It is trivial that the order doesn't matter.
(in the human dice case, at least) Please don't
waste any time trying to convince me of that.

That is not the same as saying that
color doesn't matter. The point
is that, given the same average
expectancy, the match winning
percentage will change if players
alternate white and black and
white confers a non-zero advantage.

Now I don't know if anyone has
much data on what color does for
winning and drawing probabilities
for every rating difference and rating
strength, and I doubt it is significant,
but it *does* change the match
probability.



"Tyrone Slothrop" wrote in message
oups.com...
I am not assuming symmetry. Again, you suggested the simple two
game
case. So let's look at it getting rid of symmetry:

Let's try it with it skewed toward a player.

Take the simple case of a max of two games in a match. Let's say
that
Sam wins if he can score 1.0 and Bill wins if he scores 1.5. Just
as
in the real match, the second game isnt played if it isn't
necessary.

So what are the outcomes? (W- bill wins, T - tied, L - bill loses)

WW B(ill wins)
WT B
WL S(am wins)
TW B
TT S
TL S
L S -- short match

Say that T occurs 50% of the time. Let's say that with color not
counting, W is 35% and L is 15%. Clearly things are skewed toward
Bill
winning. But when we count color, we found that the first player
has
an advantage. When Bill goes first, he finds W is 40%, and L is
10%.
When he goes second, he is still favored, but less so. W is 30% and
L
is 20%. Assuming that the players play the same number of games
going
first or second, the average still works out to 35%/15%.

What are the odds?

Say Bill goes first in game one.

WW occurs .4*.2 = 8% (B)
WT occurs .4*.5 = 20% (B)
WL occurs .4*.3 = 12%
TW occurs .5*.2 = 10% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.3 = 15%
L occurs .1*1.0 = 10% -- short match
Total = 100%

The B-favorable events add up to: 8+20+10 = 38%

Say Bill goes second in game one.

WW occurs .2*.4 = 8% (B)
WT occurs .2*.5 = 10% (B)
WL occurs .2*.1 = 2%
TW occurs .5*.4 = 20% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.1 = 5%
L occurs .3*1.0 = 30% -- short match
Total = 100%

The B-favorable events add up to: 8+10+20 = 38%

This is exactly the same as the first case. Going first in game 1
did
not change Bill's odds even a tiny bit, even if the first player is
favored over the second and the overall odds are skewed toward Bill.
No
symmetry.


David Kane wrote:
I think you are assuming equal
players, which is also not the general case
and by symmetry, means that color won't
change the match probability.

Here is a more general example.

Sloan's sample odds if color doesn't count
(assuming P(draw)=0.25):
PW PL PD
0.22048 0.52952 0.25

Now imagine that White adds 0.1 to the
expectancy, doesn't change the draw probability,
(I'm not suggesting that this is realistic, but
it is consistent with the ELO winning expectancy)

Sloan's chances as white become:
PW PL PD
0.32048 0.42952 0.25

Sloan's chances as black become:
PW PL PD
0.12048 0.62952 0.25

Calculate all the possibilities and
you'll find that this changes
Sloan's match expectancy
(from 0.3183 to 0.3158 if I did
things correctly)


"Tyrone Slothrop" wrote in message
oups.com...
OK. Let's try it then.

Take the simple case of a max of two games in a match. Let's
say
that
Sam wins if he can score 1.0 and Bill wins if he scores 1.5.
Just
as
in the real match, the second game isnt played if it isn't
necessary.

So what are the outcomes? (W- bill wins, T - tied, L - bill
loses)

WW B(ill wins)
WT B
WL S(am wins)
TW B
TT S
TL S
L S -- short match

Say that T occurs 50% of the time. To make things extreme,
let's
say
that if a player goes first, they have a big advantage, W occurs
40%
and L occurs 10%. But if a player doesn't go first, they have a
big
disadvantage, W occurs 10% and L occurs 40%. Assuming that
players
play the same number of games going first or second, the average
still
works out to 25%/25%.

What are the odds?

Say Bill goes first in game one.

WW occurs .4*.1 = 4% (B)
WT occurs .4*.5 = 20% (B)
WL occurs .4*.4 = 16%
TW occurs .5*.1 = 5% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.4 = 20%
L occurs .1*1.0 = 10% -- short match
Total = 100%

The B-favorable events add up to: 4+20+5 = 29%

Say Bill goes second in game one.

WW occurs .1*.4 = 4% (B)
WT occurs .1*.5 = 5% (B)
WL occurs .1*.1 = 1%
TW occurs .5*.4 = 20% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.1 = 5%
L occurs .4*1.0 = 40% -- short match
Total = 100%

The B-favorable events add up to: 4+5+20 = 29%

This is exactly the same as the first case. Going first in game
1
did
not change Bill's odds even a tiny bit, even if the first player
is
outrageously favored over the second.


