55% winrate means A wins 55 games out of 100, A wins 55 when B wins 45. So, A wins 55/45 times more games than B . It's true, because (55/45)*45=55.
If B wins 55% over C, B wins 55/45 times more games than C.

A wins 55/45 times more games than B, who wins 55/45 times more games than C, so A wins (55/45)*(55/45) times more games than C.
etc.
A10 wins (55/45)^10 times more games than A1
A10 wins 7.44 times more games than A1
When A1 wins 1 game, A10 wins 7.44 games, so A10 wins 7.44 out of 8.44, that's 88%

For example, with an obvious case, if An wins 50% over An-1, after 10 networks, it leads to 1^10=1, and 1 out of (1+1) is still 50%.

With 50.1%, it leads to (50.1/49.9)^10=1.04, and 1.04/2.04 =~ 51 % which looks reasonable.

With 45% , after 10 networks, we would get ~12%

If we could have 10 consecutive networks with 60% winrate over the preceding one, we'd have 98.3% winrate for A10 over A1.


I hope it's understandable, I don't really speak english (as you've probably noticed )

Interesting discussion on Github of experimental version of LZ that uses alternative logic to select the best play with the same nets. See https://github.com/leela-zero/leela-zero/issues/2282 for the details and links to code or compiled windows downloads. The experimental version is showing ~60% winning percentage in matches with various visit levels versus normal LZ.

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

It's understandable, but there's no reasons for this to be true.

It's not because you have a 1:a ratio against player A and that player A has a 1:b ratio against player B than you must have a 1:(a*b) ratio against B



It's already not true for this simple following game :

Player A roll two 8 sided dices, player B roll two 6 sided dices, and player C roll two 4 sided dices. The one with the biggest sum win, and if there's equality, the one with the smallest dices win.

In this simple game, A has 63.93% chance to win against B (1581/2304, or a 1:1.773 ratio), B has 69.10% chance to win against C (398/576 or a 1:2.236 ratio), and A has 82.42% chance to win against C (844/1024 or a 1:4.789 ratio).

But what you propose would give A 79.85% chance to win against C (1:3.964).

As I said, a little controversy...
I'm not home now, but I'm looking forward to trying your dices

Yes, it is understandable and clear. However, there is an underlying assumption that the difference between the abilities of A, B, and C to win games is reducible to a single number. (There is also the assumption of perfect accuracy of the win rate estimates, i.e., no luck, which has already been alluded to.) But as we know go requires a number of different skills, which means that skill at go may not be reduced to a single number. And that means that transitivity does not hold. Player A may beat player B more than half the time, player B may beat player C more than half the time, and player C may beat player A more than half the time.

Now, transitivity holds closely enough in go that we can have different ranks, each of which covers a range of ratings, and make pretty good predictions of the handicap between players of different ranks which will make the win rates around 50%. But, OC, for specific individual pairings the recommended handicap may not do that. One thing that makes the ranking system robust is that each player plays against a variety of different players with different levels of ability at different skills. Self play does not do that, and so, IMO, does not produce robust results.

To give a possibly related example of how multidimensionality can reduce the degree of progress, let's suppose that we are measuring progress in two independent dimensions. Suppose that B makes one unit of progress by comparison with A, and C makes the same unit of progress by comparison with B, but in the orthogonal direction to that of the progress between A and B. Then how much progress does C make with regard to A? Not 2 units, but √2 units.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

8 sided dices are common in tabletop gaming :



4, 6, 8, 10, 12 and 20 sided dices are usual



But there exist more exotic dices

@Tryss
Your dice are beautiful. I had some such.
But in your example, A-B play a certain game, B-C play another game, because the dice are different, and A-C a third different game. In this case, I'm not surprised that winrates aren't transitive.

Yes, it's true, fortunately !
It's true too, unfortunately.

