Probability Question

Hypothetical situation:

There are exactly 6 billion people on earth. 3 billion are men and 3 billion are women. All are heterosexual, monogamous and half of a married couple. So, in short, we have 3 billion married couples on earth, and no one else.

Now a spaceship comes from another planet and takes over earth. The aliens have plans to send the 3 billion couples to 3 billion different planets to test how they can assimilate in a land in which they would be the only humans.

Seems easy enough, but here’s the problem: The aliens refuse to respect the current marital pairings. So each of the 3 billion men will be paired with each of the 3 billion women in a completely random manner.

I know that the probability that Man A will end up with his wife is 1 in 3 billion. So pretty much, for any individual, your chances of ending up with your own spouse are 1 in 3 billion. But what is the probability that at least 1 man will end up with his own wife?

Part 2 of the question: Instead of 3 billion married couples, we have 3 billion monogamous couples who may be hetero- or homo-. In this case, the aliens simply pair up random people, regardless of sex or sexual preference. Now the probabilty of ending up with your monogamous partner is 1 in 5,999,999,999. Same question: what is the probability that any individual ends up with his or her original partner?

  1. As near as makes any difference, 1-e[sup]-1[/sup] = 0.6321…
    2 seems a little trickier. Let me think about it a bit.

Well I’ve got an answer but it is a rather horrible one consisting of a long sum of terms involving binomial coefficients. Just in case you want it, it’s the sum of
2[sup]r[/sup] NCr/2NCr[sym]´[/sym]2rCr[sym]´[/sym]r!
where r runs from 1 to N-1, mCn are binomial coefficients and N = 6 billion.
I’m trying to see if there is, as in case 1, a simple expression which is a good approximation to this.

I don’t know the probability, but I’m sure that in the first scenario, the odds are very high that I would be paired up with Angela Lansbury.

Kind of like taking your grandma to the prom but she does have some nice chi-chis

The solution to Part 2 is 1 - e^-0.5 = 39.35%. The mean number of correct pair-bonds has been cut in half.

I would need further convincing on that point.

The answer that there will be no matches is .367879.

This question is a form of “Montmort’s” problem and is equal to the equation:

(1/2!)-(1/3!)+(1/4)-(1/5!). . .etc.

The interesting thing about the problem is this series converges extremely quickly. For the first six digits, the solution is the same for any number greater than nine, whether the number of couples is 20, 20,000 or 3 billion.

I noticed I made a typo (1/4) should have been (1/4!).

Yes, this is the question I answered in my first post ( except that I gave the probability of at least one match, a trivial difference). What we’re really after at the moment is a decent numerical answer for problem 2.

When you have a large number of trials, with a low probability of success on each trial, the total number of “successes” will follow a Poisson distribution. The probability that the number of successes equals i is given by (e^-L)(L^i)/i!, where L is the mean number of successes.

In these examples, we are interested only in the possibility that i = 0. The question asks about the probability of at least one successful match, which is one minus the probability that i = 0. When i=0, the terms (L^i) and (i!) drop out, since both =1.

As for the mean L, in the first problem we have 3 billion “trials” (pair-bonds), each with 1 possibility in 3 billion of “success” (correct match). Because of the large number of people, each trial is effectively independent. Ergo, L=1. In the second problem, the chance of success drops to 1 in 6 billion, so L drops to 0.5.

So the solutions are:

  1. 1 - e^-1, or 63.21%.
  2. 1 - e^-0.5, or 39.35%.