market loser's model
Serkan Hosten
This is posted from a friends account. Personal replies to
ko...@orie.cornell.edu
,Bob Koca (bobk on FIBS)
To get an idea of how large/many market losers must be
to warrant a double I considered a model which is simple
enough to be analyzed exactly. I was a little surprised at how
nice the final answers are.
The Model: Suppose one is playing a doubling cube game in which all wins
score the value of the cube. (e.g. if playing backgammon disregard
gammons and backgammons) Suppose that after the next turn
the game will turn into a continuous game ( ala Zadeh and Kablisky).
Suppose that player on turn has cubeless chances x and that this turn
will place him at some new cubeless chances which will be initial state
of the continuous game. Assume that average of these equals x (so that
cubeless chances genuinely are x at start).
BACKGROUND INFO ABOUT CONTINUOUS GAMES: In a continuos game,
one should double or redouble when one reaches .8 cubeless chances. The
opponent has an optional accept/reject so for ease of calculations assume
it is a reject.
Note that if he went above .8, he lost his market, and if he goes below
.2, opponent lost market. To quantify a combination of # of and size of
the market losers define m and p as follows:
m= average over rolls of max{ 0, new cubeless chance - .8}
p= average over rolls of max{ 0, .2- new cubeless chances}
for example, if x=.6 and 50% of rolls win game immediately,
25% goes to 10%, and 25% to 30%, then
m=.1 and p=.025
If double, equity = 5x+5p-3
If own cube and hold it, equity = 2.5x-2.5m-1
If neutral cube and hold it, equity = 10(x+p-m-.5)/3
>From this it follows that:
initdouble if 2m + p > .8 - x
redouble if m + 2p > .8 - x
(Aint that nice)
Proof: Algebra after the following observation:
It is equivalent to consider instead new cubeless chances such that
goes to 100% with probability 5m
(x-5m)/(1-5m-5p) with probabiltiy 1-5m-5p
0% with probabiltiy 5p
Note that m and p are as defined and overall average is x.
Since equity calculations are linear in the 3 pieces [0,.2[ , [.2,.8] , ].8,1]
we may first dilate everythin in ].8,1] to {.8,1} keeping m constant.
Then do similar thing with [0,.2[. Now take what is remaing (including "new"
mass on .2 and .8 and dilate to a single point.
e.g.
10% to 90 5% to 100 and 5% to 80 5% to 100 and 5% to 80
70% to 60 = 70% to 60 = 70% to 60
20% to 5 20% to 5 15% to 0 and 5% to 20
5% to 100
= 80% to 42/.80
15% to 0
QED
Now the hard part is applying this to real life backgammon!
Serkan Hosten
This is posted from a friends account. Personal replies to
ko...@orie.cornell.edu
,Bob Koca (bobk on FIBS)
Magriel and Robertie both claim that it is always optimal to use a loose
1 to bear off if possible in a contactless position. Does anyone know
a nice rigorous proof of this. I tried and came close.
What I have is below with the missing step indicated.
It seems like there should be a proof other than doing massive
calculations on a computer. Besides,
1) I think Magriel's book is from before this was done on computer
2) Was only done for a dead cube and for all pieces in the home boards
THEOREM: Optimal to use a single 1 to bearoff in a contactless position
PROOF: Let Position A be that which results from using a single 1 to
bear off and position B be that which results from using the 1 differently.
Mark this piece as * in both positions.
Thus B will have an extra piece on 1 and somewhere will have a piece
advanced one more pip.
e.g.
Position A: Position B
* X X X
X X X X X X * X X X X
6 5 4 3 2 1 6 5 4 3 2 1
Suppose that an optimal player plays out B and suppose these
two games are played out with same dice.
Then the following method of playing position A shows that position A
wins whenever position B wins.
Move position A by observing B and doing following.
If B moves * then A moves *.
If B bears off from 1, then A advances * up one space.
O.W just move same pieces.
Eventually one of following must happen. Stop the above rule and
observe the position as soon as first of following happens.
