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Nov 6, 2000, 3:00:00 AM11/6/00

to

What percentage of games can one expect to win if one plays perfect

backgammon?

backgammon?

Tom

Nov 7, 2000, 3:00:00 AM11/7/00

to

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?

>

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.

Nov 7, 2000, 3:00:00 AM11/7/00

to

On Mon, 06 Nov 2000 21:40:20 -0800, Tom Tseng <tnbt...@pacbell.net>

wrote:

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

Nov 7, 2000, 3:00:00 AM11/7/00

to

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?

>

>

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%

>

> 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

Nov 7, 2000, 3:00:00 AM11/7/00

to

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

Nov 7, 2000, 3:00:00 AM11/7/00

to

Nov 7, 2000, 3:00:00 AM11/7/00

to

How do you know? Have you played EVERY BODY? ;-))

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

>

>

Nov 7, 2000, 3:00:00 AM11/7/00

to

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

"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

> >

> >

>

>

Nov 7, 2000, 3:00:00 AM11/7/00

to

It should be easy to find out. All you have to do is play perfect backgammon

for a while and keep track of your wins and losses. ;)

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...

Nov 7, 2000, 3:00:00 AM11/7/00

to

On Mon, 06 Nov 2000 21:40:20 -0800, Tom Tseng <tnbt...@pacbell.net>

wrote:

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.

Nov 7, 2000, 3:00:00 AM11/7/00

to

On Mon, 06 Nov 2000 21:40:20 -0800, Tom Tseng <tnbt...@pacbell.net>

wrote:

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,

Nov 7, 2000, 3:00:00 AM11/7/00

to

No ;-((

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

> > > >

Nov 7, 2000, 3:00:00 AM11/7/00

to

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?

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

Nov 8, 2000, 2:01:35 AM11/8/00

to

In article <3A088FFB...@math.columbia.edu>,

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?

> >

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?

> >

> > 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.

>

> > 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

Nov 8, 2000, 2:07:06 AM11/8/00

to

In article <3a0821ac...@news.cybercity.dk>,

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?

> >

> >Tom

>

>

> 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.

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

Nov 8, 2000, 3:00:00 AM11/8/00

to

In article <3A0795C3...@pacbell.net>, Tom Tseng

<tnbt...@pacbell.net> writes>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

---------------------------------------------------------------------------

Nov 8, 2000, 3:00:00 AM11/8/00

to

In article <8uau2p$vdq$1...@nnrp1.deja.com>, rintan...@my-deja.com

writes

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***

Nov 8, 2000, 3:00:00 AM11/8/00

to

On Tue, 07 Nov 2000 11:08:11 GMT, rintan...@my-deja.com wrote:

>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

Nov 9, 2000, 3:00:00 AM11/9/00

to

In article <qcph0toq970lsjp4r...@4ax.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.

Nov 10, 2000, 3:00:00 AM11/10/00

to

>Douglas Zare:

>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.

>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

Nov 10, 2000, 3:00:00 AM11/10/00

to

Sanghabum wrote:

> (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

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