[rogerclee]

NS pair A and EW pair B play a matchpoint event with a nearly infinite number of boards in a nearly infinitely large field.

Pair A has a percent score p, and Pair B has a percent score q.

If pairs A and B were to compare their results as a BAM, what would their expected percentage be?

(You can assume that their matchpoint scores per board are reasonably distributed with means p, q respectively--that is, neither pair plays an insane top or bottom style, and feel free to make reasonable simplifying assumptions).

[helene_t]

First let's try with a simplifiction, namely that there are no ties. I would assume that your mathcpoint score on a single board is beta(a,B) where (a,B) are such that they give means of p and q respectively for the two pairs. Now if the four parameters (a and b for pair A and pair B, lets call them aA, bA, aB ad bB) are known then it is simple to compute the probability that a sample from beta(aA,bA) is smaller that an independent sample from beta(aB,bB) which is the probability that pair B wins the board. This would also be the expected percentage of pair B.

With an infinity of boards it would in fact be the percentage of pair B, there would be no uncertainty of the predicition.

Without ties we have aA=bA=1 if all pair are of equal strength. It would probably be reasonable to set a equal for all pairs (a reflects the homogeneity of the field) so that the strength of the pair only affects the b parameter.

But with ties, the variance is smaller which leads to different parameter values. Also, ties must be allowed when computing the distribution of the score of a single BAM board. Now it's possible that this doesn't affect the expected score of the BAM match (although it would affect the variance of the score in a limited-length match). I am not quite sure about this. It may depend on how one models the ties. I am not sure if there is an obvious way of modeling the ties.

Assuming that there are no ties then I could give you an answer as soon as I get to a computer with some statistical software installed. If the possibility of ties does affect the expected score, though, there is no single answer to your question as it will depend on the strength and homogeneity of the field.

[matmat]

could monte carlo this


my intuition makes me think it's just going to be something like

p/(p+q) and q/(p+q)

this does break down in the case when p=1, q<1, though... but that might just be a pathological condition

[hanp]

matmat, on Jul 9 2010, 02:42 AM, said:

my intuition makes me think it's just going to be something like

p/(p+q) and q/(p+q)

Talking with Helene this morning makes me think that a better guess would be

p(1-q)/( p (1-q) + q (1-p) ) and q (1-p) / (p (1-q) + q (1-p) )

For example, if one pair gets 60% against the field and the other pair gets 40% against the field then when playing against eachother, the stronger pair gets about 69% and the weaker gets about 31%.

Using the formula matmat gave, the stronger pair would just get 60% and the weaker 40%, which doesn't make much sense.

To solve the problem you need extra assumtions. I'll wait for Phil to come up the academic hill and tell us math people what the assumptions should be.

[JLOGIC]

Am I understanding the question right?

We're wondering if pair A (NS) averages some number, and pair B (EW) averages some number, what will their BAM score be as a team generally.

Intuitively I would have thought it would be the average of their scores, like a 60 and a 40 would average 50 % at bam (13 out of 26). Since the question is being asked obviously that's wrong, or I'm misunderstanding the question.

If I'm understanding the question correctly, then why is the average of the 2 pairs not the right answer (in laymans terms if possible). Thanks in advance for your answers (especially Phil).

[hanp]

Oh I didn't read Roger's post, my post is in reaction to the question how they would do against eachother. Let me think

[hanp]

If we assume that the score of each pair on a given board is normally distributed with means A and B and standard deviations x and y, and we also assume that these distributions are independent, then we can compute the chance that the team wins the board. (by assuming normal distribution we implicitly assume that there are no ties, as the chance that the scores are exactly equal is zero).

The two pairs win a board if the sum of their scores on the board is greater than 1. The sum is normally distributed with mean A+B and standard deviation (x+y)/sqrt(2), which tells you the answer.

For example, if both pairs score 60% and their standard deviations are 20% (no idea if this is reasonable, but you could easily find out), then their sum is 120% and the standard deviation is 28.3%. So the chance that they win the board is (looking up 0.71 in normal distribution table...) 76%.

Some comments:

- Normal distributions and independence is not reasonable for bridge.
- The larger the standard deviations of the pairs, the lower their expected BAM scores.
- For two pairs that are both better than the field, the expected BAM score should be higher than the maximum of the two scores.
- For a pair with matchpoint score 60%, the largest possible standard deviation is 49% (this is achieved by 60% tops and 40% bottoms). For two such pairs their expected BAM score is 36/52 = 69%.
- BAM is a harsch game.

[JLOGIC]

I see, if you have 2 60 % pairs then you are much higher than 60 % to win the board. That makes a lot of sense intuitively, so my first post was pretty retarded. TY Hanp

[Bbradley62]

The assumed distribution of scores is the important component. You cannot, for example, assume that both A and B get 51% on every board. If this were the case, they'd never lose a board, because they would always both beat the median score, so their combined score would never be negative.

