Match length and skill advantage/World Cup format
Donald Kahn
It is obviously true that the longer the match, the more likely the
better player will win. But has anyone an idea of how to quantify
this?
For instance, suppose that against a particular opponent, I am rated
a 5 to 4 underdog in an 11 point match. What are my chances at 25
points?
What can be strictly quantified is the effect of deciding the winner
of a round as best 3 out of 5 11-point matches, as in the Dallas
'World Cup'. It gives the better player much less advantage than I
had imagined.
If X=the likelihood of the better player winning a single 11-point
match, then his likelihood to winning best 3 out of 5 =
(X**3)*[10 - 15*X + 6*(X**2)].
If, as in my example, X = 5/9, then the answer is .6033., almost
exactly 3 to 2.
If I were the better player, I would much prefer to play to 25. My
own guess is that, if you are 5 to 4 at 11, you must be somewhere
close to 2 to 1 at 25; 5 to 3 at least.
And MUCH less work.
deekay
Chuck Bower
In article <354cdf10...@news.newsguy.com>,
Donald Kahn <don...@easynet.co.uk> wrote:
>It is obviously true that the longer the match, the more likely the
>better player will win. But has anyone an idea of how to quantify
>this?
>
>For instance, suppose that against a particular opponent, I am rated
>a 5 to 4 underdog in an 11 point match. What are my chances at 25
>points?
The ratings formulas allow you to calculate this.
>
>What can be strictly quantified is the effect of deciding the winner
>of a round as best 3 out of 5 11-point matches, as in the Dallas
>'World Cup'. It gives the better player much less advantage than I
>had imagined.
>
>If X=the likelihood of the better player winning a single 11-point
>match, then his likelihood to winning best 3 out of 5 =
>
>(X**3)*[10 - 15*X + 6*(X**2)].
>
>If, as in my example, X = 5/9, then the answer is .6033., almost
>exactly 3 to 2.
>
>If I were the better player, I would much prefer to play to 25. My
>own guess is that, if you are 5 to 4 at 11, you must be somewhere
>close to 2 to 1 at 25; 5 to 3 at least.
I'll let someone else use the ratings formulas to calculate
this (since as I write this I don't have easy web access). But
there is more to the idea of multiple matches than is simply
covered by ratings formulas.
The stronger player SHOULD have a better grasp of late match
cube handling. Best three out of five (or whatever) gives him/her
more chances to excercise such knowledge.
Chuck
bo...@bigbang.astro.indiana.edu
c_ray on FIBS
Morten Daugbjerg Hansen
In <6ij6a6$qmp$1...@jetsam.uits.indiana.edu> bo...@bigbang.astro.indiana.edu (Chuck Bower) writes:
>>For instance, suppose that against a particular opponent, I am rated
>>a 5 to 4 underdog in an 11 point match. What are my chances at 25
>>points?
> The ratings formulas allow you to calculate this.
Well, I guess the ratings formula is a constructed in a way to try quantify this
and therefore these express only a few peoples expectations of how this is....
There a several ratingssystems around the world and the only real difference to
this is the way they handle this problem, so calculating from the formula's
is a bad idea...
This is one of the reasons it has been suggested that playing one-point-matches
on FIBS is better for your rating than playing longer matches cause the
formula isnt perfect....reversely in Denmark where we use a different formula
people tend to think its better to play long matches....
What the 'perfect' formula should be is hard to say but I suggested Fredrik
Dahl to try use JellyFish to help out....Unfortunately he hadnt had time yet....
What could be done is fx to have Level 6 play Level 5 in 10.000 matches of
different matchlength....from 1 to 25 ...
then again level 7 play level 6....and level 7 play level 5....and maybe even
include the lower levels....
This would produce a vast material that MAYBE could shed some light on this
subject but also MAYBE tells us its much more complex than we think....
Unfortunately this needs HUGE amount of computertime and human time....
Maybe Olivier Egger or someone else with a neural net could be interested in
spending a summer vacation on this :-)
Best regards,
Morten Daugbjerg aka md
Chuck Bower
>It is obviously true that the longer the match, the more likely the
>better player will win. But has anyone an idea of how to quantify
>this?
>
>For instance, suppose that against a particular opponent, I am rated
>a 5 to 4 underdog in an 11 point match. What are my chances at 25
>points?
>
>What can be strictly quantified is the effect of deciding the winner
>of a round as best 3 out of 5 11-point matches, as in the Dallas
>'World Cup'. It gives the better player much less advantage than I
>had imagined.
