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## The 1-in-10 Gay problem!

…a maths problem that is.

“If 1-in-10 people are gay, and there are 10 people in a room, one of the people in the room must be gay!”

Is this true?  Take a look here for the full analysis!

(!! This is the first in a series of three related posts – click to see Part II “Find a Way to my Heart” & Part III “Hallucinogenic Maths”)

Many of us, in our younger days, were in that very room when somebody brashly declared this (and often that somebody would be the duplicitous 1-in-10 themself), but what is the real probability of there being a gay person in a room of 10?  Or perhaps more pertinently (and easier to calculate), what is the probability that there is at least 1 person in the room who is gay?

You may be intrigued to know that the full analysis of this problem, which is spread over 3 posts, will in fact become so incredibly queer by the end that we’ll be imagining 3-dimensional structures that don’t even exist!  It would be far too simple and dull just to work out the answer, utter something like “oh right”, and then forget about it, which, incidentally, reflects many people’s attitudes to maths.  How boring.  Instead, I’m going to take us on a journey through mathematical abstraction and into imaginary space.

But first, back to Earth.  Let us consider a basic version of the scenario and look for patterns or characteristics that might help us solve the gay problem.

The 1-in-2 problem (Basic case)

Think about this one for a moment: “If there are 2 people in a room, what is the probability that at least one of them is female?”

We will assume that the chance of any particular individual in the room being female is 50%.  (This is a fair approximation and it’s not the only assumption we’re making here. Maths is full of approximations, but they can always be justified.)

Since the first person could be either male or female and the second person could be either male or female, there are only 4 possible combinations of males and females in the room:

The probability that there is at least one female in the room is 75% (3 out of 4) because there are three combinations that include at least one woman.  In the basic case, each of the four combinations has an equal probability, but this cannot be said for the 1-in-10 problem, so…

IMPORTANT! Here’s another way of looking at the problem: How many of the four combinations don’t include any females?  The answer is of course 1; there’s only one way of having no females in the room, and that’s to have two males.  Although this sounds obvious it is key to understanding how to tackle the 1-in-10 gay problem as we shall see in a few moments.

It’s certain that there are 2 people in the room, so the probability that any scenario occurs in which there are two people in the room is 1 (or 100%).
But the probability that neither of the people in the room are female is 0.25 (or 25%).

All other scenarios include at least one female.  Thus we have identified the following equation:

P(there is at least one female in the room) = 1 – P(there is no female in the room)

and we conclude that the probability of there being at least one female in the room is 1-0.25 = 0.75 (or 75% or ¾).

Two females in a room

Now, let us bring our attention back to the original gay problem: “If 1-in-10 people are gay, and there are 10 people in a room, what is the probability that there is at least 1 person in the room who is gay?”

We simply need to apply the formula we’ve just worked out.  With a little rewording it looks like this:

P(there is at least one gay person in the room) = 1 – P(there are no gay people in the room)

So all we need to do is work out the probability that there are no gay people in the room (which is effectively the probability that every person in the room is straight) and subtract it from 1!

If 1-in-10 people are gay then 9-in-10 people must be straight, so the probability that any particular person in the room is straight is 9/10 or 0.9.  If we need all 10 people to be straight then we just multiply 0.9 by itself 10 times, i.e.
0.9 x 0.9 x 0.9 x 0.9 x 0.9 x 0.9 x 0.9 x 0.9 x 0.9 x 0.9 = 0.910 ≈ 0.35

Then we just subtract that number from 1, i.e.  1 – 0.35 = 0.65  and that’s the answer.
The probability that there is at least one gay person in the room is 0.65

65% is roughly two-thirds, so the repressed homosexual’s original declaration is sadly false; there’s actually only about a two-thirds chance of somebody in the room being gay, but this means, given a scenario where there were, say, 30 rooms each containing 10 people, one would expect 20 of those rooms to contain at least one gay person.

To put this result into perspective, assuming the 1-in-10 chance of a person being LGBT to be roughly accurate, in a year group of 250 pupils, with 10 maths classes of 25 pupils each (much bigger classes than a room of just 10), you should expect 9 of those 10 classes to accommodate at least one LGBT pupil.

Footnote: These are all probabilities at the end of the day, but there is a point to this and one thing’s certain:  Gradually, homosexuality, after centuries of repression, is proudly taking up the place in society it deserves.  However the stigma and the bullying remain and it should be high on our agenda to provide children with a safe and free environment to discuss their sexuality openly in early life.  There is nothing more stressful for a child than thinking they are alone in a harsh world.

(At the risk of appearing like small print, I must stress that the very definitions of “gay” and “gender” are subjective and debatable and there exist no modern reliable statistics to corroborate the popular 1-in-10 estimate.  In fact, a recent survey suggests it is more like 1-in-65 people, but the survey did use some questionable strategies for collecting data.)

In the next post we will build on the mathematical concepts that have arisen here as I lead you into

1. Eldis – What is Gender?  http://www.eldis.org/go/topics/dossiers/trade-and-gender/what-is-gender

2. How Many Gay People Are There?  http://www.avert.org/gay-people.htm

1. October 2nd, 2009 at 18:05 | #1

Awesome work dude.. excellent presentation of an intriguing problem, with clear and logically laid-out arguments. Captivating and highly enjoyable reading!

2. August 4th, 2009 at 12:01 | #2

Hey Micky, really enjoyed reading the 1 in 10 gay problem. Looking forward to reading the future articles.

3. July 31st, 2009 at 00:18 | #3

RT “Hey Micky. I’ve been enjoying reading your maths blog. This one is always a favourite of mine, you might have have seen it before..

The Monty Hall Problem
http://en.wikipedia.org/wiki/Monty_Hall_problem