Janowski's formulas for cubeful equity
Joern Thyssen
Hi
I've got hold of three issues for Hoosier BG Club magazine from
1993-1994 (Volumn X, no. 6 (1993); Volumn XI, no. 1 (1994); Volume XI,
no. 2 (1994)) where Rick Janowski writes about take points in money
game.
I've tried to reproduce Rick Janowski's formulas but I have some
trouble.
W is the average points won pr. game, L is the average loss pr. game.
All the formulas are for money game without the Jacoby rule [Janowski
does give formulas for play with Jacoby w/o beavers, but let's not
complicate things more than necesary].
Take points:
Eq 1: dead cube take point: TP = (L-0.5)/(W+L)
Eq 2: live cube take point: TP = (L-0.5)/(W+L+0.5)
Eq 3: general take point: TP = (L-0.5)/(W+L+0.5x), where 0<=x<=1
(x=0 gives dead cube model, x=1 gives live cube model).
Cubeless take equity:
Eq 4: E_{take} = TP ( W + L ) - L
Equation 4 is just a special case of the cubeless equity:
E = p W - ( 1 - p ) L
= p ( W + L ) - L
with p = TP, you get Eq. 4.
Cubeful equity:
Eq 5: E_O = E_{I own cube} = Cv [ p (W+L+0.5x) - L]
Eq 6: E_{Opp own cube} = Cv [ p (W+L+0.5x) - L - 0.5x ]
Eq 7: E_{cent. cube} = 4 Cv / (4-x) [p (W+L+0.5x) - L - 0.25]
(Cv is the current value of the cube, p is the probability that you'll
win this game).
But Eqs. 5-7 is giving me trouble. Well, so far only Eq. 5, because I
didn't want to check the others before I could reproduce Eq. 5 [
Actually, Eq.
6 is easily derived from Eq. 5 (just interchance W <-> L, p <-> 1-p, and
voila!].
I would derive Eq. 5 as follows:
At p = 0: my cubeful equity is -L,
at p = TP: my cubeful equity is +1,
and do a linear interpolation inbetween.
Then I arrive at:
E_O = C_V [ p ( 1 + L ) / ( L + 0.5 + 0.5 x ) ( W + L + 0.5 x ) - L ]
whereas Janowski gets
Eq. 5: E_O = C_V [ p ( W + L + 0.5x ) - L ]
If I plug in p = 0 in Janowski's formula E_O = -L. Fine!
But with p = TP = 1 - (W-0.5)/(W+L+0.5x) (I'm using my opponents
takepoint, since I own
the cube) I get:
E_O = 0.5x + 0.5 (Janowski) != 1 (for x != 1 ).
E_O = 1 ("my" formula).
So, does any of you know how Janowski derives his Eq. 5?
I appreciate any help.
Joern Thyssen
(dirac at fibs)
Michael Crane
Also, see
http://www.msoworld.com/mindzine/news/classic/bg/match_equities.html
regarding the same subject from Nigel Merrigan n.mer...@cableinet.co.uk ,
who seems to understand it all.
Michael Crane
Joern Thyssen <j...@chem.ou.dk> wrote in message
news:39881471...@chem.ou.dk...
Phill Skelton
> Take points:
>
> Eq 1: dead cube take point: TP = (L-0.5)/(W+L)
> Eq 2: live cube take point: TP = (L-0.5)/(W+L+0.5)
> Eq 3: general take point: TP = (L-0.5)/(W+L+0.5x), where 0<=x<=1
> (x=0 gives dead cube model, x=1 gives live cube model).
>
> Cubeful equity:
>
> Eq 5: E_O = E_{I own cube} = Cv [ p (W+L+0.5x) - L]
> Eq 6: E_{Opp own cube} = Cv [ p (W+L+0.5x) - L - 0.5x ]
> Eq 7: E_{cent. cube} = 4 Cv / (4-x) [p (W+L+0.5x) - L - 0.25]
>
> (Cv is the current value of the cube, p is the probability that you'll
> win this game).