David Kane wrote:
Actually, color selection would matter, even if people were
dice.

You are correct that it doesn't matter who get white *first*,
but
it will change the overall match probability if you calculate
essentially 2 matches of 2 games, each with their own
distribution
that gives the correct average winning expectancy.

Try it!



"Tyrone Slothrop" wrote in
message
oups.com...
Of course you are right on your point on change of strategy
and
game
theory.

I was addressing Paul's assertion that mathematically,
something
changes if one player gets white on game 1. This doesn't
happen
if
we
assume that the expectation of the players for each new game
remains
the same no matter how many they win or lose previously.

wrote:
Color selection does make a difference, because humans are
not
dice.
The loser in game one has to change strategy.

How it makes a difference is a matter for game theory.

Hmmm, I thought the question was what is the probability of Sloan
getting a score of 2 or better. To do so he will need a performance of
2042 or better. My calculation comes up with the probability of that
happening. It strikes me as a elegant solution. But, given I'm not in
the field of probability, I can't say with confidence that what I have
done is correct. But, I have yet to see a convincing, IMHO, rebuttal
to it so far.

I did pass this along to an expert in the field. I will let you know
if I hear anything from him about this.

Best regards,

George John

You are making a distinction I don't understand. I suppose I just
don't understand what you are trying to say.

You agree that order doesn't matter. So the fact that in the real
match, Sam takes the first move as white in game #1 will not change the
expectancy in the slightest compared with Bill taking the first move as
white in game #1. It is understood that players alternate in color. In
the first case, Sam would play white, then black, then white, then
black. In the second case, it would be Bill doing this. I claimed both
of these cases have the same expectancy. This is the alternation of
color you are talking about, but it seems not.

Your distinction concerns "the match winning percentage will change if
players alternate white and black and white confers a non-zero
advantage." How does this differ from what is laid out above? How
does it differ from my last example where white confered a 10%
advantage?

Can you explain to me what you mean?

David Kane wrote:
It is trivial that the order doesn't matter.
(in the human dice case, at least) Please don't
waste any time trying to convince me of that.

That is not the same as saying that
color doesn't matter. The point
is that, given the same average
expectancy, the match winning
percentage will change if players
alternate white and black and
white confers a non-zero advantage.

Now I don't know if anyone has
much data on what color does for
winning and drawing probabilities
for every rating difference and rating
strength, and I doubt it is significant,
but it *does* change the match
probability.



"Tyrone Slothrop" wrote in message
oups.com...
I am not assuming symmetry. Again, you suggested the simple two
game
case. So let's look at it getting rid of symmetry:

Let's try it with it skewed toward a player.

Take the simple case of a max of two games in a match. Let's say
that
Sam wins if he can score 1.0 and Bill wins if he scores 1.5. Just
as
in the real match, the second game isnt played if it isn't
necessary.

So what are the outcomes? (W- bill wins, T - tied, L - bill loses)

WW B(ill wins)
WT B
WL S(am wins)
TW B
TT S
TL S
L S -- short match

Say that T occurs 50% of the time. Let's say that with color not
counting, W is 35% and L is 15%. Clearly things are skewed toward
Bill
winning. But when we count color, we found that the first player
has
an advantage. When Bill goes first, he finds W is 40%, and L is
10%.
When he goes second, he is still favored, but less so. W is 30% and
L
is 20%. Assuming that the players play the same number of games
going
first or second, the average still works out to 35%/15%.

What are the odds?

Say Bill goes first in game one.

WW occurs .4*.2 = 8% (B)
WT occurs .4*.5 = 20% (B)
WL occurs .4*.3 = 12%
TW occurs .5*.2 = 10% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.3 = 15%
L occurs .1*1.0 = 10% -- short match
Total = 100%

The B-favorable events add up to: 8+20+10 = 38%

Say Bill goes second in game one.

WW occurs .2*.4 = 8% (B)
WT occurs .2*.5 = 10% (B)
WL occurs .2*.1 = 2%
TW occurs .5*.4 = 20% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.1 = 5%
L occurs .3*1.0 = 30% -- short match
Total = 100%

The B-favorable events add up to: 8+10+20 = 38%

This is exactly the same as the first case. Going first in game 1
did
not change Bill's odds even a tiny bit, even if the first player is
favored over the second and the overall odds are skewed toward Bill.
No
symmetry.


David Kane wrote:
I think you are assuming equal
players, which is also not the general case
and by symmetry, means that color won't
change the match probability.

Here is a more general example.

