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Oct 20, 1997, 3:00:00 AM10/20/97

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13 14 15 16 17 18 19 20 21 22 23 24

+------------------------------------------+ X: Jellyfish - score: 5

| X X X | | X X X X X |

| X | | X X X |

| | | X |

| | | X |

| | | |

v| |BAR| | 9-point match

| | | |

| O | | |

| O | | O O O |

| O | | O O O O |

| X O | | O O O O |

+------------------------------------------+ O: Me - score: 5

12 11 10 9 8 7 6 5 4 3 2 1

Pipcount O:83 X:95

JF3.0 Level 7 Evaluation:

Wins

JF 19.7

Me 80.3

Equity Me: 0.605

Volatility: 0.189

Cube Action: Double/Take

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

I guess I am bit confused over this. How can this be a take with such

slim winning chances?

Oct 21, 1997, 3:00:00 AM10/21/97

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MJR <hac...@ptd.net> writes:

> 13 14 15 16 17 18 19 20 21 22 23 24

> +------------------------------------------+ X: Jellyfish - score: 5

> | X X X | | X X X X X |

> | X | | X X X |

> | | | X |

> | | | X |

> | | | |

> v| |BAR| | 9-point match

> | | | |

> | O | | |

> | O | | O O O |

> | O | | O O O O |

> | X O | | O O O O |

> +------------------------------------------+ O: Me - score: 5

> 12 11 10 9 8 7 6 5 4 3 2 1

(You don't show the cube; I assume it's centred at 1. If you were holding a

2 cube, it would be a clear redouble/drop).

> Pipcount O:83 X:95

>

> JF3.0 Level 7 Evaluation:

>

> Wins

> JF 19.7

> Me 80.3

>

> Equity Me: 0.605

> Volatility: 0.189

> Cube Action: Double/Take

>

> I guess I am bit confused over this. How can this be a take with such

> slim winning chances?

At 4-away, 4-away, the take (of a 2 cube) is a little bit easier because

of the excellent recube vig at this score -- Jellyfish can redouble you later

and give you a dead cube. Admittedly with only 19.7% winning probability

the take looks pretty borderline regardless, but here's the maths:

If Jellyfish drops, it ends up at 4-away, 3-away with match equity of

42% (depending whose tables you use). If it takes, it should redouble to

kill the cube the moment its game equity is greater than its match equity

if it plays without doubling (to bring the game to 4-away, 2-away with

match equity 68% to the leader). Since its redoubles won't be perfectly

efficient, let's assume it's forced to redouble when it has 2:1 odds for

winning, to keep the maths simple. In that case, since it has 19.7% winning

chances, we can assume it reaches its redoubling point in 29.6% of the

games. The results then are expected to be 70.4% losses; 9.9% redoubled

losses; and 19.7% redoubled wins. So its total match equity is:

Takes and loses: match equity 0.32 x 0.704 = 0.225

Takes, redoubles and loses: match equity 0.00 x 0.099 = 0.000

Takes, redoubles and wins: match equity 1.00 x 0.197 = 0.197

For a total match equity of 42%... exactly the same as if it had dropped!

Oh well...

(In practice the match equities will vary somewhat depending on the gammon

rate you assume, the relative strengths of the players, etc., and also the

probability of "takes, redoubles and loses" will depend on how efficiently

Jellyfish can redouble. But when it comes down to it, the situation is a

very close take/drop with little equity lost either way). So (assuming

Jellyfish's evaluation of the game equity is reasonable), it's a clear

double, and borderline take/drop.

Cheers,

Gary (GaryW on FIBS).

--

Gary Wong, Computer Science Department, University of Auckland, New Zealand

ga...@cs.auckland.ac.nz http://www.cs.auckland.ac.nz/~gary/

Oct 22, 1997, 3:00:00 AM10/22/97

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In article <ieyhgab...@cs20.cs.auckland.ac.nz>,

Gary Wong <ga...@cs.auckland.ac.nz> wrote:

>MJR <hac...@ptd.net> writes:

>> 13 14 15 16 17 18 19 20 21 22 23 24

>> +------------------------------------------+ X: Jellyfish - score: 5

>> | X X X | | X X X X X |

>> | X | | X X X |

>> | | | X |

>> | | | X |

>> | | | |

>> v| |BAR| | 9-point match

>> | | | |

>> | O | | |

>> | O | | O O O |

>> | O | | O O O O |

>> | X O | | O O O O |

>> +------------------------------------------+ O: Me - score: 5

>> 12 11 10 9 8 7 6 5 4 3 2 1

>

>(You don't show the cube; I assume it's centred at 1. If you were holding a

>2 cube, it would be a clear redouble/drop).