[hanp]

Well my post was equally retarded as I said that the sum of two normal distributions is normal. Let's say I was talking about identical distributions as in the example. I should leave these problems to the people who understand statistics.

[Free]

Is there a particular reason to use a nearly infinite number of boards instead of infinite?

[Siegmund]

hanp either meant to say means A,B variances x,y, yielding combined mean A+B and combined variance x+y... or standard deviations x,y and combined standard deviation sqr(x^2+y^2).
But the general idea of finding P(sum>1) is correct.

His worked example came out right accidentally, since when x=y, sqr(x^2+y^2) and (x+y)/sqr(2) both happen to equal x*sqr(2).


While we're on the subject, I have always felt that "matchpointed teams" -- score every board across the whole field, and your team's score on the board is the sum of your EW and NS scores -- would be an interesting way of scoring a game, lower-variance than a BAM, though the people who think "teams is better because you can point your finger at the only 8 people who influenced your result" would dislike that.

[hanp]

It's Roger's favorite number (see watercooler).

[hanp]

Siegmund, on Jul 9 2010, 05:21 AM, said:

hanp either meant to say means A,B variances x,y, yielding combined mean A+B and combined variance x+y... or standard deviations x,y and combined standard deviation sqr(x^2+y^2).
But the general idea of finding P(sum>1) is correct.

Thanks for the correction! I'm glad that it is still true that the standard deviation is sqrt(2) times as much when you have two identical distributions, I guess this is what I remembered from highschool.

[awm]

Regardless of the mathematical answer, there are definite correlations between "wins" and "losses" in a real BAM field. Typically what happens is that stronger pairs tend to have stronger teammates (and vice versa). When playing a weak team it is fairly likely to produce a "win" at both tables (compressing the wins), whereas playing a strong team the opposite is likely. In other words, obtaining a good result at each table against a particular opposing team is highly correlated (not independent). This tends to move the scores towards the average of the two pairs percentages (i.e. if wins/losses were totally correlated you would expect to score the average of the two pairs percentages). In reality neither "totally correlated" nor "totally independent" will hold, but the result should be in between the extremes.

[matmat]

hanp, on Jul 9 2010, 03:12 AM, said:

p(1-q)/( p (1-q) + q (1-p) ) and q (1-p) / (p (1-q) + q (1-p) )

That looks better
Hey. I was tired and I'm dumb

[Phil]

hanp, on Jul 9 2010, 03:12 AM, said:

I'll wait for Phil to come up the academic hill and tell us math people what the assumptions should be.

Wow, things are getting bad for me around here when I get called out on a thread I haven't participated in.

[rogerclee]

hanp, on Jul 9 2010, 01:23 AM, said:

Oh I didn't read Roger's post, my post is in reaction to the question how they would do against eachother. Let me think

The questions should be equivalent. The expected BAM score should be equal to the expected matchpoint result of pair A (strength p) against the rest of the field (strength 1-q).

[gwnn]

I'm quite confused now rogerclee.

For me there are two possible interpretations and my impression is that this is what people have been thinking.

Interpretation number one: suppose pair A has, after 100 rounds of play, a percentage of p and pair B has, after 100 rounds of play, a percentage of q.

If they met in round 101 at a table, what % would be the result on average?

Interpretation number two: suppose EW pair A has, after 100 rounds of play, a percentage of p and NS pair B has, after 100 rounds of play, a percentage of q. If now we imagine that they were on a BAM team all along, how many boards have they won out of the 100?
Say on board 1 NS got a good +430 and EW got an average +400, then "team AB" won this board.

I don't see how the two interpretations can ever be equivalent. In interpretation number one if they both have an equal % before they meet, the expected result of their round will clealy be 50%. In interpretation number two, if they both have an equal %, the expected result of their match will go sharply up as the common matchpoint score increases. Am I misunderstanding something?

[pooltuna]

gwnn, on Jul 10 2010, 07:33 AM, said:

I'm quite confused now rogerclee.

For me there are two possible interpretations and my impression is that this is what people have been thinking.

Interpretation number one: suppose pair A has, after 100 rounds of play, a percentage of p and pair B has, after 100 rounds of play, a percentage of q.

If they met in round 101 at a table, what % would be the result on average?

Interpretation number two: suppose EW pair A has, after 100 rounds of play, a percentage of p and NS pair B has, after 100 rounds of play, a percentage of q. If now we imagine that they were on a BAM team all along, how many boards have they won out of the 100?
Say on board 1 NS got a good +430 and EW got an average +400, then "team AB" won this board.

I don't see how the two interpretations can ever be equivalent. In interpretation number one if they both have an equal % before they meet, the expected result of their round will clealy be 50%. In interpretation number two, if they both have an equal %, the expected result of their match will go sharply up as the common matchpoint score increases. Am I misunderstanding something?

I suspect you don't have enough information as the std deviation for each of their %ages is probably critical to generating a sum of the normal curves of each