Match Series odds
Hank Youngerman
answer to the following question:
A best 3-of-5 series of A-point matches is equivalent to a single
N-pont match in terms of the odds of the stronger player winning
You can get a reasonable approximation as follows:
1) Define A and the rating difference between the two players
2) Determine, by the FIBS formula, the odds of the better player
winning a match of length A
3) Determine, by binomial expansion, the odds of the better player
winning the series
4) Solve this for N, again using the FIBS rating formula
But this assumes that the FIBS formula is accurate to the degree of
precision required. Does it necessarily follow that if I am a 2-1
favorite in a 9pt match (implying a rating difference of 200 points)
that I am a 3.16-1 favorite in a 25pt match? I'm not at all sure.
If I could get two bots to play against each other on different skill
settings I might be able to solve this, but I don't know how to do
that. Does anyone else have any ideas?
Back4U2 BBL
"Hank Youngerman" <Red...@RedTopBG.com> wrote in message
news:3d024b6e...@news.charter.net...
Tony Lezard wrote a program, Dueller.
URL:
http://www.jobstream.com/~tony/backgammon/
Pretty sure he wants to help you with this.
Nardy
Adam Stocks
"Hank Youngerman" <Red...@RedTopBG.com> wrote in message
news:3d024b6e...@news.charter.net...
Although some server rating systems are good, the relationship between MWC,
skill difference, and match length is too complex to come up with a *very*
highly accurate mathematical formula (as yet), so for now, I think you will
have to 'solve' it empirically, with the bot v bot approach. But I would
add that the approximation you suggested would be pretty darned good in
reality.
Adam
Maik Stiebler
> settings I might be able to solve this, but I don't know how to do
> that. Does anyone else have any ideas?
You can do this with GnuBG. Settings -> Players will let you configure
both sides independently.
Nis Jorgensen
Youngerman) wrote:
>I'm wondering if anyone has a bright idea about finding a definitive
>answer to the following question:
>
>A best 3-of-5 series of A-point matches is equivalent to a single
>N-pont match in terms of the odds of the stronger player winning
False. Next question.
--
Nis Jorgensen
Amsterdam
Back4U2 BBL
"Nis Jorgensen" <n...@dkik.dk> wrote in message
news:hnr8guc29bb25616s...@4ax.com...
Possible to explain 'false'?
ty
___
Nardy
(follows what Hank wrote in his post:)
(*start*)
( You can get a reasonable approximation as follows:
( 1) Define A and the rating difference between the two players
( 2) Determine, by the FIBS formula, the odds of the better player
( winning a match of length A
( 3) Determine, by binomial expansion, the odds of the better player
( winning the series
( 4) Solve this for N, again using the FIBS rating formula
(*end*)
>
> --
> Nis Jorgensen
> Amsterdam
Nis Jorgensen
<nardy.p...@skynet.be> wrote:
>> >A best 3-of-5 series of A-point matches is equivalent to a single
>> >N-pont match in terms of the odds of the stronger player winning
>>
>> False. Next question.
>
>Possible to explain 'false'?
>ty
A best-of-five series of A-point matches is not equivalent to a single
N-point match for any N 'in terms of the odds of the stronger player
winning'. Both your suggestions (using FIBS rating and using empiric
bot vs bot data) will give you good approximations. Note that they
will both give different N, depending on the skill difference. Also
note that you do not need to do any rating calculations for the
imperic data (you probably already know this).
--
Nis Jorgensen
Amsterdam
Hank Youngerman
rating difference, there IS some value for N for which the statement
is true. The problem is to find it. I am fairly convinced that the
range of N will not be highly sensitive to the rating difference - for
example, it won't be 25 points for a 100-rating point difference and
35 points for a 200-point difference.
Many thanks to the person who directed me to Dueller. I am using that
now, but it may take months to get useful results.
Murat Kalinyaprak
Hank Youngerman wrote 3d024b6e...@news.charter.net
> But this assumes that the FIBS formula is accurate to the degree
> of precision required. Does it necessarily follow that if I am a 2-1
> favorite in a 9pt match (implying a rating difference of 200 points)
> that I am a 3.16-1 favorite in a 25pt match? I'm not at all sure.
Hank, I am really pleased to finally see people like you are having
the intelligence and guts to ask such queations...
What are you trying to do to me...? I had given up any hope on the
human species long time ago but maybe there is still a hope...?
Have a nice (FIBSless) day... :))
MK
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jthyssen
> Well, it can't be "false." My point is that if you know A and the
> rating difference, there IS some value for N for which the statement
> is true. The problem is to find it. I am fairly convinced that the
> range of N will not be highly sensitive to the rating difference - for
> example, it won't be 25 points for a 100-rating point difference and
> 35 points for a 200-point difference.
>
> Many thanks to the person who directed me to Dueller. I am using that
> now, but it may take months to get useful results.
I may be easier/faster with gnubg, as you may let gnubg play itself
(e.g., with player 0 playing at 0-ply and player 1 playing at 2-ply).
A question: does it change your doubling strategy when playing
multiple matches? You will now have a MWC (current match winning
chance), but also a TMWC (total match winning chance). I would most
likely tend to use MWC for my doubling strategy, but wouldn't it be
more correct to use TMWC? Of course, if the match is tied 1-1 best of
3, we have MWC = TMWC, but for all other scores MWC <> TMWC.
Jørn
Nis Jorgensen
>A question: does it change your doubling strategy when playing
>multiple matches? You will now have a MWC (current match winning
>chance), but also a TMWC (total match winning chance). I would most
>likely tend to use MWC for my doubling strategy, but wouldn't it be
>more correct to use TMWC? Of course, if the match is tied 1-1 best of
>3, we have MWC = TMWC, but for all other scores MWC <> TMWC.
Since TMWC = MWC * P1 + (1-MWC) * P2 = P2 + (P1-P2) MWC
with P1 being TMWC after winning the current match, P2 being TMWC
after losing this match, and (P1-P2) obviously positive, yoiu maximize
TMWC by maximizing MWC
For øvrigt: Jeg har et par spørgsmål ang. GNU og MatchID ... de følger
i en email senere.
--
Nis Jorgensen
Amsterdam