>

>> Pipcount O:83 X:95

>>

>> JF3.0 Level 7 Evaluation:

>>

>> Wins

>> JF 19.7

>> Me 80.3

>>

A nice analysis by Gary. There are a couple things worth questioning,

though. First, it looks like Gary has assumed a "continuous model". This

means that a player with access to the cube will either NEVER get to a cash,

or land exactly on the cash point (that is, NEVER LOSE a market). Actually

Gary made a slight adjustment: instead of landing exactly on the "cash

point", X lands exactly on O's "take point" in the cases where s/he turns

the game around. But this still assumes very high cube efficiency, and

is virtually equivalent to the continuous model. I now assume a purely

continuous model (equivalently, perfect cube efficiency):

O's drop point for the 4-cube is 68% cubeless (as seen by X). X is now

at 19.7%, so s/he must go 68 - 19.7 = 48.3 forward before going 19.7

backward (which would be a loss!). The chances of that happening in the

continuous model is 19.7 / (19.7 + 48.3) = 29.0%. So 29% of the time

X will reach the drop/take point. Since by definition, if this happens

the win/loss result will be the same regardless of whether or not O

takes or drops, we can assume "drop". (Of course the remainder of the

time, that is 71%, X's chances decrease to zero before ever getting the

chance to cash.)

29% * 0.68 + 71% * 0.32 = 42.4%

(Compare this with 42.2% in Gary's model, almost the identical result.)

Gary asssumed that a -4, -3 score was 42%, so you can see that WITH PERFECT

CUBE EFFICIENCY it's pretty much of a coin toss (as Gary concluded).

Can we do better than assuming perfect cube efficiency? Yes. One

way is to assume a semi-continuous model and determine the drop/take

point as somewhere between "zero cube efficiency" (that is, assume if

X were to take, then s/he will NEVER redouble, no matter how good) and

"perfect cube efficiency". For races of this length, 60% cube efficiency

seems about right. Let's do that, assuming the same match winning chances

as Gary did:

X drops cube -4, -3 42%

compare to:

X takes and wins -2, -4 68%

X takes and loses -4, -2 32%

WITH ZERO CUBE EFFICIENCY: X risks 42 - 32 = 10 to gain 68 - 42 = 26.

Winning chances required to take would then be 10 / (10 + 26) = 27.8%.

WITH PERFECT CUBE EFFICIENCY: X only has to get to 68% winning chances.

So 68% * 0.28 = 19%. (NOTE THAT THIS IS CONSISTENT WITH THE ABOVE

CONTINUOUS MODEL ANALYSIS. DROP/TAKE POINT IS VERY CLOSE TO THE "ACTUAL"

GAME WINNING CHANCES OF 19.7%, AND IT FALLS ON THE "TAKE" SIDE OF THE

FENCE.)

60% cube efficiency means go 60% of the way from "ZERO CUBE" (27.8%) to

"PERFECT CUBE" (19.7%) or 23% game winning chances. (NOTE now that this

is very close to the money drop/take point!)

This "semi-continuous" match cube model therefore says that the above

position is really a pass.

But, can we do better? How about using Jellyfish level-5 limited

cube rollouts! If these are ever of value, they ought to work here!!

First thing, though, is to CAREFULLY choose the "settlement limit".

At or above the "settlement limit" and cube holder will cash the game.

We already said (many times!) that X will be able to cash 2 points when

s/he gets to 68% game winning chances. Because real backgammon isn't

continuous, a player is better off doubling a little early than a little

late (and "losing the market"). To offset this, usually one goes slightly

lower than the true drop/take line to determine the settlement limit.

Gary said 2::1 was reasonable. This may still be a tad high, but heck-if-

I-know, so let's go with that. Of course "settlement limit" is in cubeless

equity units ==> 0.6667 - 0.3333 = 0.333. So that is what you plug in to

JF level-5.

I ran 20,736 games at level-5 and got the following results:

X (player not on roll) owns cube:

X wins 23.6%

O wins 76.4%

(standard deviation is 0.3%)

So, assuming a take, X wins the match:

23.6% * 0.68 + 76.4% * 0.32 = 40.5% of the time.

If you compare this to the assumed 42% MWC by dropping, then you

see that this is a close pass.

Normally it's a good idea to also run level-6 cubeless and compare

with the level-5 cubeless. I did 180 trials and got O winning 80.8%

cubless (with a standard deviation of 0.25%). The level-5 cubeless

result also had O winning 80.8% of all games, with a std. dev. of 0.27%.

BTW, if we use these cubeless results in the continuous model, then

the take point would be 42.2%, not much different than Gary and I got

with that model and JF level-7 evaluation.

Finally we get back to hacksaw's original question: "How can this

be a take" or reworded: "why does JF take?" I don't know. Maybe

Fredrik can answer (he knows JF's match cube algorithm!). Possibly JF

uses a different match equity table than Gary and I assumed. Maybe

JF assumes a higher cube ownership equity. In any case, it is a pretty

close call, but my $ is on "pass".

Chuck

bo...@bigbang.astro.indiana.edu

c_ray on FIBS

Oct 22, 1997, 3:00:00 AM10/22/97

to

SNIP...

> Normally it's a good idea to also run level-6 cubeless and compare

>with the level-5 cubeless. I did 180 trials and got O winning 80.8%

>cubless (with a standard deviation of 0.25%). The level-5 cubeless

>result also had O winning 80.8% of all games, with a std. dev. of 0.27%.

>BTW, if we use these cubeless results in the continuous model, then

>the take point would be 42.2%, not much different than Gary and I got

>with that model and JF level-7 evaluation.

>

Would someone please explain why it's a good idea to compare level-5 to

level-6 to level-7? Since JF is supposed to be strongest at level-7, why

bother with level-5 & 6?

Thanks in advance,

Ed Mooney

Oct 26, 1997, 2:00:00 AM10/26/97

to

Nice post! Thanks.

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