>
> I would derive Eq. 5 as follows:
>
> At p = 0: my cubeful equity is -L,
> at p = TP: my cubeful equity is +1,
> and do a linear interpolation inbetween.
>
> Then I arrive at:
>
> E_O = C_V [ p ( 1 + L ) / ( L + 0.5 + 0.5 x ) ( W + L + 0.5 x ) - L ]
But look at Jankowski's formula with x = 0:
E_O = Cv [ p W - (1-p) L]
which is the cubeless equity. This is different to what you do;
evidently
Jankowksi's work assumes that the quantity x also represents something
to do with the probability that you get to turn the cube. Your formula
assumes that you can offer a perfectly efficient cube and then your
opponent has a cube efficiency of x. Whereas if you both have an
efficiency
of x then in the dead cube situation (x = 0) you never get to offer a
double in the first place and so you end up with the cubeless equity.
However I have no idea how Jankowksi factors in the cube efficiency for
your double, but then I've only just spent the last 5 minutes looking
at it so that's no surprise. Hope that helps.
Phill
Phill Skelton
Jankowski's formulae:
> Take points:
>
> Eq 1: dead cube take point: TP = (L-0.5)/(W+L)
> Eq 2: live cube take point: TP = (L-0.5)/(W+L+0.5)
> Eq 3: general take point: TP = (L-0.5)/(W+L+0.5x), where 0<=x<=1
> (x=0 gives dead cube model, x=1 gives live cube model).
>
> Cubeful equity:
>
> Eq 5: E_O = E_{I own cube} = Cv [ p (W+L+0.5x) - L]
> Eq 6: E_{Opp own cube} = Cv [ p (W+L+0.5x) - L - 0.5x ]
> Eq 7: E_{cent. cube} = 4 Cv / (4-x) [p (W+L+0.5x) - L - 0.25]
>
> (Cv is the current value of the cube, p is the probability that you'll
> win this game).
Part of the problem is that your cube efficiency and your opponents may
well not be the same, or even close to each other. As I mentioned
before,
sticking x = 0 into the equations evidently implies that you can't
double
at all as it reproduces the cubeless equity. OTOH if you are able to
double, your opponent has a dead cube and p = 0.75 (i.e. you are at his
dead cube take-point, when W = L = 1) then your equity is clearly 1, and
you are offering a perfectly efficient double. The assumptions Jankowski
uses evidently means that the formula is not applicable to this
situation.
OTOH the approach taken by Joern makes a different assumption (that the
player holding the cube has a perfectly efficient double and that the
parameter x only applies to your opponent). In the real world x is
different
for each player and varies depending on the situation - you might try
to model it as a function of the volatility and the distance from the
ideal doubling point, but that begins to introduce even more variables
into the equations.
I think that the bottom line is that it's just not possible to come up
with a formula for cubeful equities that is generally applicable and
sufficiently simple to be useful, but perhaps the 'special case'
equities
may be good enough for many situations.
Phill
Michael Manolios
possibilities for you winning and losing the game? I 'm afraid I don't
understand that "W is the average points won pr. game, L is the average
loss pr. game". An arithmetic example would be of great help.
Could you also give the equities with Jacoby rule and Beavers? Is it
possible to find this stuff somewhere online?
And finally, I don't see any gammons taken into consideration. Does
these formulae imply that no gammons are involved or am I missing
something?
Thanks in advance,
In article <39881471...@chem.ou.dk>,
Joern Thyssen <j...@chem.ou.dk> wrote:
>
> Hi
>
> I've got hold of three issues for Hoosier BG Club magazine from
> 1993-1994 (Volumn X, no. 6 (1993); Volumn XI, no. 1 (1994); Volume
XI,
> no. 2 (1994)) where Rick Janowski writes about take points in money
> game.
>
> I've tried to reproduce Rick Janowski's formulas but I have some
> trouble.