>
>If X=the likelihood of the better player winning a single 11-point
>match, then his likelihood to winning best 3 out of 5 =
>
>(X**3)*[10 - 15*X + 6*(X**2)].
I also derived this. Not that I don't trust you, Donald. Well, OK,
I don't. Because I don't trust ANYONE if I can fairly easily verify
what has been said. That's why I'm called a scientist. ;)
>
>If, as in my example, X = 5/9, then the answer is .6033., almost
>exactly 3 to 2.
>
>If I were the better player, I would much prefer to play to 25. My
>own guess is that, if you are 5 to 4 at 11, you must be somewhere
>close to 2 to 1 at 25; 5 to 3 at least.
>
>And MUCH less work.
I found the ratings formula disussion by Kevin Bastian at:
http://www.northcoast.com/~mccool/fibsrate.html
I'll let the interested readers go there and try it out. If you're
just one of those people who wants answers (and you TRUST me!), I
get the following:
Assume player A is a 4::5 favorite (i.e. wins 5/9) in an 11 point match
against player B.
What is their ratings difference? 58.44
What is the chance that A will win a 25 point match over B? 58.33%
(don't be fooled by the 58.... it's just a coincindence.
REALLY. Trust me....)
How long of a match need A and B play for A to be a 60.33%
favorite (which is A's winning chances in a best 3/5 11 point
matches as Donald has already said)? 39 points.
How long of a match to be a 3::5 favorite? 58 points.
How long of a match to be a 1::2 favorite? 106 points.
At least that is what the (FIBS) rating point formula says. And
before you go and start bashing FIBS, I think Marvin just used the same
formula that Kent Goulding used. (One of the early volumes of "Inside
Backgammon" has an article about backgammon ratings systems, BTW. Also,
see discussions in Jacobs/Trice book "Can a Fish Taste Twice as Good?")
So, you don't mind bashing KG, Inside BG article author (sorry, I forgot
his name and my copies are at home), Jake Jacobs, Walter Trice,...?
Some people have no scruples.
Hank Youngerman
Standard statistical theory says that the deviation from expected
outcome is proportional to the square root of the sample size.
Translated into English, that means that, roughly speaking, if you are
55.6% likely to win an 11-point match, you are 5.6% more likely than
even - so in a 44-point match, you would be 11.2% more likely than
even.
But we can do this more scientifically. Let's ignore games worth more
than 1 point. If you are 51.51% likely to win each game, you are
55.56% likely to reach 11 before your opponent does. If you are
51.51% likely to win each game, you are also 58.39% likely to reach 25
before your opponent does.
It's certainly true that not every point is independent, but this
formulation is probably valid since it starts from a conclusion about
the likelihood of winning an 11-point match and works backward. I
would also say that playing multiple matches DOES bring into play more
skill in late-match cube handling. It also diminishes the effect of
individual large cubes. While cubes of 8 and 16 are somewhat rare in
match play, they DO happen. Say you are redoubled to 16. You
estimate your match equity with a take at 20% and with a drop at 19%,
so you take. Still, in effect the whole match could be on the line on
the next roll or two of the dice. Multiple shorter matches prevent
this.
I don't play high-level tournament backgammon so I personally couldn't
give a rat's behind about which way it's played, but this is my
mathematical analysis.
Chuck Bower
In article <6il8cd$101$1...@jetsam.uits.indiana.edu>,
Chuck Bower <bo...@bigbang.astro.indiana.edu> wrote:
(snip)
>And before you go and start bashing FIBS, I think Marvin just used the same
>formula that Kent Goulding used. One of the early volumes of "Inside
>Backgammon" has an article about backgammon ratings systems,... (whose)
author (sorry, I forgot his name and my copies are at home)...
Volume 1, Number 5, Sept-Oct, 1991 "Backgammon Ratings" by Larry Kaufman
("...one of the world's leading authorities on rating games of skill...
and former consultant to the U.S. Chess Federation on its rating system.")
alan...@my-dejanews.com
In article <354cdf10...@news.newsguy.com>#1/1,
don...@easynet.co.uk (Donald Kahn) wrote:
>
> It is obviously true that the longer the match, the more likely the
> better player will win. But has anyone an idea of how to quantify
> this?
>
> For instance, suppose that against a particular opponent, I am rated
> a 5 to 4 underdog in an 11 point match. What are my chances at 25
> points?
