win percentage
Tom Tseng
backgammon?
Tom
rintan...@my-deja.com
Tom Tseng <tnbt...@pacbell.net> wrote:
> What percentage of games can one expect to win if one plays perfect
> backgammon?
>
Could be anything between 50-100%. If the expert is playing against
an other expert winning percentage is 50%. If he plays against a
person who does not want to win then winning percentage is 100%.
:)
Sent via Deja.com http://www.deja.com/
Before you buy.
Robert-Jan Veldhuizen
wrote:
>What percentage of games can one expect to win if one plays perfect
>backgammon?
Well, we don't know of course but one could make some sort of an
educated guess. It depends on the skill of the opponent of course. If he
also plays perfect backgammon, than you can't win more than 50% assuming
it's not decided yet who gets the first move.
Based on FIBS rating system, and what ratings the best bots have
achieved there compared to the average rating, I estimate that for
simple money play, a perfect playing bot would have a rating somewhere
between 2200 and 2500. Let's say it's 2350. Against an average FIBS
player with a rating of 1550, we get:
Rating difference D = 2350 - 1550 = 800
Probability of underdog to win a 1pt match: 1/(10^(800/2000)+1)= 28.5%
Idem for a 2pt match: 1/(10^(800*sqrt(2)/2000)+1)= 21.4%
It's not clear how matchplay and FIBS rating system for it translate to
money, but I think the winning percentage for the underdog is somewhere
between these values, and closer to the two point match, since money
games involve (back-)gammons and full cube usage.
So, my estimation would be something like "a perfect player would win
around 77% of the points (probably also games, but not necessarily) in a
money game against an average FIBS player, translating into an average
turn-out of 0.77-0.23= 0.54 points per game."
--
Robert-Jan/Zorba
rintan...@my-deja.com
R.J.Vel...@cable.b2000(b=a).nl wrote:
> On Mon, 06 Nov 2000 21:40:20 -0800, Tom Tseng <tnbt...@pacbell.net>
> wrote:
>
> >What percentage of games can one expect to win if one plays perfect
> >backgammon?
>
>
> achieved there compared to the average rating, I estimate that for
> simple money play, a perfect playing bot would have a rating somewhere
> between 2200 and 2500. Let's say it's 2350. Against an average FIBS
> player with a rating of 1550, we get:
>
> Rating difference D = 2350 - 1550 = 800
>
> Probability of underdog to win a 1pt match: 1/(10^(800/2000)+1)= 28.5%
> Idem for a 2pt match: 1/(10^(800*sqrt(2)/2000)+1)= 21.4%
>
If these are the winning probabilities underdog is not an average FIBS
player because he would double immediatly and then his winning
probability would be 28.5%. Right?
> It's not clear how matchplay and FIBS rating system for it translate
to
> money, but I think the winning percentage for the underdog is
somewhere
> between these values, and closer to the two point match, since money
> games involve (back-)gammons and full cube usage.
>
http://www.bkgm.com/rgb/rgb.cgi?view+566
Matti
Tapio Palmroth
Tom Tseng <tnbt...@pacbell.net> kirjoitti
viestissä:3A0795C3...@pacbell.net...
> What percentage of games can one expect to win if one plays perfect
> backgammon?
>
>
But there is nobody who plays perfect backgammon.
t
rintan...@my-deja.com
michael.a.crane
Michael
"Tapio Palmroth" <ta...@palmroth.com> wrote in message
news:8u8pnl$chv$1...@tron.sci.fi...
>
> Tom Tseng <tnbt...@pacbell.net> kirjoitti
> viestissä:3A0795C3...@pacbell.net...
> > What percentage of games can one expect to win if one plays perfect
> > backgammon?
> >
> > Tom
> >
> But there is nobody who plays perfect backgammon.
>
> t
>
>
rintan...@my-deja.com
"michael.a.crane" <michael...@ntlworld.com> wrote:
> How do you know? Have you played EVERY BODY? ;-))
>
> Michael
And would you know if you have played EVERY BODY? :))
>
> "Tapio Palmroth" <ta...@palmroth.com> wrote in message
> news:8u8pnl$chv$1...@tron.sci.fi...