No, it's the same game : roll your dices, the one with the better score win

Player A is just a stronger player than B or C

I suppose that the faces of each die are numbered consecutively from 1 to the number of faces. Let's suppose that each player rolls only one die. Then A has an 9/16 chance (56.25%) to beat B, with odds of 9:7, and B has a 7/12 chance (58.33%) to beat C, with odds of 7:5. Multiplying the odds gives A odds of 9:5 to beat C, or 9/14 of the time (64.29%). But A beats C 11/16 of the time (81.25%), with odds of 11:5.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Out of curiosity I tested my method: 55% winrate means ~0.178 sd distribution distance, and 1.78 sd gives 89% - no surprise here. Transitivity is OC debatable but I doubt that would be the larger effect in this case.

Just retesting those 55% promotions with more games may reduce most to lower winrates. This is, afterall, how "55% for 400 games" were chosen: a statistical mass that makes it hard to pass on luck ALONE (in a few dozen tries), so new nets are at least slightly better - but nothing more. And those 400 samples are not even really independent: the first few moves, joseki choices are often identical, which further reduces the statistical weight.

There is a simple scientific point here, as well. Suppose that B beats A more often than not, and C beats B more often than not, and D beats C more often than not, etc., and we want to know how much more often, say, K beats A, our preferable method is not to try to figure it out based upon our estimates of how often B beats A and C beats B, etc., but to have A and K play against each other. Unless it is prohibitively costly or there are other reasons for not doing so.

One possible reason for not doing so is that both J and K beat A almost 100% of the time, so the answer is uninteresting. But maybe how often K beats D would be interesting. We really should not be arguing about the pluses and minuses of an inferior method.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

40 game match LZ0.16_#213 v. LZ0.16_ELFv2
at time parity (--visits=1601 for #213, --visits=3201 for ELFv2)
twogtp 1.5.0, 3 duplicate games, 37 games used.

Result : ELFv2 wins 19-18

The stats : The games (#213 is B in the even numbered games.games n° 12, 22 and 24 are duplicates)Next time, I'll use -m 20 to avoid duplicates.

several matches 25x25, nets received by board_resize.py.txt vs
LZ 40x256 #205 by ChangeBoardSizeOfWeight.cpp, 10sec/move, cpuonly, gogui-twogtp:
(https://github.com/leela-zero/leela-zero/issues/2240)

LM 192x15 GX89 - LZ 40x256 #205 13:27
LZ 192x15 f438268e - LZ 40x256 #205 5:35
elf v2 256x20 - LZ 40x256 #205 12:28
converted minigo(25x25) 000990-cormorant works, did not test
and LM 192x15 GX89(by ChangeBoardSizeOfWeight.cpp) - LM 192x15 GX89(by board_resize.py.txt) 37:3 (White 20:0)

Time parity match.
LZ0.16 XXX and LZ0.16 Elfv2
2x1080ti, 60s per move.

5). #211

6). #213

7). #214
in progress...

50 game match at time parity#214 v. ELFv2
LZ0.16, twogtp 1.5.0
-v 1601 for #214 and -v 3201 for Elf, -m 20 for both.
no duplicate game, no error

ELFv2 wins 28-22 (56%)
The games : (#214 is B in the even numbered games):Command line and stats:

This is for selfplay I think, it may be too random for matches. If you just want to avoid duplicates you could look into --randomtemp (and/or check if there are no weird edge moves).

You're right, maybe it's too much random.
I've looked at the first 20 games, there is no obviously weird move that I can see. In one game, Elf is caught in a ladder before resigning .
Anyway, I'll try -m 20 --randomtemp=0.xxx

I've tried another 50 game match #214 v. ELF v2
Same parameters, but for -m 20 --randomtemp=0.3
Average game length and average times are almost the same as before, no duplicate.
The games look "normal", but in one case (THIS GAME, n°40) , it's #214 (B) which gets caught in a ladder, and the last W moves look weird, but maybe it's because the winrate was near 100% for W.


Command line and stats : The games (#214 is B in the even numbered games)

Maybe it has a preference for moves on the first line when the game is nearly over.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Full disaster:



I think "-v 1601" is too small for l0.