, that each envelope contains a valid check that you can cash for a certain amount of money. One envelope contains a check for twice as much as the other one. You get to choose an envelope, open it, and then decide whether to take the other envelope. You choose an envelope and inside you find a check for $100. Should you take the other envelope? Logically, it seems that it does not matter. Half the time you will get more money, half the time you will get less. How much money it is should be irrelevant.

But there is an argument for switching. Half the time the other envelope will contain $200, half the time it will contain $50. So your expected payoff for switching is $125. Should you switch?

As against that there is the argument that if you compare the strategy of always keeping the envelope, no matter what is in it, with the strategy of always switching, their expected payoffs are the same. Should you switch?

Making this a finite game can spoil it. For instance, if the maximum payoff is $100 you should obviously not switch, if the minimum payoff is $100 you should switch. For the infinite game there are any number of good strategies that are better than always switching and never switching, but none known to be best. The first one I came up with was to switch if the amount was less than $1,000, otherwise not to switch.

[Bill Spight]

Why do you think that many would be sceptical about local gote versus sente positions?

Experience.

Did you notice in our recent discussion how when I talked about the value of a position some people interpreted me as talking about the value of a play? Years ago on SL even good players with mathematical backgrounds insisted that sente and gote only applied to plays, not positions.

[EdLee]

Hi Bill,

Thanks for the write-up.

In two of the three cases switching gets the prize, in one it loses it, so switching is the superior strategy. Once you have found the superior strategy you can work out the probabilities.

But this seems a brute forcing approach, no ?

As usual, it turns into work

[Bill Spight]

Hi Bill,

Thanks for the write-up.

In two of the three cases switching gets the prize, in one it loses it, so switching is the superior strategy. Once you have found the superior strategy you can work out the probabilities.

But this seems a brute forcing approach, no ?

As usual, it turns into work

Sure, it's brute force. You are not relying upon any concepts, just strategies. What works? Once a person knows what works, they are open to learning the concept.

Edit: I realized a classroom demo for the two envelopes problem. Ask the students how much is a fair price for the privilege of switching. Then get two students who are willing to pay to switch, give each of them an envelope and tell them before they reveal the amount in their envelopes that if they are still willing to pay to switch you will give them the amount in the other student's envelope, minus their payment, even if the other student does not switch. It should be obvious to the whole class that the overall payoff is more if they do not switch.

[RobertJasiek]

Many questions, which I will gloss over, if you don't mind too much.

Ok - but there is a reason why I pose them so carefully and detailed. In the current state, we cannot teach it as a general method. Somebody (i.e., you or I) would need to do the mathematical research to possibly create it as a general method. If I did it, I would need about 4 to 6 months (which I don't have) for formulating and proving the theorems. Therefore, I have hoped that you might already have completed the research. You haven't. No problem, except see above.

[Bill Spight]

Bill, you have shown the following two multiples:





You rely on the CGT definitions of local temperatures and mean. Can you explain the relation between the definitions and their application, please?

I'm not sure what you are asking.

You show application for a local endgame with one player's simple follow-up and without kos. For such, application is straightforward. How about local endgames with more complicated follow-ups and without kos? Is the method exactly the same, except for needing more multiples?

You need a strategy at each decision point. So you end up with several comparisons.

Can you write down your general method as a procedure applicable to all examples of a class?

Not sure what you are asking. It's a general method for comparing strategies and classifying positions, but more work needs to be done to verify means and temperatures. The method of multiples is inefficient.

Do we know a priori how many multiples we need at least? How do we find out the minimal necessary number of multiples? (Why) is it always 4 for a local endgame with one player's simple follow-up and without ko?

The minimum is 2^d, where d is the depth of the tree.

How would you define as a procedure the gote strategy and the sente strategy a) for a local endgame with one player's simple follow-up and without ko and b) for an arbitrary local endgame possibly with follow-ups and without ko?

Each decision point requires a different strategy, so the eventual strategy must say what to do at each node. Finding the best strategy for an arbitrary tree can take a lot of work.

For a local endgame with both players' follow-ups, how to determine for whom to test a sente strategy? Must a procedure be more complicated by possibly having to test either player's gote strategy and either player's sente strategy?

Yes. In these cases we could eliminate a possible sente strategy for Black, but, OC, that is not always possible. Also, the final node is always a gote, so we don't have to test it for sente.

Suppose we test one player's gote versus sente strategies with the resulting counts G versus S. Does the more favourable count for the starting player determine the correct strategy?

Each player must have made the same number of plays with each strategy, so that when you double the number of multiples the result will be the same ad infinitum. Also, you must have compared every possible strategy to find out which is correct. But you can compare strategies one by one.

Is the count of one local endgame the resulting count of the more favourable strategy divided by the number of multiples?

If you have found the correct strategy, yes. But in the arbitrary case the correct strategy can be arbitrarily complex.

Why is this so explained by the CGT definitions? If we determine a local sente, is the reverse sente gain S - G (with counts calculated in favour of the starting player)? Why is this so explained by the CGT definitions?

Thermography determines mean values and temperatures. These mean values must be the same (for non-ko games) as those found by the method of multiples, by the mean value theorem.

Suppose we have a local endgame with follow-ups and without ko and have to test both players' gote versus sente strategies with the resulting counts Gb, Sb, Gw and Sw. How to determine the correct strategy, calculate the correct count, explain by the CGT definitions, possibly calculate a reverse sente gain and explain by CGT definitions?

For all cases, how to calculate the move value and explain it by CGT definitions?

In summary, your examples look convincing but how and why can they be generalised as methods?

I think your questions here can be answered by the fact that the method of multiples finds mean values, and temperatures can be derived from them. As for why generalize, why indeed? Thermography is more efficient.

[RobertJasiek]

You rely on the CGT definitions of local temperatures and mean. Can you explain the relation between the definitions and their application, please?

I'm not sure what you are asking.

Your answer "the method of multiples finds mean values, and temperatures can be derived from them" goes in the intended direction of my question.

You need a strategy at each decision point. So you end up with several comparisons.
[...] The method of multiples is inefficient. [...]
Each decision point requires a different strategy, so the eventual strategy must say what to do at each node. Finding the best strategy for an arbitrary tree can take a lot of work.
[...] you must have compared every possible strategy to find out which is correct [...]
in the arbitrary case the correct strategy can be arbitrarily complex [...]
why generalize, why indeed? Thermography is more efficient.

OIC. Looking at your examples, I thought that it was a 1 ply decision-making at the top level. You explain that it is iterative strategic decision-making bottom-up. If not thermography, we may use methods for value comparisons to distinguish gote from sente for short or long sequences worth playing successively. I also suspect that this would be more efficient than methods involving multiples, at least when we would need 8+ multiples.

Can you write down your general method as a procedure applicable to all examples of a class?

Not sure what you are asking. It's a general method for comparing strategies and classifying positions, but more work needs to be done to verify means and temperatures.

If it is a general method, state its procedure explicitly - otherwise, it is not general:)

For the class of all examples with one player's simple follow-up, stating the procedure is straightforward. Can you also state it so that it applies to iterative follow-ups? (If you have the time for doing so and think it is worthwhile.)