> >
> > Tom Tseng <tnbt...@pacbell.net> kirjoitti
> > viestissä:3A0795C3...@pacbell.net...
> > > What percentage of games can one expect to win if one plays
perfect
> > > backgammon?
> > >
> > > Tom
> > >
> > But there is nobody who plays perfect backgammon.
> >
> > t
> >
> >
>
>
Andrew Grant
for a while and keep track of your wins and losses. ;)
Andrew Grant.
Tom Tseng <tnbt...@pacbell.net> wrote in message
news:3A0795C3...@pacbell.net...
Daniel Murphy
wrote:
>What percentage of games can one expect to win if one plays perfect
>backgammon?
>
>Tom
If Perfect Player played Perfect Player's Twin, either of player could
expect to win half of the time. If PP played anyone less imperfect,
he'd win more often. How much more often would depend on how
imperfectly his opponent played.
Online and national rating systems are based on the idea that
differences in rating are a more or less accurate predictor of winning
chances. And this they more or less accurately do.
What would a perfect player's rating be? That's hard to say, but we
observe players as perfect as they come these days (but still far from
perfect) boasting online ratings of about 2000 or so. So God might be
able to maintain a 2200 rating, maybe higher. In which case, He could
expect to win about 55% of one-point matches against a 2000 player,
and about 60% of matches against an 1800 player.
Daniel Murphy
wrote:
>What percentage of games can one expect to win if one plays perfect
>backgammon?
>
>Tom
If Perfect Player played Perfect Player's Twin, either player could
expect to win half of the time. If PP played anyone less perfect,
michael.a.crane
Michael
<rintan...@my-deja.com> wrote in message
news:8u8tn9$a7t$1...@nnrp1.deja.com...
> In article <RcSN5.82444$hk2.1...@news6-win.server.ntlworld.com>,
> "michael.a.crane" <michael...@ntlworld.com> wrote:
> > How do you know? Have you played EVERY BODY? ;-))
> >
> > Michael
>
> And would you know if you have played EVERY BODY? :))
>
>
> >
> > "Tapio Palmroth" <ta...@palmroth.com> wrote in message
> > news:8u8pnl$chv$1...@tron.sci.fi...
> > >
> > > Tom Tseng <tnbt...@pacbell.net> kirjoitti
> > > viestissä:3A0795C3...@pacbell.net...
> > > > What percentage of games can one expect to win if one plays
> perfect
> > > > backgammon?
> > > >
> > > > Tom
> > > >
Douglas Zare
> In article <3A0795C3...@pacbell.net>,
> Tom Tseng <tnbt...@pacbell.net> wrote:
> > What percentage of games can one expect to win if one plays perfect
> > backgammon?
When Snowie says that I have played perfectly for an entire game,
usually this means that I have dropped a double after a few moves or
spent a long time dancing on the bar. Perhaps this suggests that perfect
play loses almost all of the time. (Ok, sometimes I offer the double,
too.)
I have developed some mathematical techniques which can estimate how far
from perfect play Snowie (or any other bot which estimates match winning
chances) is. It may be counterintuitive that this is possible without
knowing the actual equities, but mathematics can be surprisingly
effective. The rigorous estimates are not very good, but I hope to have
decent statistical estimates soon under some assumptions which are
plausible to backgammon players but far from provable (just as it should
be plausible that a knight handicap in chess is a handicap). Basically,
one can use the inconsistencies of Snowie's evaluations, how much Snowie
whines about the dice, to estimate how much equity it gives up.
Without a complete set of data, let me just say that Snowie 3 seems to
play within 200 elo points of perfect. One might be able to extract a
greater advantage from Snowie by steering towards positions it
misevaluates greatly which are relatively rare when it plays against
perfect play. A source of this, in a 1-point match, is positions which
would probably be played much differently in the future if one is trying
to win/save a gammon versus trying to win the game, hence in which
Snowie's cubeless equities are flawed.
> Could be anything between 50-100%. If the expert is playing against
> an other expert winning percentage is 50%. If he plays against a
> person who does not want to win then winning percentage is 100%.
