## Tuesday, 8 January 2013

### Maths puzzle: Gender based seating preferences on Tube trains.

As I was about to get off the Tube this morning, I noticed that there were six women on the row of six seats opposite me and five men and one woman on the row on which I was sitting. None of the people concerned appeared to be travelling together, it was twelve individuals.

I see this quite often, and I wonder to myself, is this because women have a preference for sitting next to other women rather than next to men (perfectly understandable, women use less space etc) or is this purely a random thing? (Let's ignore the possibility that men prefer sitting next to other men, because this is merely the equal and opposite).

Now, if we have a fixed pool of five men and seven women, and they sit down at random, we can calculate the probability of there being six women on one row*, it is 16.3% (Method A).

But this is a meaningless calculation because there is no fixed and limited pool: hundreds of people will get on and off each Tube train in a forty-minute journey. Just because the last passenger to sit down was a women does not materially affect the odds of the next passenger being a man or a woman (like tossing coins rather than pulling coloured balls out of a hat).

So the other way of calculating the odds of having six women in a row is 0.5^5 = 3.125% (Method B).

COUNTERPOINT: as Kj in the comments says "The objections to Method A are correct, but then again, if there is a 0.5 chance of every passenger being either sex, on average the pool of potential sitters is also roughly 50/50".

I definitely see this phenomenon (all women on one row, all men on the opposite row) quite often, certainly about once a week (ten journeys) and more often than once or twice a month (forty journeys).

There again, I only really notice it if there is a clear pattern, which might only exist for part of the journey. I'll have to start running a tally, but my question(s) to the great minds reading this is:

Which is the better way of calculating the odds of six women (each travelling alone) ending up sitting next to each other on a six-seat row? Method A or Method B, or is there an even better way, Method C, and if so, what is Method C?

Supplementary question: if you knew how often this "all women" pattern emerged, would you be able to work out the marginal preference of women for sitting next to women?

* There are 7 permutations of having 6 women on one side and one on the other, 21 permutations of 5 and 2; and 15 permutations of 4+3 = 43. 6/43 = 16%.

Kevin the Chimp said...

You would need to account for the preference of the male to stand or for the gentleman to give up his seat for a lady...

The filling order is also critical - a handful of men boarding first would naturally avoid one another and prevent a six-women-in-a-row emerging.

KTC, nobody stands up for anybody on the Tube, ignore that element.

The filling is initially not random but after a dozen stops it is, because people standing will take the first available seat - the point is, I think that if a woman standing has a choice between a (a) seat between two women or (b) a seat between two men, she will prefer the former (a). And presumably, if the choice is (c) remain standing or (d) sit down between two men, enough women will prefer (c).

SumoKing said...

It's just left over pavlovian response isn't it.

Everyone goes to school and spends about 13 years being split into lines of boys and girls, or seating which is boy girl or attedning schools which are single sex so it is ingrained to match your gender and sit that side of tube etc

either that or it is simply women want the boobs oggled and men want to oggle boobs and tube seating allows this mutually beneficial activity to occur without danger or social stigma.

SK, yes and amen to all that*, but you didn't answer the question(s).

* Depending on your preference between the "sideways and downwards" versus "straight ahead" oggling techniques.

SumoKing said...

I think camoflage is key to playing the game as it were

I tend to sort of yawn and glance or rub at my eyes (in the hope of presenting a loveable sleepy puppy appearance) and glance or just pretend to be reading while probably quite obviously staring over the top of book, metro, phone etc

Lola said...

The words 'life' and 'getting one' spring to mind...

Highland Cooncil said...

WHY does not all the people decide to get the 9.30 to Inverness bus?

Why is there always the same amount of people at any given bus stop?

Even the weather does not make any diff............

SK, the best camouflage (for a man) is to remain standing, or so I have been told, allegedly.

L, my life is ten 45-minute journeys a week = 7 or 8 hours a week. And these infuriating maths puzzles are how I keep my mind occupied (once I've done the Sudoku and reformed the UK tax system).

HC, sorry, off topic.

Kj said...

Which is the better way of calculating the odds of six women (each travelling alone) ending up sitting next to each other on a six-seat row? Method A or Method B, or is there an even better way, Method C, and if so, what is Method C?

The objections to Method A are correct, but then again, if there is a 0,5 chance of every passenger being either sex, on average the pool of potential sitters is also roughly 50/50, so the straight method of using permutations on a known pool is probably pretty close to correct IMO.
From the behavioural side, I'd be interested in seeing if the probabilities of the row filling up with same-sex sitters were to increase with each of the one sex sitting down. If the first sitter is a woman, there is probably a much smaller chance that the whole row will be woman-filled than if five out of six seats is filled with women.

Providing there is a 50/50 distribution of men having a sideways or frontal oogling preference ofcourse...

The only clear pattern I've noticed, being scandinavian, is that all obvious scandinavian tube-travellers will try to sit as so the whole wagon is to be seated as evenly distributed in space as possible if it's a half-empty car, if it's crowded, sit wherever you are least likely to be touched. The obviously non-integrated will lump together by the exit ;)

Lola said...

MW. 8 hours commuting....sounds awful. That's one day a week doing nothing much. no wonder your mind wanders into these esoteric areas...

Kj, thanks, I have updated the post. The seating pattern you refer to is known as The Urinal Protocol.

