A host of issues

[lamford]

Our club has a host each week, who only plays if there is an odd number of players without partners. Assuming I know the mean number of players who turn up without partners, based on attendance records, can I work out the probability of the host getting a game?

 

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.

[aguahombre]

Actually, he only plays if there are an even number showing up without partners. He shows up without a partner, so he counts as one. Wait, maybe you don't consider him a player; have you told him that?

[lamford]

Actually, he only plays if there are an even number showing up without partners. He shows up without a partner, so he counts as one. Wait, maybe you don't consider him a player; have you told him that?

Yes, it should read "an odd number of players not including him or her".

[gwnn]

Unless the mean or the standard deviation is very small, the probability surely is very close to 50%.

 

edit: if you have the attendance records, can't you just count the "odd" and "even" evenings and compare?

Edited September 6, 2013 by gwnn

[Fluffy]

But your assumptions are not there Csaba, the probability of 0 unpartnered buddies showing up apart from the host is greater than the probabilty of 1, which is greater than the probability of 2 etc, but since 0 is the bigger it will always be below 50%.

 

There could be other factors like a player who nobody wants to play with who always shows up, that would change everything obviously.

[gwnn]

It depends on the club, Fluffy. In my club the probability distribution would look something like:

 

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.

[Fluffy]

I was thinking of the distribution to be something similar to a team in soccer scoring goals, even when Barcelona's average goals is above 2 now, if you look in history chances of scoring 0 is still greater than anything else, although perhaps right now it is not the biggest one.

[gwnn]

I was thinking of the distribution to be something similar to a team in soccer scoring goals, even when Barcelona's average goals is above 2 now, if you look in history chances of scoring 0 is still greater than anything else, although perhaps right now it is not the biggest one.

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.

Edited September 6, 2013 by gwnn

[Trinidad]

I have always considered that the scoring of a goal in soccer is close to a Poisson event. Goals are rare and come up suddenly, typical for Poisson events. I do think that there is a strong relation between the score in the match and the probability that there will be a goal in the next minute, though(*). This makes it different from a Poisson distribution.

 

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.

[Trinidad]

The problem, of course, is that while Aunt Millie getting ill and Uncle Bob getting ill are both Poisson events, they are correlated Poisson events: Uncle Bob usually gets the flue one week after Aunt Millie gets it.

 

Rik

[gwnn]

The problem, of course, is that while Aunt Millie getting ill and Uncle Bob getting ill are both Poisson events, they are correlated Poisson events: Uncle Bob usually gets the flue one week after Aunt Millie gets it.

 

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? :)

[StevenG]

I'm not sure whether Lamford's question is meant as a practical or theoretical one. Unless you assume the distribution can be deduced solely from the mean, i.e. Poisson or similar, I don't see that there can be a theoretical answer.

 

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.

[Vampyr]

Lamford: would it be too much trouble for you to give us the actual mean? :)

 

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.

[mikeh]

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.

 

 

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 :P

[gwnn]

Hence my word 'usually.' For a lawyer you don't read the fine print very carefully :)

[mikeh]

Hence my word 'usually.' For a lawyer you don't read the fine print very carefully :)

 

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 :D

 

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

[gwnn]

OK if it is a Poisson distribution, the numbers are (l(ambda) is the mean, p is the probability of an even number of people showing up):

 

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%).

[kenberg]

I will leave this as a testament to inanity, but see comment at the end.

 

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 ( :) ) simplifies to (1+E^(-2m))/2.

 

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.

[gwnn]

Capital E for Euler's number? My eyes bleed...

[kenberg]

Well, yes. Ugly but useful. If all constants are in caps, you need not remember which are and which aren't.