Really? I don't think the 100% is exactly correct. Suppose I choose the
dice, and you play legally (no resignations) for both sides, trying to
make side A win. I try to make side B win using my control over the
rolls. Side A, get ready for 5-5 after 5-5 for a while, while side B
gets nothing but 5's and 6's. Side B will win (a backgammon if one is
more careful, I believe), which means that there is a positive
probability that perfect play against perfectly bad play will still lose
because there is a chance that the dice will behave as though called by
me. This gives a lower bound on the expected ratings of players who only
play 1-point matches against a player with a fixed rating.
Douglas Zare
rintan...@my-deja.com
Douglas Zare <za...@math.columbia.edu> wrote:
> rintan...@my-deja.com wrote:
>
> > In article <3A0795C3...@pacbell.net>,
> > Tom Tseng <tnbt...@pacbell.net> wrote:
> > > What percentage of games can one expect to win if one plays
perfect
> > > backgammon?
> >
> > an other expert winning percentage is 50%. If he plays against a
> > person who does not want to win then winning percentage is 100%.
>
> Really? I don't think the 100% is exactly correct. Suppose I choose
the
> dice, and you play legally (no resignations) for both sides, trying to
> make side A win. I try to make side B win using my control over the
> rolls. Side A, get ready for 5-5 after 5-5 for a while, while side B
> gets nothing but 5's and 6's. Side B will win (a backgammon if one is
> more careful, I believe), which means that there is a positive
> probability that perfect play against perfectly bad play will still
lose
> because there is a chance that the dice will behave as though called
by
> me. This gives a lower bound on the expected ratings of players who
only
> play 1-point matches against a player with a fixed rating.
>
ok. I agree. The number I gave is not *exact* but rounded. I suppose
that the exact winning percentage is over 99.5%, but I'm not going to
prove that ;)).
Matti
rintan...@my-deja.com
rac...@best.com (Daniel Murphy) wrote:
> On Mon, 06 Nov 2000 21:40:20 -0800, Tom Tseng <tnbt...@pacbell.net>
> wrote:
>
> >What percentage of games can one expect to win if one plays perfect
> >backgammon?
> >
>
>
> What would a perfect player's rating be? That's hard to say, but we
> observe players as perfect as they come these days (but still far from
> perfect) boasting online ratings of about 2000 or so. So God might be
> able to maintain a 2200 rating, maybe higher. In which case, He could
> expect to win about 55% of one-point matches against a 2000 player,
> and about 60% of matches against an 1800 player.
No! God can control dices and maybe also His opponents moves. There
is no upper limit for His rating, see the link below.
http://www.bkgm.com/rgb/rgb.cgi?view+627
Julian Hayward
>What percentage of games can one expect to win if one plays perfect
>backgammon?
With my dice, about 20%... 8-)
--
Julian Hayward 'Booles' on FIBS jul...@ratbag.demon.co.uk
+44-1480-210097 http://www.ratbag.demon.co.uk/
---------------------------------------------------------------------------
"A witty saying proves nothing" - Voltaire
---------------------------------------------------------------------------
Julian Hayward
writes
>No! God can control dices and maybe also His opponents moves. There
>is no upper limit for His rating, see the link below.
And He has the perfect remedy for anyone who drops on Him... ***zap***
Robert-Jan Veldhuizen
>In article <5nlf0tkei0ovqh5p6...@4ax.com>,
> R.J.Vel...@cable.b2000(b=a).nl wrote:
>> On Mon, 06 Nov 2000 21:40:20 -0800, Tom Tseng <tnbt...@pacbell.net>
>> wrote:
>>
>> >What percentage of games can one expect to win if one plays perfect
>> >backgammon?
>>
>>
>> Based on FIBS rating system, and what ratings the best bots have
>> achieved there compared to the average rating, I estimate that for
>> simple money play, a perfect playing bot would have a rating somewhere
>> between 2200 and 2500. Let's say it's 2350. Against an average FIBS
>> player with a rating of 1550, we get:
>>
>> Rating difference D = 2350 - 1550 = 800
>>
>> Probability of underdog to win a 1pt match: 1/(10^(800/2000)+1)= 28.5%
>> Idem for a 2pt match: 1/(10^(800*sqrt(2)/2000)+1)= 21.4%
>>
>
>If these are the winning probabilities underdog is not an average FIBS
>player because he would double immediatly and then his winning
>probability would be 28.5%. Right?