Sloan's sample odds if color doesn't count
(assuming P(draw)=0.25):
PW PL PD
0.22048 0.52952 0.25

Now imagine that White adds 0.1 to the
expectancy, doesn't change the draw probability,
(I'm not suggesting that this is realistic, but
it is consistent with the ELO winning expectancy)

Sloan's chances as white become:
PW PL PD
0.32048 0.42952 0.25

Sloan's chances as black become:
PW PL PD
0.12048 0.62952 0.25

Calculate all the possibilities and
you'll find that this changes
Sloan's match expectancy
(from 0.3183 to 0.3158 if I did
things correctly)


"Tyrone Slothrop" wrote in message
oups.com...
OK. Let's try it then.

Take the simple case of a max of two games in a match. Let's
say
that
Sam wins if he can score 1.0 and Bill wins if he scores 1.5.
Just
as
in the real match, the second game isnt played if it isn't
necessary.

So what are the outcomes? (W- bill wins, T - tied, L - bill
loses)

WW B(ill wins)
WT B
WL S(am wins)
TW B
TT S
TL S
L S -- short match

Say that T occurs 50% of the time. To make things extreme,
let's
say
that if a player goes first, they have a big advantage, W occurs
40%
and L occurs 10%. But if a player doesn't go first, they have a
big
disadvantage, W occurs 10% and L occurs 40%. Assuming that
players
play the same number of games going first or second, the average
still
works out to 25%/25%.

What are the odds?

Say Bill goes first in game one.

WW occurs .4*.1 = 4% (B)
WT occurs .4*.5 = 20% (B)
WL occurs .4*.4 = 16%
TW occurs .5*.1 = 5% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.4 = 20%
L occurs .1*1.0 = 10% -- short match
Total = 100%

The B-favorable events add up to: 4+20+5 = 29%

Say Bill goes second in game one.

WW occurs .1*.4 = 4% (B)
WT occurs .1*.5 = 5% (B)
WL occurs .1*.1 = 1%
TW occurs .5*.4 = 20% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.1 = 5%
L occurs .4*1.0 = 40% -- short match
Total = 100%

The B-favorable events add up to: 4+5+20 = 29%

This is exactly the same as the first case. Going first in game
1
did
not change Bill's odds even a tiny bit, even if the first player
is
outrageously favored over the second.


David Kane wrote:
Actually, color selection would matter, even if people were
dice.

You are correct that it doesn't matter who get white *first*,
but
it will change the overall match probability if you calculate
essentially 2 matches of 2 games, each with their own
distribution
that gives the correct average winning expectancy.

Try it!



"Tyrone Slothrop" wrote in
message
oups.com...
Of course you are right on your point on change of strategy
and
game
theory.

I was addressing Paul's assertion that mathematically,
something
changes if one player gets white on game 1. This doesn't
happen
if
we
assume that the expectation of the players for each new game
remains
the same no matter how many they win or lose previously.

wrote:
Color selection does make a difference, because humans are
not
dice.
The loser in game one has to change strategy.

How it makes a difference is a matter for game theory.

"George John" writes:
The USCF system uses a Boltzmann distribution and FIDE a Gaussian.
And, the two are very close to each other. Are they both wrong?

They are close to each other when the ratings are close together and
the parameters are chosen appropriately. I don't think what the USCF
uses is the same as the Boltzmann distribution, but it's pretty close
too. The big difference between the USCF formula and Elo's original
one is that the USCF uses a much larger standard deviation.

I know how to compute We (score expectancy). I do NOT know how to
compute the win expectancy or draw expectancy. We know that Sloan will
tend to score about .35 points for every game played or 35 points in
ever 100 games. What we don't have a clue about is what the
distribution of draws and wins will be that add up to 35. BTW, this is
something that I have long been interested in knowing more about in
general.

The draw probability is to some extent up to the players. For
example, some openings are more likely to lead to draws than others.
So if one player really really wants a draw, he can choose drawish
openings and so forth. Assuming constant draw probability is a
somewhat bogus approximation since the draw probability depends (among
other things) on the match equity at any round: if Sloan wins game 1,
he might then choose play drawishly in all three remaining rounds.
But if he draws games 1 and 2 and loses game 3, he has to go all-out
for a win in game 4, draws be damned. Similar considerations hold for
Brock.

"Tyrone Slothrop" writes:
Therefore, without loss of generality, we can take the case that all
four games are played all the way through. Each player will get white
twice and black twice.

Sigh, this is correct of course. It took a little while to convince
myself that it still held when the ratings were unequal.

Anyway I think we're all agreed that the match odds are pretty close to:

{'Sloan': 0.32266175328797803, 'Brock': 0.67733824671202181}

given 25% draw probability at each round, and white worth 50 points.