1) B bears off from 1, in this case our rule has generated exactly
the same position.
2) B bears off * which results in A having * on 1. Again our rule has
generated exactly the same position.
3) B bears off * which results in A bearing off * also.
Now the positions are the same except that B has one additional
piece on 1 instead of off. (This position is clearly superior
for A) **** PROOF?!?!******
thus position A is at least as good as position B.
,Bob Koca
Christopher Yep
Anthony R Wuersch <a...@world.std.com> wrote:
>In article <1994Mar18....@cs.cornell.edu>,
>Serkan Hosten <ser...@cs.cornell.edu> wrote:
>>
>>This is posted from a friends account. Personal replies to
>>ko...@orie.cornell.edu
>>,Bob Koca (bobk on FIBS)
>
Hmm, well some of the implied conclusions may be useful.
>The .8 and .2 bounds for market losers and takes assume
>a continuous game, but a continuous game means that there are *no* market
>losers.
Agreed.
>
>A well known cubeless 18.75% take position is where the doubler and the
>taker each have a single checker on their respective six points.
Ah! But it's also a drop... see below...
>
>If the cube is for all intents dead, we cannot take below 25%.
>
Agreed.
>So a 20% bound for continuous takes can be lower or greater in the real
>discontinuous world.
>
>Zadeh and Kobliska introduce a "discontinuity adjustment," a small fudge
>factor which should increase as pip totals decrease. It causes the drop
>point to rise from its continuous 20% level up to a maximum 25%.
>
>But the 18.75% take position contradicts this increasing trend.
The position in which both the doubler and his opponent each have a single
checker on their respective six points is either a drop or a take -- the
equity for the doubler's opponent will be the same in both cases. So, if
we define all take/drop cube decisions to be "drops" in the cases where
the equity is equal after both a "take" or a "drop," then what can we say
about the take/drop point in general?
In particular, for gammonless positions, will it ever be less than 20%?
In particular, is there any gammonless position in which Player 1 has
cube access , Player 2 has < 20% winning chances, and if Player 1
doubles, Player 2 is actually _worse_ off by dropping? I don't think so,
although I could be wrong. The lowest winning chance mandatory take
position that I can think of is the position in which both players have
two men on their respective two points (260/1296 = 20.06% winning chance
for the player being doubled).
>
>We're still left with the question of what is the take (eg. market loser)
>point for a concrete position, before we can make any pretty calculations
>about market losers. All I know is that it's not always 20%, and it will
>never exceed 25% in a money game.
>
>We're also still left with a question of how to estimate the probability
>that a cube taker will *really* gammon the doubler. Cubeless probability
>of gammoning cannot be enough (for me at least) to answer this question.
>
>How Zadeh and Kobliska estimated their fudge factor is also a question.
>
>Estimates which incorporate position features (blot hit prob, degree of
>contact, prime blockage, etc) not just pip count is what we really need
>to answer questions of take point, gammon prob, and thereby doubling or
>redoubling points.
>
>Cheers,
>Toni
>--
>Toni Wuersch
>a...@world.std.com {uunet,bu.edu,bloom-beacon}!world!arw
I believe that while the "market loser model" posted a few days ago can
never be used as "book" in any game of backgammon, that some of the
implicit conclusions are useful. One is that, in general, the more
volatile a position is, the less of an edge one needs to justify doubling
(this is a slight simplification). (Last-roll positions would be an
example of taking volality to an extreme).
The model is reasonably useful for positions in which the volality on the
next roll is very high, but comparatively low for all rolls after that.
Of course, this doesn't represent very many positions at all.
The model could be changed to assume that the next 5-6 rolls are very
volatile, with the game drifting into an "almost-continuous" game after that.
I agree that this still wouldn't be that accurate for most positions, and
that it would be less of a model and more of a brute-force
number-crunching algorithm. Oh well... Maybe Bob's model would have
some use outside of backgammon.
Bye,
Chris