FIBS rating system doesn't account for this "automatic" double in a 2pt
match. It just sees a 2pt match as somewhere in between a 1 and a 3pt
match. If both players know the correct strategy a 2pt match is
essentially a 1pt match, and therefore a way to change one's rating
faster.
BTW, in my experience 1550 rated FIBS players usually don't know about
this "automatic" double.
>> It's not clear how matchplay and FIBS rating system for it translate
>to
>> money, but I think the winning percentage for the underdog is
>somewhere
>> between these values, and closer to the two point match, since money
>> games involve (back-)gammons and full cube usage.
>>
>
>http://www.bkgm.com/rgb/rgb.cgi?view+566
According to David Montgomery's estimate, you'd get to +1.6ppg which
seems a bit too much here. His method might not be accurate for such
large rating differences.
--
Robert-Jan/Zorba
rintan...@my-deja.com
I see. I though these errors could happen on low rated players but
not to average FIBS player :(. That's very low cube handling
skill level indeed!
> >> It's not clear how matchplay and FIBS rating system for it
translate
> >to
> >> money, but I think the winning percentage for the underdog is
> >somewhere
> >> between these values, and closer to the two point match, since
money
> >> games involve (back-)gammons and full cube usage.
> >>
> >
> >http://www.bkgm.com/rgb/rgb.cgi?view+566
>
> According to David Montgomery's estimate, you'd get to +1.6ppg which
> seems a bit too much here. His method might not be accurate for such
> large rating differences.
Why? I don't think so. In average in money game you win 2.2 points.
If you win about 80% of the games your average win, ignoring
differences in cube handling and gammon factors, would be
(.8-.2)*2.2 = 1.32ppg
If taking account also differences in cube handling and gammon factors,
average win is higher than 1.3ppg. How to reach correct number, see the
link above.
Sanghabum
>When Snowie says that I have played perfectly for an entire game,
>usually this means that I have dropped a double after a few moves or
>spent a long time dancing on the bar. Perhaps this suggests that perfect
>play loses almost all of the time.
I've also noticed this with Snowie, and used it to devise two ego-boosting ways
of getting Snowie to assess my play as extra-terrestrial.
Way 1. Play a 9-point match. Resign the first three games as a backgammon
*before* making any moves. Your play will be perfect, and your rating
extra-terrestrial.
Way 2: Assuming you don't want to so blatantly cheat a bot as above, try this.
Set up this proposition. Snowie has all 15 men ready to bear off in an ideal
distribution (3 each on 5,6,4 points, 2 each on 3,2,1). You have all 15 men on
the bar. Play naturally. Unless Snowie messes up her bear-off, you'll get
assessed as extra-terrestrial.
Once you've got such a rating remember to leave it casually on the screen if
any BG-playing buddies are around.
(Serious point here for Oasya: Snowie's assessment criteria need changing. A
forced move does not change equity and so should not be quality assessed.And a
resignation should be assessed as either sensible, an error, or a blunder
depending on how much equity it loses).
–Colin
Douglas Zare
> (Serious point here for Oasya: Snowie's assessment criteria need changing. A
> forced move does not change equity and so should not be quality assessed.And a
> resignation should be assessed as either sensible, an error, or a blunder
> depending on how much equity it loses).
I agree with the latter, but I believe the forced moves should count as perfectly
played. Getting into a situation in which one's moves are forced is no different
from getting into a position in which the moves are really easy to make but not
forced (say, sitting on an anchor as you build your board, with a shot a few moves
in the future). Either gives you an equity greater than the theoretical equity if
your opponent continues to make errors at the usual rate. The average error per
move is not particularly meaningful, but in order to compare the average errors
between the two players the number of moves should be roughly the same, or else
the stronger player could show up as worse over the long run.
Douglas Zare