>
> W is the average points won pr. game, L is the average loss pr. game.
> All the formulas are for money game without the Jacoby rule [Janowski
> does give formulas for play with Jacoby w/o beavers, but let's not
> complicate things more than necesary].
>
> Take points:
>
> Eq 1: dead cube take point: TP = (L-0.5)/(W+L)
> Eq 2: live cube take point: TP = (L-0.5)/(W+L+0.5)
> Eq 3: general take point: TP = (L-0.5)/(W+L+0.5x), where 0<=x<=1
> (x=0 gives dead cube model, x=1 gives live cube model).
>
> Cubeless take equity:
>
> Eq 4: E_{take} = TP ( W + L ) - L
>
> Equation 4 is just a special case of the cubeless equity:
>
> E = p W - ( 1 - p ) L
> = p ( W + L ) - L
>
> with p = TP, you get Eq. 4.
>
> Cubeful equity:
>
> Eq 5: E_O = E_{I own cube} = Cv [ p (W+L+0.5x) - L]
> Eq 6: E_{Opp own cube} = Cv [ p (W+L+0.5x) - L - 0.5x ]
> Eq 7: E_{cent. cube} = 4 Cv / (4-x) [p (W+L+0.5x) - L - 0.25]
>
> (Cv is the current value of the cube, p is the probability that you'll
> win this game).
>
> But Eqs. 5-7 is giving me trouble. Well, so far only Eq. 5, because I
> didn't want to check the others before I could reproduce Eq. 5 [
> Actually, Eq.
> 6 is easily derived from Eq. 5 (just interchance W <-> L, p <-> 1-p,
and
> voila!].
>
> I would derive Eq. 5 as follows:
>
> At p = 0: my cubeful equity is -L,
> at p = TP: my cubeful equity is +1,
> and do a linear interpolation inbetween.
>
> Then I arrive at:
>
> E_O = C_V [ p ( 1 + L ) / ( L + 0.5 + 0.5 x ) ( W + L + 0.5 x ) - L ]
>
> whereas Janowski gets
>
> Eq. 5: E_O = C_V [ p ( W + L + 0.5x ) - L ]
>
> If I plug in p = 0 in Janowski's formula E_O = -L. Fine!
>
> But with p = TP = 1 - (W-0.5)/(W+L+0.5x) (I'm using my opponents
> takepoint, since I own
> the cube) I get:
>
> E_O = 0.5x + 0.5 (Janowski) != 1 (for x != 1 ).
> E_O = 1 ("my" formula).
>
> So, does any of you know how Janowski derives his Eq. 5?
>
> I appreciate any help.
>
> Joern Thyssen
>
> (dirac at fibs)
>
--
Michael Manolios (mann on FIBS, Glass on GG)
We play one and only money game session through our whole life...
Sent via Deja.com http://www.deja.com/
Before you buy.
Oystein Johansen
Phill Skelton wrote:
> In the real world x is
> different
> for each player and varies depending on the situation - you might try
> to model it as a function of the volatility and the distance from the
> ideal doubling point, but that begins to introduce even more variables
> into the equations.
This is the real interesting part. Joern, does your article say anything
about this?
I have done some small experimets to calculate a good value for x, as
function off the position. This following values are based on Jellyfish
evaluations and conversations with a strong player.
Pure race: x ~ 0.65
Holding games: x ~ 0.5
Deep backgames: x ~ 0.4
Faceing an attack: x ~ 0.9
Last roll positions: x = 0.0 (obviously)
It is hard to estimate a good value for x in late bearoff positions. I
think maybe it should be increasing with the expected number of rolls
left for the accepting/droping player. Take a "~one-roll-each" position.
Then if this player expects to get off in 1.25 rolls, (27 good rolls - 9
bad rolls), the value of x should be 1.0.
Of course you can try to make this values more sophisticated. They're
far for perfect.
