[gwnn]

I wonder if there's a better IMP scale (i.e. it would reproduce the ideas of the original IMP scale truer) to use if you have to compare to the datum score. Imagine you're stranded in a jungle and you have no computers available to calculate x-imp easily (but you can quickly average all results). Comments? Ideas on how to calculate something like this?

Note that there is no element in this post that suggests that such a scale should be used, or it wouldn't be confusing to many people, or it would be readily accepted. It would be awesome if we didn't debate these points here.

[phil_20686]

I thought about this briefly a few years ago.

Suppose that the IMP scale is roughly logarithmic, then for score x_(1), butlering and cross imping yield:

\Log(x_{1}-\frac{1}{n}\sum_{i}x_{i}) and \frac{1}{n}\sum_{i}\Log(x_{1}-x_{i}) respectively.

Now if it is roughly logarithmic, then its inverse is roughly exponential, so if I exponential these two results I get x_{1}-\frac{1}{n}\sum_{i}x_{i} for both, which is why cross imping and bultering always basically give the same results. Of course, the imp scale is not perfectly smooth, and the binning introduces some errors, and I have no idea how close they are to perfectly exponential, but if the imp scale was smooth and perfectly logarithmic, then butlering and cross imping are identical.

PS apologise for latex-speak, but assume most who care about this can read latex speak......

[phil_20686]

So to answer your question, having worked out that to a first approximation it didnt matter, I opened a physics book instead.

[WellSpyder]

 gwnn, on 2012-February-14, 10:54, said:

I wonder if there's a better IMP scale (i.e. it would reproduce the ideas of the original IMP scale truer) to use if you have to compare to the datum score. Imagine you're stranded in a jungle and you have no computers available to calculate x-imp easily (but you can quickly average all results). Comments? Ideas on how to calculate something like this?

 phil_20686, on 2012-February-14, 12:25, said:

I thought about this briefly a few years ago....

Anyone else think in their less sane moments that this deserves a nomination for post of the year????

[phil_20686]

 WellSpyder, on 2012-February-14, 12:35, said:

Anyone else think in their less sane moments that this deserves a nomination for post of the year????

lol. I stared at this for ages before I realised what was funny. Now I am almost wetting myself.

[gwnn]

I don't understand what you're doing phil, I did the same but to me it looks like
e^{Butler} \approx (arithmetic mean of differences), and
e^{XIMP} \approx (geometric mean of differences)

This assumes that all differences are positive, i.e. we have the best or worst score in the whole field.

Is your 'first approximation' that arithmetic~geometric mean? That is very brutal, since in the most common situation most results are in clusters (e.g. 30% around +620 30% around -100 and 40% around +170). The arithmetic and geometric means are only similar if most values are very close to each other.

[Fluffy]

I am still wondering, need more practice with english humor

[gwnn]

 Fluffy, on 2012-February-15, 06:27, said:

I am still wondering, need more practice with english humor

The point was that my post asked a math question but then I started talking about the jungle and some such nonsense. Phil replied "oh yea I've been wondering about that too" but you could read it like "yea I've been wondering about how I'd direct an IMP pairs bridge tournament in a jungle".

[gwnn]

OK let's assume there is a XIMP board where x% of the pairs got 0, and (100-x)%, including us, got a score of +A (good for us that we don't play those super-conservative openings). We also assume of course that the imp scale is logarithmic. Our XIMP result would be:

XIMP=x/100*ln(A) (we assume there's a lot of pairs).

Our modified version of the imp scale (jmp from jungle match points) would also be logarithmic but slightly scaled:

jmp(A)=a*ln(A)+b

Maybe it is simpler to use only a, or only b. Not sure.

the Butler score with jmps would be:

BJMP=a*(ln(A)+ln(1-x/100))+b

I guess if I try to fit XIMP(A,x) with BJMP(A,x) for various (common) values of A, x, I could get the optimal coefficients a and b. Is this on the right track?

[WellSpyder]

 Fluffy, on 2012-February-15, 06:27, said:

I am still wondering, need more practice with english humor

I suppose really I thought it just said something awesome about us bridge players that not only could someone ask what is at heart a completely pointless question (no one is ever likely really to use a different imp scale for Butlers, are they?) but that they could then find that someone else had already spent some time looking at the maths of it. (And just to be clear, I'm just as capable of spending time thinking about completely pointless things as the next bridge player, so I didn't intend anything personal about the specific people involved here.)

[gwnn]

Oh but mine was funny too.

[gwnn]

 gwnn, on 2012-February-15, 06:45, said:

OK let's assume there is a XIMP board where x% of the pairs got 0, and (100-x)%, including us, got a score of +A (good for us that we don't play those super-conservative openings). We also assume of course that the imp scale is logarithmic. Our XIMP result would be:

XIMP=x/100*ln(A) (we assume there's a lot of pairs).

Our modified version of the imp scale (jmp from jungle match points) would also be logarithmic but slightly scaled:

jmp(A)=a*ln(A)+b

Maybe it is simpler to use only a, or only b. Not sure.

the Butler score with jmps would be:

BJMP=a*(ln(A)+ln(x/100))+b

I guess if I try to fit XIMP(A,x) with BJMP(A,x) for various (common) values of A, x, I could get the optimal coefficients a and b. Is this on the right track?

No. I tried for two different ln(A)'s (5 and 10) and two different domains for x (10..90 and 40..60) and the resulting coefficients are very different.

But the fits are very bad. I should look a little harder. Any statisticians out there, hint hint.