>
> What can be strictly quantified is the effect of deciding the winner
> of a round as best 3 out of 5 11-point matches, as in the Dallas
> 'World Cup'. It gives the better player much less advantage than I
> had imagined.
>
> If X=the likelihood of the better player winning a single 11-point
> match, then his likelihood to winning best 3 out of 5 =
>
> (X**3)*[10 - 15*X + 6*(X**2)].
>
> exactly 3 to 2.
>
> If I were the better player, I would much prefer to play to 25. My
> own guess is that, if you are 5 to 4 at 11, you must be somewhere
> close to 2 to 1 at 25; 5 to 3 at least.
>
> And MUCH less work.
>
>
The FIBS rating formula purports to compute the probability of player x
beating player y given their ratings and the length of the match. If player
x is rated 1800 and player y is rated 1859 the probability of y winning is
.556 according to the formula. The probability of y winning a match of 25 is
computed as .584.
alanback
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Claes Thornberg
Just a few thoughts on the ratingsystem on fibs ...
Why is it that you (on FIBS) gain more from a 2ptr than from a 1ptr,
when we all know that in a perfect world a 2ptr is not different from
a 1ptr? (see kit woolsey monograph 'How to play tournament backgammon')
In a ratingsystem for matches of a fixed length, for instance, 1ptrs,
the ratings of individual players are adjusted so that the probabilies
of the rating formula becomes 'true'. That is, if players play
infinitely many matches, then a ratingdifference of 100 really means that
the better player has 52.8751% chance of winning. But can we design a
ratingsystem that works for all different kinds of matchlengths? (FIBS
rating system does not work, see above). Or is it better to have
different ratings for different match lengths, and only allow matches
of certain lengths?
--
______________________________________________________________________
Claes Thornberg Internet: cla...@it.kth.se
Dept. of Teleinformatics URL: NO WAY!
KTH/Electrum 204 Voice: +46 8 752 1377
164 40 Kista Fax: +46 8 751 1793
Sweden
Kate McCollough
Claes Thornberg <cla...@cuchulain.it.kth.se> wrote in article
<yvkhg34sh...@cuchulain.it.kth.se>...
: Just a few thoughts on the ratingsystem on fibs ...
:
: Why is it that you (on FIBS) gain more from a 2ptr than from a 1ptr,
: when we all know that in a perfect world a 2ptr is not different from
: a 1ptr? (see kit woolsey monograph 'How to play tournament backgammon')
I guess the reality is, we don't live in a perfect world. In this world, it
happens all the time: double (maybe too late, maybe not,) drop (maybe
correctly, maybe not,) start the second game, lose, start the third game.
It may not be ideal play, but it happens enough. Why shouldn't this be
worth more in experience than a one pointer which will never be longer than
one game?
Kate
--
Kate McCollough
mcc...@northcoast.com http://www.northcoast.com/~mccool
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~"Fate laughs at probabilities." ~~~ "I dwell in Possibility-" ~
~ ~Bulwer-Lytton ~~~ ~Emily Dickinson ~
~ Eugene Aram ~~~ ~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Chuck Bower
In article <354e3c78...@news.mindspring.com>,
Hank Youngerman <hankyou...@mindspring.com> wrote:
>Standard statistical theory says that the deviation from expected
>outcome is proportional to the square root of the sample size.
>Translated into English, that means that, roughly speaking, if you are
>55.6% likely to win an 11-point match, you are 5.6% more likely than
>even - so in a 44-point match, you would be 11.2% more likely than
>even.
>
>But we can do this more scientifically. Let's ignore games worth more
>than 1 point.
OK, but how realistic is that? (See below for more discussion.)
If you are 51.51% likely to win each game, you are
>55.56% likely to reach 11 before your opponent does. If you are
>51.51% likely to win each game, you are also 58.39% likely to reach 25
>before your opponent does.
>
>It's certainly true that not every point is independent, but this
>formulation is probably valid since it starts from a conclusion about
>the likelihood of winning an 11-point match and works backward. I
>would also say that playing multiple matches DOES bring into play more
>skill in late-match cube handling. It also diminishes the effect of
>individual large cubes. While cubes of 8 and 16 are somewhat rare in
>match play, they DO happen. Say you are redoubled to 16. You
>estimate your match equity with a take at 20% and with a drop at 19%,
>so you take. Still, in effect the whole match could be on the line on
>the next roll or two of the dice. Multiple shorter matches prevent
>this.
>
>I don't play high-level tournament backgammon so I personally couldn't
>give a rat's behind about which way it's played, but this is my
>mathematical analysis.
