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## Re: [Bug-gnubg] EMG article

 From: Massimiliano Maini Subject: Re: [Bug-gnubg] EMG article Date: Mon, 10 Dec 2007 16:30:24 +0100

> Hi all;
> Hi Max,
>
> Thanks for providing the link to my paper:
>

>> Hi all,
>> you can find here an interesting article on EMG (Equivalent to Money
>> Game) equities.
>>
>>                         http://www.fortuitouspress.com/emg.html

>
> I had posted to GammonU and was about to post here but you beat me
> to it.  Thanks also for your introduction and summary of the issue.
> I hope people on this list get interested in the problem.
>
>
>> I've made a suggestion which wouldn't be too complicate to put in place:
>>
>> 1- let's call W1/2/3 (L1/2/3) the MWC at the scores of a
>> single/gammon/backgammon win (loss)
>> respectively. They are associated to NE (Normalized Equities)of
>> +1/2/3 (-1/2/3) respectively.
>> The six points [L3,-3], [L2,-2], ... , [W3,+3] form a poly-line with
>> 5 segments (at most,
>> at some scores two point may be identical because
>> gammons/backgammons may not count).
>>
>> 2- draw the poly-line, then use it to convert MWC to NE.
>>
>> It's like having a different interpolation depending on the
>> magnitude of the error you're
>> trying to normalize.
>>
>> Three examples:
>>
>> - I'm leading 3-0 to 5 cube at 1, what can happen ? With a
>> simple/gammon/backgammon win I go
>> to 4-0/5-0/5-0 while with a simple/gammon/backgammon loss I go to 3-1/3-2/3-3.
>>
>> - I'm leading 4-1 to 5 post-Crawford (I owe the cube at 2), what can
>> happen ? With a simple/
>> gammon/backgammon win I go to 5-1/5-1/5-1 while with a
>> simple/gammon/backgammon loss I go to
>> 4-3/4-5/4-5.
>>
>> - I'm leading 3-0 to 5 owing the cube at 2, what can happen ? With a
>> simple/gammon/backgammon
>> win I go to 5-0/5-0/5-0 while with a simple/gammon/backgammon loss I
>> go to 3-2/3-4/3-5.
>>
>> In any of the above situation, just associate the w/wg/wb scores
>> with NNE +1/+2/+3 and the
>> l/lg/lb scores with NNE -1/-2/-3, reads the MWC of the different
>> MET, put the points on a graph and draw the poly-line (attention: in
>> some cases you have to
>> use post-Crawford METs).
>>
>> Upside:
>>         - it solves the issue above: all the 3 errors wil have the
>> same normalized equity
>>         - for "small errors" (leading to MWC that are in the
>> interval [single loss, single win]),
>>           my suggestion would return the good old EMG.

>>
>>
> Can you please provide a full numerical example of your proposal?
> You say that "it solves the issue above:  all the 3 errors wil have
> the same normalized equity," but you don't say what that equity will
> be.  Can you show the calculation?

Hi Jeremy, I will take your example from the beginning, to make this post self-contained.
I'll use G11 MET, like in your article, to get the very same figures.

You presented 3 situations where black faces a take/pass decision:

Situation A: black is led 3away-2away and is doubled to 2

Situation B: black is 3away-3away and is redoubled to 4

Situation C: black leads 3away-5away and is redoubled to 8

In all your 3 cases, black has cubeless GWC of 19.822% (no gammons, no backgammons),
and the cube is dead after this decision (in situation A black has a mandatory double
the next turn). White having doubled, the alternatives for black are:

- pass and go 3away-1away (crawford), with 24.923% MWC
- take and play for the match with 19.822% GWC, i.e. 19.822% MWC

Hence, the error in all three situations is (19.822%-24.923%) = -5.101% MWC.

Now in order to normalize this, the usual procedure (EMG) is the following:

- draw the line passing via the points (ML,-1) and (MW,+1) where ML and MW are,
respectively, the MWC of a simple loss and win
- use the drawn line to obtain the EMG of the decision

If you put on X axis the MWC and on Y axis the EMG, the equation of the line passing
via the points (ML,-1) and (MW,+1) is:

y = ((+1 - (-1))/(MW-ML))*(x-ML) - 1 = (2/(MW-ML))*(x-ML) - 1

The above line depends on the current score:

Situation A (black is led 3away-2away and is doubled to 2):
- a simple loss will give black a score of 3away-1away, or 24.923% MWC = ML
- a simlpe win will give black a score of 2away-2away, or 50.000% MWC = MW
Situation B (black is 3away-3away and is redoubled to 4):
- a simple loss will give black a score of 3away-1away, or 24.923% MWC = ML
- a simlpe win will give black a score of 1away-3away, or 75.077% MWC = MW
Situation C (black leads 3away-5away and is redoubled to 8):
- a simple loss will give black a score of 3away-1away, or 24.923% MWC = ML
- a simlpe win will give black a win in the match, or 100.000% MWC = MW

Substituting the ML and MW values in the equation and computing for x = 19.822% MWC
you will find the following EMG:

Sit. A:        EMG(Take) = -1.4068 ==> EMG error = -0.4068
Sit. B:        EMG(Take) = -1.2034 ==> EMG error = -0.2034
Sit. C:        EMG(Take) = -1.1359 ==> EMG error = -0.1359

The above is exactly what you exposed in your article. My point is that, in all three
cases, the take decision is so bad that the corresponding MWC are outside the interval
used for the normalization. In fact the 3 cases have the same left end-point for the
normalization interval (MWC of a simple loss = 24.923% in all cases) while the MWC of
a take are 19.822% < 24.923%.

The idea is then to use another interval, the one with end-points MWC of a double loss
(or gammon loss) and MWC of a simple loss, associated respectively to -2 and -1
"new normalized equities" (NNE). Notice that the rightmost point of this new segment
is the leftmost point of our original EMG segment.

In all 3 cases, if black loses a gammon *at the current cube level* (i.e. 1 in situation
A, 2 in B and 4 in C), black loses the match. This means that, in all three cases, the
interval used for the normalization is given by the two points:
- 00.000% MWC, -2 NNE
- 24.923% MWC, -1 NNE

Since the segment is the same in all three cases, the NNE of the take will be the same.
To compute its value you need to write the equation of of the line pasing via the points
(MLg,-2) and (ML,-1), where MLg are the MWC after a gammon loss:

y = ((-1 - (-2))/(ML-MLg))*(x-MLg) - 2 = (1/(ML-MLg))*(x-MLg) - 2

For ML = 24.923%, MLg = 00.000% and x = 19.822%, the NNE is (in all 3 cases) -1.2047,
which says that the normalized error wrt a pass is -0.2047.

I've attached a plot of the poly-line in the 3 cases (X and Y axes are swapped, I have MWC
on Y and normalized equities on X, because Excel sucks): this graphically represents the
computations.

MaX. NNE.png
Description: Binary data