[phil_20686]

 gwnn, on 2012-February-15, 06:22, said:

I don't understand what you're doing phil, I did the same but to me it looks like
e^{Butler} \approx (arithmetic mean of differences), and
e^{XIMP} \approx (geometric mean of differences)

This assumes that all differences are positive, i.e. we have the best or worst score in the whole field.

Is your 'first approximation' that arithmetic~geometric mean? That is very brutal, since in the most common situation most results are in clusters (e.g. 30% around +620 30% around -100 and 40% around +170). The arithmetic and geometric means are only similar if most values are very close to each other.

Yes, csaba you are right, I only vaguely remembered my proof and was tired - you are meant to get the geometric mean and the arithmetic mean. But remember you are comparing the logs of these means, not the means themselves, so the difference is exponentially suppressed. Even should the means differ by a factor of two, the cross-imp butlers will differ only by ln(2). (well, log to base imp (2))

If you want to do it properly you need to add a step function S(a) = O(x-a)-O(a-x), and you can use this to see that:

exp(cross imps) = product sum(x-a)/product sum(mod(x-a)) where the scores on the bottom would give negative numbers, using it for scores in the middle improves the correspondence, as you can basically split the scores into positive and negative, and since you are subtracting the weighted arithmetic mean and comparing to the ratio of the geometric means the inequality now works in your favour, reducing the difference. You can also see this trivially as the cross imp is always smaller variance than the butler, so they must cross over somewhere in the middle.

Clustering actually improves the correspondence. The easiest way to see this is that if all the numbers are the same then the inequality is satisfied. Clustering means taking a weighted mean of a handful of averages for which the inequality is satisfied. A random collection of numbers between 1 and 1000 will have a bigger difference between the arthithemitc and geometric mean than a clustered one. (try 9*620 and 1*-100 compared to 5*620 and 5*-100).

But honestly, none of it matters that much, because once you take the log of the inequality the difference is basically never more than one imp for a score.

[phil_20686]

 gwnn, on 2012-February-15, 06:45, said:

OK let's assume there is a XIMP board where x% of the pairs got 0, and (100-x)%, including us, got a score of +A (good for us that we don't play those super-conservative openings). We also assume of course that the imp scale is logarithmic. Our XIMP result would be:

XIMP=x/100*ln(A) (we assume there's a lot of pairs).

Our modified version of the imp scale (jmp from jungle match points) would also be logarithmic but slightly scaled:

jmp(A)=a*ln(A)+b

Maybe it is simpler to use only a, or only b. Not sure.

the Butler score with jmps would be:

BJMP=a*(ln(A)+ln(1-x/100))+b

I guess if I try to fit XIMP(A,x) with BJMP(A,x) for various (common) values of A, x, I could get the optimal coefficients a and b. Is this on the right track?

You can certainly improve it this way, but there will be a variance in the difference between the means that you cannot get rid of completely. You might be able to reduce the difference a little, but I think it basically amounts to dividing the cross imp score by a number.

Also, to scale a logarithm while keeping it logarithmic you should just change the base, that means that you have an a = lod_1/log_2 or whatever, but no b. alternatively you can see this that if they all had the same score b=0 for matching the zeros.

[jallerton]

 gwnn, on 2012-February-14, 10:54, said:

I wonder if there's a better IMP scale (i.e. it would reproduce the ideas of the original IMP scale truer) to use if you have to compare to the datum score. Imagine you're stranded in a jungle and you have no computers available to calculate x-imp easily (but you can quickly average all results). Comments? Ideas on how to calculate something like this?

Note that there is no element in this post that suggests that such a scale should be used, or it wouldn't be confusing to many people, or it would be readily accepted. It would be awesome if we didn't debate these points here.

I think you are asking slightly the wrong question. The anomalies of Butler scoring arise from the use of the "datum" score in the IMPing, not because the IMP scale being used is wrong per se.

For example, consider a 20 table Butler Pairs event, in which the datum is calculated by throwing out the top and bottom scores and then averaging the rest.

On Board 2, a routine cold 4♠ contract, 19 North/South pairs score +620. The other N/S declarer has a blindspot in the play, 4♠-1 and scores -100. The datum is +620, the first 19 pairs score 0 IMPs and the pair in 4♠-1 ends up with -12 IMPs.

Board 4 is also a routine 4♠ contract, reached by every North/South pair. This time 10 Norths make the contract exactly, scoring +620. The other 10 Norths misguess a 2-way finesse and record 4♠-1 for -100. This time, the datum is +260, the first 10 pairs score +8 IMPs and the pairs in 4♠-1 ends up with -8 IMPs.

So going off in 4♠ on board 4 costs 16 IMPs, whilst going off on board 2 costs 'only' 12 IMPs.

If you normalise/fiddle the IMPs scale to reduce the swing on board 4 to (say) the intuitive vulnerable game swing figure of 12 IMPs then you will also reduce the swing on board 2. Ideally the swings for going off in 4♠ should be the same on the two boards.

My solution would be to use a real score, not a mean average score, for the 'datum'. The real score could be the modal (most common) score on the board, the median score or even double dummy par. Whichever of these is used, the IMPs will be more akin to the IMP swings one might expect at teams-of-four, the form of scoring for which the IMPs scale was devised. It does not matter if the modal scores are duly favourable or unfavourable to N/S, as each N/S pair is comparing its IMP scores against the IMP scores of the other pairs in the same direction.

[Vampyr]

 jallerton, on 2012-March-15, 17:33, said:

My solution would be to use a real score, not a mean average score, for the 'datum'. The real score could be the modal (most common) score on the board, the median score or even double dummy par. Whichever of these is used, the IMPs will be more akin to the IMP swings one might expect at teams-of-four, the form of scoring for which the IMPs scale was devised.

This would also make your task much easier. While you're stuck in the jungle.