How many "rat's behinds" do you have to wager? ;)
This thread has taken on a new emphasis. Donald was wondering
about the relative merits of best 3/5 11 point matches versus a single
long (25 point) match but now we are onto ratings formulas.
In fact, the FIBS/Goulding/Kaufman formula does appear to be
based upon the assumption that Hank mentions above: a match of length
N is assumed to be a series of one point games. If the cube is used,
then the expectation for a game of value 2^m is equivalent to the
expectation of 2^m single point games. Is that a good assumption??
I don't know the answer to this. BUT, I can say fairly confidently
that the volatility associated with a match where the cube is used is
MUCH higher than a match of equivalent 'length' where the cube is not
used.
The above opinion has parallels in the "underdog strategy" ideas
from gambling (which are discussed in "Oswald Jacoby on Gambling" and
in the Jacobs/Trice 'Fish book' and many other places, I'm sure): if
you are the underdog, go out of your way to put all of your chances on
one toss, as opposed to trying to "grind it out" with lots of small
bets. The underdog in a long match should try to get the cube elevated
so that the dice can play a bigger factor than skill. Does the standard
ratings formula ignore this with corresponding deleterious results? I
open this up to the newsgroup for discussion.
Claes Thornberg
"Kate McCollough" <mcc...@northcoast.com> writes:
> I guess the reality is, we don't live in a perfect world. In this world, it
> happens all the time: double (maybe too late, maybe not,) drop (maybe
> correctly, maybe not,) start the second game, lose, start the third game.
> It may not be ideal play, but it happens enough. Why shouldn't this be
> worth more in experience than a one pointer which will never be longer than
> one game?
But the astute underdog, when reaching 2-away,2-away does then (almost
always or always?) gain from doubling immediately. For instance, a
difference in rating of 100 has the underdog at 47.1249% chance of
winning a 1-point match (1-away,1-away). Whereas the same rating
difference makes the underdog 45.9385% of winning a 2-pt match
(2-away,2-away). This is a (large?) flaw in the FIBS rating system.
(As an aside: I once wrote a program, which is still available for
free, which calculated match equity tables from two parameters: gammon
rate and winning percentage in each game for the better player. The
calculations are based on the method described in
http://www.bkgm.com/articles/met.html. The match equity tables
produced by this program/method all have the same equities at
1-away/1-away and 2-away/2-away.)
CT
Tom Keith
Claes Thornberg wrote:
>
> Just a few thoughts on the ratingsystem on fibs ...
>
> Why is it that you (on FIBS) gain more from a 2ptr than from a 1ptr,
> when we all know that in a perfect world a 2ptr is not different from
> a 1ptr? (see kit woolsey monograph 'How to play tournament backgammon')
>
> ratingsystem that works for all different kinds of matchlengths? (FIBS
> rating system does not work, see above). Or is it better to have
> different ratings for different match lengths, and only allow matches
> of certain lengths?
I've looked into this question a little bit. One idea I
had was to use a different way of measuring the opportunity
for skill in the match. FIBS just uses the match length.
I figure a better indication of the amount of work involved
is the total number of dice rolls in the match. Or, even
better, the number of dice rolls in "contact" positions.
So how many "contact position rolls" are there in typical
matches of various lengths? To answer this, I scanned
through the Big_Brother database and counted up the rolls.
Here are the results.
Average Rolls Per Game in Big_Brother Matches
rolls matches
----- -------
Match length 1: 17329 / 359 = 48.27
Match length 3: 54800 / 639 = 85.76
Match length 5: 116451 / 907 = 128.39
Match length 7: 86353 / 498 = 173.40
Match length 9: 9992 / 44 = 227.09
Match length 11: 8230 / 31 = 265.48
Note that there are less than twice as many rolls in a
3-point match as there are in a 1-point match. So, if
counting rolls is a good indicator, it suggests that
opportunity for skill is not really proportional to the
length of the match.
The other idea I had was to do exactly what Claes suggested
in another post: Compute a match equity table for players of
different ability. This method is not perfect -- the
computation assumes that both players always make efficient
doubles. This will produce some error, but I think the
results are still useful.
The following table is constructed assuming the stronger player
wins 51% of the time and that 25% of all wins are gammons.
