FYI, saying that "Sam was right" doesn't work for me as a convincing
rebuttal, nor does "This really just is freshman probability".

The expert I contacted said he would look at this thread and post
something.

Best regards,

George John

Tyrone Slothrop wrote:

[SNIP]

Because continuous distributions model large numbers of events. They
are not accurate for small sample sizes. If the continuous
distribution was accurate for any sample size, no matter how small,
then there would never be a need to use a discrete distribution!

I appreciate your taking the time to answer my questions. For now, I
will be mostly content to wait for the expert on the subject to post
here. I am very interested in hearing what he has to say.

I do understand that my suggested method may have a good deal of
uncertainty associated with it. I have a couple questions please. If
we assume a *very large number* of four game matches between players
with a rating difference of 111, do you agree with the following:

1) The lower rated player will need to *on average* have a performance
rating greater than or equal to 111 points higher than his or her
rating to attain a score of 2.0 or higher.

2) Assuming the rating performance of a player conforms to a Normal
Distribution, and the standard deviation is 200, the probability of a
player having a performance = their rating + 111 is roughly 29%

Even if my analysis is inferior to other systems, I want to verify that
my basic premise is correct please.

[SNIP]

However,
you can verify for yourself that draw frequency does matter for small
sample sizes. Ergo, the continuous distribution does not model small
sample sizes well!

I understand the lack a granularity of the system. I understand that
the We formula does not take into account the distribution of draws and
wins. I do understand it would be helpful to take into account the
distribution of draws and wins (a number I have no good feel for). I
agree a more detailed analysis than what I did will likely yield a more
confident prediction of the outcome.

I make NO assumptions. You have not understood my example. You think I
am calculating the odds for the match?

Sorry, I was thinking about Paul's work when I wrote this.

Best regards,

George John

"George John" wrote in message
oups.com...


I appreciate your taking the time to answer my questions. For now,
I
will be mostly content to wait for the expert on the subject to post
here. I am very interested in hearing what he has to say.

I do understand that my suggested method may have a good deal of
uncertainty associated with it. I have a couple questions please.
If
we assume a *very large number* of four game matches between players
with a rating difference of 111, do you agree with the following:

1) The lower rated player will need to *on average* have a
performance
rating greater than or equal to 111 points higher than his or her
rating to attain a score of 2.0 or higher.

False. He has to have a performance higher than Brock's, which
can also vary. These are not identical.

2) Assuming the rating performance of a player conforms to a Normal
Distribution, and the standard deviation is 200, the probability of
a
player having a performance = their rating + 111 is roughly 29%

False. A correctly done calculation of that type might be relevant
for a single game, not a match.

Even if my analysis is inferior to other systems, I want to verify
that
my basic premise is correct please.

It is somewhere between completely wrong and incomplete.
Not knowing probability isn't a crime. However, many have
laid out the problem correctly and yet you seem to be
doing everything in your power to keep your mind shut
tight.

"George John" writes:
1) The lower rated player will need to *on average* have a performance
rating greater than or equal to 111 points higher than his or her
rating to attain a score of 2.0 or higher.

Performance rating makes the most sense in a multi-player event or at
least a multi-day event. In a 1-day match like this, it doesn't mean
much. The combinatorial approach is the most straightforward way to
do this, and the result is exact.

"samsloan" writes:
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.

This can't happen, because of the Bronstein delay.

David Kane wrote:
"George John" wrote in message
oups.com...


I appreciate your taking the time to answer my questions. For now,
I
will be mostly content to wait for the expert on the subject to post
here. I am very interested in hearing what he has to say.

I do understand that my suggested method may have a good deal of
uncertainty associated with it. I have a couple questions please.
If
we assume a *very large number* of four game matches between players
with a rating difference of 111, do you agree with the following:

1) The lower rated player will need to *on average* have a
performance
rating greater than or equal to 111 points higher than his or her
rating to attain a score of 2.0 or higher.

False. He has to have a performance higher than Brock's, which
can also vary. These are not identical.

I understand that Brock's performance can vary, too. But, on average
it will be 2042. So, the performance will need to equal or exceed *on
average* 2042. Sloan has draw odds, so he "wins" the match if he gets
a score of 2.0.

2) Assuming the rating performance of a player conforms to a Normal
Distribution, and the standard deviation is 200, the probability of
a
player having a performance = their rating + 111 is roughly 29%

False. A correctly done calculation of that type might be relevant
for a single game, not a match.

Why does it work for a game and not a match?

Even if my analysis is inferior to other systems, I want to verify
that
my basic premise is correct please.

It is somewhere between completely wrong and incomplete.
Not knowing probability isn't a crime. However, many have
laid out the problem correctly and yet you seem to be
doing everything in your power to keep your mind shut
tight.

I'm perfectly willing to consider a superior system, and nowhere have I
said elsewhere that anything anyone has written is wrong. I'm only
wondering why what I proposed is so dead wrong. It doesn't require
simulation; although, simulation very well way yield a better answer,
possibly much better.

I will now study the other posts in detail. I haven't done that so
far. They all seemed to use simulation, which I have been trying to
avoid. Maybe it can't be avoided. I can accept that.

Best regards,

George John

"samsloan" writes:
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?

According to my program, your match equity if you accept the draw is
0.32419540845288114 if you have white in the first round and
0.39795776320644605 if you have black in the first round. It's higher
if you have black because that means of the remaining three games, you
get white in two of them.

If you refuse the draw, your match equity is 0.37713138626306819 if
you have white in the first round or 0.43893685785773762 if you have
black in the first round.

So in both cases, you're slightly better off refusing the draw.

Note that if you overestimate your winning chances and they're not
really 50% after all, refusing the draw can be a big error. And with
30 seconds left, maybe you can't estimate so accurately. That's the
usual reason for accepting draws in situations like this.

Note also that in later rounds, the numbers change completely.

Paul Rubin wrote:
"samsloan" writes:
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?

According to my program, your match equity if you accept the draw is
0.32419540845288114 if you have white in the first round and
0.39795776320644605 if you have black in the first round. It's higher
if you have black because that means of the remaining three games, you
get white in two of them.

If you refuse the draw, your match equity is 0.37713138626306819 if
you have white in the first round or 0.43893685785773762 if you have
black in the first round.

So in both cases, you're slightly better off refusing the draw.

Note that if you overestimate your winning chances and they're not
really 50% after all, refusing the draw can be a big error. And with
30 seconds left, maybe you can't estimate so accurately. That's the
usual reason for accepting draws in situations like this.

Note also that in later rounds, the numbers change completely.

I agree. I have also concluded that in that situation, I should refuse
the draw.

Of course, that exact situation will probably never arise.

Sam Sloan

"George John" wrote in message
oups.com...


False. A correctly done calculation of that type might be relevant
for a single game, not a match.

Why does it work for a game and not a match?

For the same reason that the odds of rolling at least
a 6 on a die (1/6) are not the same as the odds
for rolling at least a 12 on two dice (1/36)

The point is we already know the expectancy
from the rating point difference. The question
is how to apply that information to the match
conditions. Several have shown how that can
be done.

Color selection has no effect.

Wrong. Sloan gets white twice and can clinch the match with two wins as
white without Brock getting white twice himself.

That's about as fair as the coin-flip in NFL overtime (why don't they just
continue play, with the next score winning the game?).