Increasing draw probability to .35 by my program gives:

{'Sloan': 0.30629506119989242, 'Brock': 0.69370493880010753}

Decreasing it to zero gives:

{'Sloan': 0.43437259760118008, 'Brock': 0.56562740239881981}

"Tyrone Slothrop" wrote in message
oups.com...
You are making a distinction I don't understand. I suppose I just
don't understand what you are trying to say.

You agree that order doesn't matter. So the fact that in the real
match, Sam takes the first move as white in game #1 will not change
the
expectancy in the slightest compared with Bill taking the first move
as
white in game #1. It is understood that players alternate in color.
In
the first case, Sam would play white, then black, then white, then
black. In the second case, it would be Bill doing this. I claimed
both
of these cases have the same expectancy. This is the alternation of
color you are talking about, but it seems not.

Your distinction concerns "the match winning percentage will change
if
players alternate white and black and white confers a non-zero
advantage." How does this differ from what is laid out above? How
does it differ from my last example where white confered a 10%
advantage?

Can you explain to me what you mean?

Sure. An easier calculation is to assume that the
same probability applies to every game,
irrespective of color. A refinement is to
include the effect of color. The refinement
in general leads to a different match
expectancy. I thought you were disputing
that, but I guess you were only making
a claim that the order of the colors
doesn't matter. I guess I didn't understand
that because, to me, this is really
an assumption we make when we
postulate a winning probability
that is not a function of when in
the match the game occurs.
(An assumption quite likely
questionable in practice, but that
is another issue)

Under this assumption, there is
no need to even alternate colors, so
long as each side is allocated equal
numbers of Whites.

I would check exactly how you implemented "White worth 50 points"
You still want to preserve the overall winning expectancy, I think.

"Paul Rubin" wrote in message
...
"Tyrone Slothrop" writes:
Therefore, without loss of generality, we can take the case that
all
four games are played all the way through. Each player will get
white
twice and black twice.

Sigh, this is correct of course. It took a little while to convince
myself that it still held when the ratings were unequal.

Anyway I think we're all agreed that the match odds are pretty close
to:

{'Sloan': 0.32266175328797803, 'Brock': 0.67733824671202181}

given 25% draw probability at each round, and white worth 50 points.

Increasing draw probability to .35 by my program gives:

{'Sloan': 0.30629506119989242, 'Brock': 0.69370493880010753}

Decreasing it to zero gives:

{'Sloan': 0.43437259760118008, 'Brock': 0.56562740239881981}

"George John" wrote in message
oups.com...
Hmmm, I thought the question was what is the probability of Sloan
getting a score of 2 or better. To do so he will need a performance
of
2042 or better. My calculation comes up with the probability of
that
happening. It strikes me as a elegant solution. But, given I'm not
in
the field of probability, I can't say with confidence that what I
have
done is correct. But, I have yet to see a convincing, IMHO,
rebuttal
to it so far.

"Elegant" is not the word that comes to mind. Sam was
right. This really just is freshman probability.

"David Kane" writes:
I would check exactly how you implemented "White worth 50 points"
You still want to preserve the overall winning expectancy, I think.

Yes, changing the white-pieces advantage from 50 points to another
number changes the match odds slightly. It's not that sensitive,
given reasonable values for the advantage. I've always heard it's
about 50 points.

There is another possible source of asymmetry though, which is that
the draw probability really isn't constant between rounds. It will
depend on player choices which in turn will depend on the match equity
up to that game. For example, let's say the score is 1.5 each after 3
games. Sloan's best chance is to try forcing a draw in game 4. Maybe
his expected score is .4 with white, but (in the extreme) he gets to
choose between (winprob=.6, loseprob=.4, drawprob=0) and (winprob=0,
loseprob=.2, drawprob=.8). With black in that same extreme case,
Sloan is sunk.

I didn't try modelling that draw strategy because this calculation has
been error-prone enough without it. But it's important effect and
I may try to code something.

Paul Rubin wrote:
"David Kane" writes:
I would check exactly how you implemented "White worth 50 points"
You still want to preserve the overall winning expectancy, I think.

Yes, changing the white-pieces advantage from 50 points to another
number changes the match odds slightly. It's not that sensitive,
given reasonable values for the advantage. I've always heard it's
about 50 points.

I have just made a bunch of new calculations. I will not tell you what
my results are, because this would reveal to my opponent what my match
strategy is. I will continue this discussion when I get back.

However, here is a new question:

Suppose in the first game, we reach a complex position, each player has
only 30 seconds left on the clock, and each player has exactly a 50%
chance of winning.

Suddenly, Brock offers me a draw. Should I accept?

It seems obvious that I should accept because I have draw odds in the
match.

However, I feel that it would be best to refuse his offer of a draw.

Work this out and tell me if you think that I am correct or not.

Sam Sloan