Match Equity Table between Players of Unequal Ability
-1 -2 -3 -4 -5 -6 -7 -8 -9
----- ----- ----- ----- ----- ------ ----- ----- -----
-1: .5100 .6974 .7636 .8300 .8549 0.9011 .9186 .9431 .9528
-2: .3226 .5100 .6067 .6770 .7496 0.8061 .8461 .8788 .9045
-3: .2563 .4158 .5124 .5849 .6602 0.7237 .7717 .8129 .8474
-4: .1871 .3441 .4394 .5123 .5883 0.6543 .7077 .7529 .7940
-5: .1611 .2717 .3651 .4381 .5144 0.5816 .6390 .6890 .7351
-6: .1121 .2128 .3003 .3717 .4475 0.5150 .5754 .6280 .6779
-7: .0934 .1715 .2513 .3179 .3903 0.4555 .5162 .5700 .6220
-8: .0664 .1367 .2082 .2714 .3393 0.4026 .4625 .5167 .5694
-9: .0558 .1093 .1722 .2292 .2925 0.3525 .4108 .4647 .5177
If this table is accurate, then a player who beats his opponent
51% of the time in a 1-point match can expect to win 51.77%
of the time in a 9-point match.
There are some interesting features in this table. It says a
2-point match gives the stronger player no more advantage than
a 1-point match. This is true if the weaker player always
doubles at his first opportunity. It also says a 4-point match
is no better for the stronger player than a 3-point match.
This may be an exaggeration (coming from the assumption of both
players making efficient doubles), but it is something to think
about.
To summarize, let me compare the relative skill levels
estimated by each of these three methods:
1. The FIBS method.
2. The counting dice rolls method.
3. The match equity table method.
We'll start by assuming that a 1-point match gives the
stronger player 1.00 units of advantage. Here are the
relative advantages predicted by each method for longer
length matches:
FIBS Rolls MatEq
---- ----- -----
Match length 1: 1.00 1.00 1.00
Match length 2: 1.41 1.00
Match length 3: 1.73 1.33 1.24
Match length 4: 2.00 1.23
Match length 5: 2.23 1.63 1.44
Match length 6: 2.45 1.50
Match length 7: 2.64 1.89 1.62
Match length 8: 2.83 1.67
Match length 9: 3.00 2.16 1.77
Match length 10: 3.16 1.84
Match length 11: 3.31 2.34 1.92
Both the counting-rolls and match-equity-table methods
give less credit to longer length matches than FIBS does.
Or, another way of saying this, they give more credit to
1-point matches.
The question of how much greater advantage a stronger
player has at longer length matches is an interesting one.
I'd like to hear other people's ideas on this.
Tom
Julian
In article <yvkhg34sh...@cuchulain.it.kth.se>, Claes Thornberg
<cla...@cuchulain.it.kth.se> writes
>Why is it that you (on FIBS) gain more from a 2ptr than from a 1ptr,
>when we all know that in a perfect world a 2ptr is not different from
>a 1ptr? (see kit woolsey monograph 'How to play tournament backgammon')
A 2ptr is only the same as a 1ptr to players who have read Kit's book,
or numerous posts passim. If your opponent is unaware of the correct
cube strategy, you have a big advantage...
--
Julian Hayward 'Booles' on FIBS jul...@ratbag.demon.co.uk
+44-1344-640656 http://www.ratbag.demon.co.uk/
---------------------------------------------------------------------------
"A monk is expected to be awarded the contract for a 12.2 mile stretch
of the M4 motorway..." - Constructor's World
---------------------------------------------------------------------------
Chuck Bower
In article <3550B3...@bkgm.com>, Tom Keith <t...@bkgm.com> wrote:
(snip)
> Average Rolls Per Game in Big_Brother Matches
>
> rolls matches
> ----- -------
> Match length 1: 17329 / 359 = 48.27
> Match length 3: 54800 / 639 = 85.76
> Match length 5: 116451 / 907 = 128.39
> Match length 7: 86353 / 498 = 173.40
> Match length 9: 9992 / 44 = 227.09
> Match length 11: 8230 / 31 = 265.48
>
>Note that there are less than twice as many rolls in a
>3-point match as there are in a 1-point match. So, if
>counting rolls is a good indicator, it suggests that
>opportunity for skill is not really proportional to the
>length of the match.
(snip)
An interesting study. One thing, though, is that more
rolls give more chances for CHECKER PLAY skill (or lack of
skill) to be manifested. Longer matches give more chances
for CUBE PLAY skill. In the extreme (one point match) there
is obviously ZERO cube play skill involved. Could it be that
the Kaufman/Goulding/FIBS system takes this into account?