Best regards,
Øystein Johansen
Phill Skelton
> Joern could you define more clearly the W and L? Are they the
> possibilities for you winning and losing the game? I 'm afraid I don't
> understand that "W is the average points won pr. game, L is the average
> loss pr. game". An arithmetic example would be of great help.
>
> <snip>
>
> And finally, I don't see any gammons taken into consideration. Do
> these formulae imply that no gammons are involved or am I missing
> something?
AIUI the probability for winning the game is p, and by extension
the probability for losing it is 1-p. If I win, then the average
number of points per win is W (so if 20% of my wins are gammons and
80% are normal wins then W = 1.2). If I lose then on average I lose L
points.
So to grab an example from another thread, suppose I win a normal
game in 60% of games, win a gammon in 20% and lose a normal game 20%
of the time, the overall I win 80% of games, so p = 0.8. A quarter
of my wins are gammons (80% wins of which 20% are gammons) so
W = (0.75 * 1) + (0.25 * 2) = 1.25.
All my losses are normal losses so L = 1.
HTH
Phill
Joern Thyssen
>
> Phill Skelton wrote:
> > In the real world x is
> > different
> > for each player and varies depending on the situation - you might try
> > to model it as a function of the volatility and the distance from the
> > ideal doubling point, but that begins to introduce even more variables
> > into the equations.
>
> This is the real interesting part. Joern, does your article say anything
> about this?
I quote:
"Finding accurate values for x is a difficult, almost impossible task.
However, we can make estimates of /typical/ values for /typical/
situations. In my opinion, for the majority of /typical/ positions, x
will commonly be between 1/2 and 3/4, with 2/3 being a /normal/ value."
>
> I have done some small experimets to calculate a good value for x, as
> function off the position. This following values are based on Jellyfish
> evaluations and conversations with a strong player.
>
> Pure race: x ~ 0.65
> Holding games: x ~ 0.5
> Deep backgames: x ~ 0.4
> Faceing an attack: x ~ 0.9
Sound reasonable. They are (almost) in the range given in the quote
above.
>
> Last roll positions: x = 0.0 (obviously)
:-)
>
> It is hard to estimate a good value for x in late bearoff positions. I
> think maybe it should be increasing with the expected number of rolls
> left for the accepting/droping player. Take a "~one-roll-each" position.
> Then if this player expects to get off in 1.25 rolls, (27 good rolls - 9
> bad rolls), the value of x should be 1.0.
>
> Of course you can try to make this values more sophisticated. They're
> far for perfect.
It's probably reasonable to use x = 2/3, since a n-ply (n>0) evaluation
would "catch" last-roll, almost last-roll, up to (n-1) last roll
situations.
Joern Thyssen
Oystein Johansen
Michael Manolios wrote:
>
> Joern could you define more clearly the W and L? Are they the
> possibilities for you winning and losing the game? I 'm afraid I don't
> understand that "W is the average points won pr. game, L is the average
> loss pr. game". An arithmetic example would be of great help.
> Could you also give the equities with Jacoby rule and Beavers? Is it
> possible to find this stuff somewhere online?
> And finally, I don't see any gammons taken into consideration. Does
> these formulae imply that no gammons are involved or am I missing
> something?
Hi Michael,
Take a look at this article:
http://x70.deja.com/[ST_rn=ps]/getdoc.xp?AN=520127604&CONTEXT=965308405.525336606&hitnum=1
I think this should explain the most of what Joern is talking about!
Regards,
Øystein Johansen
Joern Thyssen
>
> Joern could you define more clearly the W and L? Are they the
> possibilities for you winning and losing the game? I 'm afraid I don't
> understand that "W is the average points won pr. game, L is the average
> loss pr. game". An arithmetic example would be of great help.
W = p(win) + p(win gammon/backgammon) + p(win backgammon)
-----------------------------------------------------
p(win)
Analogous for L.
