A host of issues
#1
Posted 2013-September-05, 17:07
I have no idea what the distribution of players without partners would be. My guess is that it would approximate to a Poisson distribution, but I could be quite wrong.
#2
Posted 2013-September-05, 18:28
#3
Posted 2013-September-06, 03:02
aguahombre, on 2013-September-05, 18:28, said:
Yes, it should read "an odd number of players not including him or her".
#4
Posted 2013-September-06, 03:07
edit: if you have the attendance records, can't you just count the "odd" and "even" evenings and compare?
This post has been edited by gwnn: 2013-September-06, 03:13
George Carlin
#5
Posted 2013-September-06, 03:20
There could be other factors like a player who nobody wants to play with who always shows up, that would change everything obviously.
#6
Posted 2013-September-06, 03:30
0: 99.3%
1: 0.6%
2: 0.1%
In other clubs it could look like:
0: 10%
1: 30%
2: 40%
3: 10%
4: 5%
5: 3%
...
In a very big club maybe the mean number of unpaired players is 7 or 8. I think you are talking about a mean number of 0.1-0.5 or so, in that case Lamford would not have opened this thread, methinks.
George Carlin
#7
Posted 2013-September-06, 04:49
#8
Posted 2013-September-06, 06:24
Fluffy, on 2013-September-06, 04:49, said:
What is the connection between the two? When a team has a bad game, it will often score 0 goals and when they have a good game, it will score more. But there is no such thing as a good or a bad time for the whole city to produce unpaired bridge players during bridge season, other than perhaps very bad weather or the outbreak of an epidemic. If Messi is injured, Barcelona are more likely than normal to score 0 goals. But if Aunt Millie is feeling under the weather, Uncle Johnny might still want to go to the bridge club. If the only connection you mean is that both events (a football team scoring and an unpaired bridge player appearing at your club) are relatively rare, I disagree: it depends on the club.
This post has been edited by gwnn: 2013-September-06, 07:23
George Carlin
#9
Posted 2013-September-06, 06:42
The occurance of Aunt Millie getting ill is also a Poisson event, as long as she isn't chronically ill.
Rik
(*) At 2-2, this probability will be lower than at 4-0. In the first case both teams will play carefully, because they don't want to lose. In the latter case, the stronger team will start to play for fun where they might neglect the defense whereas the weaker team might want to score a goal to save their honor. Of course, I never investigated this - I have better things to do (like posting on the internet) - but it might be fun to do.
The most exciting phrase to hear in science, the one that heralds the new discoveries, is not “Eureka!” (I found it!), but “That’s funny…” – Isaac Asimov
The only reason God did not put "Thou shalt mind thine own business" in the Ten Commandments was that He thought that it was too obvious to need stating. - Kenberg
#10
Posted 2013-September-06, 11:01
Rik
The most exciting phrase to hear in science, the one that heralds the new discoveries, is not “Eureka!” (I found it!), but “That’s funny…” – Isaac Asimov
The only reason God did not put "Thou shalt mind thine own business" in the Ten Commandments was that He thought that it was too obvious to need stating. - Kenberg
#11
Posted 2013-September-06, 11:25
Trinidad, on 2013-September-06, 11:01, said:
Rik
Sorry, wrong of me to call them this way. I meant for the two of them to be unrelated old people at the club, i.e. usually uncorrelated Poisson events.
I still don't get why 0 would be more common than 1. I think in some clubs it is normal that you just show up and find a partner. Of course some people will have arranged for one from home/have regular partners, but up to half the pairs may be formed on the spot. I think Fluffy just has presuppositions based on the clubs he's been to. Lamford: would it be too much trouble for you to give us the actual mean?

George Carlin
#12
Posted 2013-September-06, 12:06
In practice, the number of players willng to turn without a partner can vary significantly due to other factors. If someone partnerless is an unsuitable partner (bad-tempered, say, or with early dementia) then other players will stay away rather than risk being forced to play with that person, and the numbers will tumble. Similarly, if the host's name is published beforehand, then players might be more likely to turn up if the host is fun to play with (and try to manipulate it so they partner the host).
I used to have an arrangement with a regular partner that she'd host, and I'd stay at home. If she ended up partnerless, I'd get a phone call about 5 minutes before the scheduled start; I lived close enough to be there before the first board was completed. With that arrangement the host always got a game.
#13
Posted 2013-September-06, 12:11
gwnn, on 2013-September-06, 11:25, said:

I think that it is only possible to guess, unless lamford has done some research of which I am unaware. I would say that the mean is near 2.
#14
Posted 2013-September-06, 15:27
gwnn, on 2013-September-06, 11:25, said:
Well, if they frequently play the same game at the same club, then if one of them gets the flu, the chances that the other will get the flu within a short time is increased beyond the random, because the sick one might have played while contagious.
Or, if the club provides food, the two of them might suffer a food Poisson issue

#15
Posted 2013-September-06, 15:52

George Carlin
#16
Posted 2013-September-06, 16:08
gwnn, on 2013-September-06, 15:52, said:

Hence my merely providing an example of when your 'usually' wouldn't apply, which is not at all the same as saying that your 'usually' was incorrect

Besides, the main reason I wrote my post was to indulge in my taste for bad puns...the food poissonning line.
I usually make very good puns. However, this may be another example in which 'usually' doesn't apply
#17
Posted 2013-September-07, 00:58
l p
1.0 56.8%
1.5 52.5%
2.0 50.9%
2.5 50.3%
3.0 50.1%
(I just got it from Excel by adding up all the probabilities between 0 and 10. the total probabilities added up to at least 99.97%).
George Carlin
#18
Posted 2013-September-07, 12:13
Inspired by gwnn, I decided to try for an exact answer from mathematica. No problem. Letting m be the mean the probability of getting an even number is E^(-m) times the hyperbolic cosine of m, which as we all know (

For example, m=3 gives a probability of 0.501239 just as gwnn says.
I can say that if I never read bbo wc, I quite likely would have gone to my grave without knowing this fact.
Comment at the end: Of course we get the hyperbolic cosine divided by the exponential, just write out the formula for the Poisson and sum it.
As has often been observed, we are all becoming way to quick too grab a calculator.
#19
Posted 2013-September-07, 12:49
George Carlin
#20
Posted 2013-September-07, 13:27