Robert-Jan Veldhuizen
On 06-mei-98 19:59:25, Tom Keith wrote:
[snip]
TK> Average Rolls Per Game in Big_Brother Matches
TK> rolls matches
TK> ----- -------
TK> Match length 1: 17329 / 359 = 48.27
TK> Match length 3: 54800 / 639 = 85.76
TK> Match length 5: 116451 / 907 = 128.39
TK> Match length 7: 86353 / 498 = 173.40
TK> Match length 9: 9992 / 44 = 227.09
TK> Match length 11: 8230 / 31 = 265.48
TK> Note that there are less than twice as many rolls in a
TK> 3-point match as there are in a 1-point match. So, if
TK> counting rolls is a good indicator, it suggests that
TK> opportunity for skill is not really proportional to the
TK> length of the match.
One can ask WHY the number of rolls in a 3ptr is less than TWICE that of
a 1ptr...My own estimate is that a 3ptr would take a little more than
two games on average to complete.
The answer is easy of course: dropped doubles. With f.i. a blitz attack
you can sometimes cash after only a few rolls have been played.
With longer matches, cubes will sometimes get turned to four or even
higher, reducing the number of rolls considerably, whether dropped or
not.
[snip]
TK> The question of how much greater advantage a stronger
TK> player has at longer length matches is an interesting one.
TK> I'd like to hear other people's ideas on this.
I think it's very difficult to generalise here. From my own experience,
there's a big difference between a match with players rated 1600 and
1800, and one with player rated 1400 and 1600.
Players from 1600 and up usually know how to use the cube, in theory at
least. They might often misjudge positions, but they know the basics
mostly.
OTOH, players with a rating of 1400 (and lower) sometimes move their
checkers relatively good, but don't understand the cube. Many play on
for the gammon, in almost every position for instance. Others take
every cube, or don't understand the importance of gammon threat when
doubled. And again others will drop every cube when they think they're
behind, even if it's only 60%/40% f.i.
Not trying to make a fool out of these people, but if you play a longer
match and notice this in the first game, you can often make a huge
advantage out of this, an advantage which seems to be much higher than
FIBS calculates.
If you just play a 1ptr however, you miss the doubling cube AND the
importance of (back)gammons (especially going for them is something
lower rated players almost never do) and your advantage is sometimes
very little, if they happen to move their checkers adequately.
In a match between a 1800 rated player and a 1600 rated player, the
difference probably will show up most in difficult match score
(re)cubes and prime battles or blitzes with sometimes (very) difficult
checker play and cubes.
However, the 1800 player will not always be able to "steer" the game
that way and may even be punished sometimes for "steering" too much! If
the dice dictate a more or less "standard game", he will probably not
have such a big advantage, both with his checker play and with doubles.
My own feeling is that a 1600 vs. 1400 will win more when playing 3ptrs
and up than a 1800 vs. 1600 will win.
With increased match length, something like 11ptrs and up, it might not
make much difference anymore, because of the increased chance at complex
games for the 1800 player and high cubes.
Just some thoughts... :)
--
Zorba/Robert-Jan
Jim Williams
> The above opinion has parallels in the "underdog strategy" ideas
> from gambling (which are discussed in "Oswald Jacoby on Gambling" and
> in the Jacobs/Trice 'Fish book' and many other places, I'm sure): if
> you are the underdog, go out of your way to put all of your chances on
> one toss, as opposed to trying to "grind it out" with lots of small
> bets. The underdog in a long match should try to get the cube elevated
> so that the dice can play a bigger factor than skill. Does the standard
> ratings formula ignore this with corresponding deleterious results? I
> open this up to the newsgroup for discussion.
This is an interesting topic. My intuition has always been that as the
weaker
player in a match, it made sense to be quite aggressive with the cube,
both offering early and taking late. This would seem to increase
volitility
and favor the weaker player. Likewise when playing a weaker player,
cautious
cube play would tend to prolong the match and allow more opportunity to
use skill and grind it out.
My experience when playing jellyfish (where I am a major underdog) seems
to indicate just the opposite. I have a much better chance of scoring
an
upset when being quite cautious with the cube. (This is with regard to
2 cubes. 4 and 8 cubes are another matter.) Although cautious cube
play
tends to increase the number of games in a match, the average length of
a game is reduced because there are more drops.
Also I find that the pre-cube positions tend to be much more familiar
and a lot easier to play correctly. Post-cube positions tend to be much
trickier, and I often get wildly out played by the fish.
- Jim