L, you get used to it after ten years :-)

Lola said...

MW 8 x 48 (4 weeks hols) X 10 years = 3840 hrs. = 160 day, or say 44% of a year...

You could be 6 months younger...

Bayard said...

From the xkcd blog: "Discussion question: This is obviously a male-specific issue.  Can you think of any female-specific experiences that could benefit from some mathematical analysis, experiences which — being a dude — I might be unfamiliar with?"
Go on Mark, make his day.

Jonathan Bagley said...

Hi Mark,

"Which is the better way of calculating the odds of six women (each travelling alone) ending up sitting next to each other on a six-seat row?"

This question is not precise enough. Referring to the earlier part of your post: I think your method A should have given you 6.06%. The number of ways of choosing a subset size 6 from a set size 12 is 12!/6!x6! =924. The number of ways of choosing 6 from twelve including 6 women is 7. Hence 7/924. Then x2 as the women could be in either row. Alternatively 7/12x6/11x5/10x4/9x3/8x2/7, then x2.
In the situation where there is no fixed pool, but 12 arrivals, and the population ratio is 50:50, there is now the possibility that all 12 seats will be occupied by women, so you maybe should ask the probability of at least one row of women. This is 2x(0.5)^6 = 3.125%, which is what you got.
If you now assume the population ratio is 7:5, you get 2x(7/12)^6 = 7.880%. I think method B is more correct, because it correctly accounts for the population gender ratio; but, it does not take into account passengers arriving and leaving.
You might be wanting to know the probability that after some large number of changes, there is at least one row of women. This will be the same as given by method B, assuming the seats are always full. The person who gets off will be equally likely male or female as the person who gets on, so we are always in the earlier situation where 12 people have arrived.

L, who says I want to go through the last 6 months of my life again? Well, apart from theXmas holiday bit.

JB, agreed, I reworked it for a 6 men-6 women combination and the chance of all women being on one row is 1 in 42 = 2.4% (using Method A) and 3.1% (using Method B) so whichever is correct, "about 3%" is the right answer.

The Cowboy Online said...

I'll always choose to sit next to a woman over sitting next to a man, it's true - they do take up less space, mainly due to them not feeling the need to spread their legs in an exaggerated manner to prove their manliness. Men with widely spread legs, maybe the psychology driving that as the big powerful sports car?

The Cowboy Online said...

Whoops, slightly mangled, the last part should have read;

"maybe the same psychology driving that as the big powerful sports car?"

TCO, yes, that is what I was alluding to (as an aside, this whole "sticking your knees out as far as you can" habit is absolutely appalling behaviour, I go for the "straight ahead and parallel" configuration.)

Steven_L said...

What a load of rot, from all of you.

1) Pick 6 seats on the tube in a row.

2) There are 64 possible combinations of m/f on them, one of which is 6 females.

3) Count the number of stops the bet runs for. If it is 2 the base odds to work from are 32/1, if 4 it's 16/1 etc.

4) If you can get better odds, it's perhap a good value bet.

5) If you're going say Vauxhall to Oxford Circus it's a better bet there will be more women as they all get on at Victoria and work in Green Park or are going shopping.

6) If you're headed toward the City or Canary Wharf then bet on 6 men.

PS. Make sure you bet with someone who will pay you of course.

Steven_L said...

Method A is the probability of 6 females OR 6 men in a row. You have to divide the answer by 2 you bozos!

Steven_L said...

And that's assuming a full underground, if you bring empty seats into play there are much more than 64 combinations.

Look, the method is:

mmmmmm - 1 combination
mmmmmf - 6
mmmmff - 15
mmmfff - 20
mmffff - 15
mfffff - 6
ffffff - 1

= 64

Now maybe there is one of these really clever, expensive accountants or IFA's around these parts who can work it out properley with men, women AND empty seats.

SL, good work, thanks.

But it's 1 in 32, not 1 in 64 (as you said) or 1 in 42 (as I said) because assuming 6 of each, if you have 6 men on one row you must also have 6 women on the other, they are the same thing.

Out of interest, would you consider your last workings to be more akin to Method A or Method B?

Please note, I applied Method A to the trickier issue of 7 women and 5 men.

Bayard said...

"women use less space etc"

Are there still armrests beween the seats? If not, I found out in my student days, when we often had to cram three people into the back seat of a Herald convertible, that, although on the whole women are smaller than men, their hips are wider, which is the dimension that counts. OTOH none of us men were fatties, so perhaps we weren't a representative sample of yer average London commuter.

View from the Solent said...

I came late to this one. KJ is on the right track.
Looks like you’re assuming (pseudo)random occurrences. But they ain’t randoms, they’s peoples.
Let’s assume an attractive young lady sits down. What is the most likely seat (assuming it’s available) to be chosen by a following male – adjacent and facing away, or opposite and facing towards?
I posit the second, with reinforcement if the female subject is wearing a short skirt. The adjacent opposite seats would also be preferred by subsequent male subjects. Leaving lots of empty seats on the other side, so more likely to be filled by newly-arriving females, and so on. It’s a positive feedback.
And it’s changing all the time with a few nonconformers thrown in to break the pattern, a lovely dynamical system.
What you need is a software model to simulate it. Get one of the CAGW modellers to write one for you. Don’t forget to tell them what you want the outputs to be.