For example, in a pure race (with no contact):
W = p(win) = 1
------
p(win)
Suppose a evaluation gives:
win 50%
g/bg 11%
bg 0%
(a typical opening situation)
W = 50 + 11 = 1.22
-------
50
> Could you also give the equities with Jacoby rule and Beavers?
Nope :-( Because they are not given in the articles. However, the
article does give formulae for take, beaver, racoon, initial double,
redouble, cash, and too good points. Only the initial double points are
affected by Jacoby/no Jacoby and beaver/no beaver options.
> Is it
> possible to find this stuff somewhere online?
I don't know. I've asked the author. It is possible to order magazines
where the articles appeared. You can email me if you want to know which
persons to contact. They cost a few dollars each plus postage.
> And finally, I don't see any gammons taken into consideration. Does
> these formulae imply that no gammons are involved or am I missing
> something?
It's included in W and L.
Joern
Joern Thyssen
[snip]
> Cubeful equity:
>
> Eq 5: E_O = E_{I own cube} = Cv [ p (W+L+0.5x) - L]
> Eq 6: E_{Opp own cube} = Cv [ p (W+L+0.5x) - L - 0.5x ]
> Eq 7: E_{cent. cube} = 4 Cv / (4-x) [p (W+L+0.5x) - L - 0.25]
>
> (Cv is the current value of the cube, p is the probability that you'll
> win this game).
After personal communication with Rick Janowski I now know how to derive
these formulae:
E = (1-x) (E_dead) + x (E_live)
where
E_dead = p * (W+L) - L (equal to the cubeless equity)
E_live = p * (W+L+0.5) - L
(by doing linear interpolation between (p=0,E=-L) and (p=opponent take
point=1-(W-0.5)/(W+L+0.5)=(L+1)/(W+L+0.5), E=+1).
This gives
E_O = (1-x) ( p (W+L) - L ) + x ( p (W+L+0.5) - L )
= p (W+L+0.5x) - L
Joern
Michael Crane
word of it.
Michael
Joern Thyssen <j...@chem.ou.dk> wrote in message
news:398977C6...@chem.ou.dk...
Øystein Johansen
>
> I quote:
>
> "Finding accurate values for x is a difficult, almost impossible task.
> However, we can make estimates of /typical/ values for /typical/
> situations. In my opinion, for the majority of /typical/ positions, x
> will commonly be between 1/2 and 3/4, with 2/3 being a /normal/ value."
>
> > I have done some small experimets to calculate a good value for x, as
> > function off the position. This following values are based on Jellyfish
> > evaluations and conversations with a strong player.
> >
> > Pure race: x ~ 0.65
> > Holding games: x ~ 0.5
> > Deep backgames: x ~ 0.4
> > Faceing an attack: x ~ 0.9
>
> Sound reasonable. They are (almost) in the range given in the quote
> above.
I think the values for pure race and holding game is quite accurate. I
was actually thinking of writing = instead of ~ for pure race, but I
changed my mind because I think it can be as high as .70 for longer
races. (>110 pips) Maybe a linear regression based on the pips whould be
good? This result is based on JF analyses and expert comments.
Holding game is also quite accurate. The value must of course be less
than the value for a race, since in many single shot situation, you
can't recube intil you hit a shot, and when hit it's a drop.
I'm much less certain about the values presented for deep backgames and
attacks, and this are the values outside the range of the quoted text. I
have not checked anyone of this with JF (just based on expert comments),
but I don't think this will have much value anyway, since backgames is
not the strong side of JF. The recube in such positions is also usually
many many rolls away so the value is very hard to estimate, and maybe my
consulting expert is wrong.
> It's probably reasonable to use x = 2/3, since a n-ply (n>0) evaluation
> would "catch" last-roll, almost last-roll, up to (n-1) last roll
> situations.
Will such evaluation "catch" the right cube action in the Jacoby paradox
position? My doubling routine does not, but that may be because I
evaluate the volatility based cubeless equities.
> Joern Thyssen
Oystein Johansen