# Thread: Adam and Eve

1. ## Adam and Eve

There are 125,000 people on earth a few centuries after the initial Adam and Eve. 50% are male, 50% female. 1 in every 500 people is called Adam and 1 in every 500 people is called Eve (no males are called Eve and no females are called Adam).
If all the people are put randomly into male and female pairs, what is the most likely number of Adam and Eve couples that will occur?

Very hard without a method!

2. Originally Posted by Obsidantion
There are 125,000 people on earth a few centuries after the initial Adam and Eve. 50% are male, 50% female. 1 in every 500 people is called Adam and 1 in every 500 people is called Eve (no males are called Eve and no females are called Adam).
If all the people are put randomly into male and female pairs, what is the most likely number of Adam and Eve couples that will occur?

Very hard without a method!
Lets assume that this means that the probability that an arbitary male is an Adam is $1/250$, and a femail is $1/250$. Then the probability that an arbitary couple is an Adam and Eve is $p=(1/250)(1/250)=0.000016$.

Now in a sample of $62500$ couples the number of Adam and Eves has a binomial distribution $B(62500,0.000016)$ , which is approximatly Poisson with mean $1$, so the probability of $k$ such couples is $e^{-1}/k!$, which has a maximum at $k=0$ and $k=1$. So we have to repeat the calculation using the binomial distribution for these, and we find:

$P(0)=(1-p)^{62500}$

and:

$P(1)=(62500 \times p)(1-p)^{62500-1}=(1-p)^{62500-1}$

So $P(1)>P(0).$

RonL

3. Originally Posted by CaptainBlack

So $P(1)>P(0).$

RonL
Thanks for taking the time to work on this puzzle.
Although I had little idea of what you were saying in the previous post, what I gather is that you're saying that the probability of 1 Adam & Eve couple occurring is greater then that of none occuring, but what is your answer? Could the probability of any one number above 1 amount of Adam & Eve couples occurring be greater?

Originally Posted by CaptainBlack

so the probability of $k$ such couples is $e^{-1}/k!$, which has a maximum at $k=0$ and $k=1$.

RonL
Does this part mean that the most likely number can only be 0 or 1?

4. Originally Posted by Obsidantion
Thanks for taking the time to work on this puzzle.
Although I had little idea of what you were saying in the previous post, what I gather is that you're saying that the probability of 1 Adam & Eve couple occurring is greater then that of none occuring, but what is your answer? Could the probability of any one number above 1 amount of Adam & Eve couples occurring be greater?

Does this part mean that the most likely number can only be 0 or 1?
The Poisson approximation tell us that the probability of zero and one such couples is equal and greater than any other possibility, then we revert to greater precission and work out the exact probabilities of these two cases to see if they are resolvable when we use the actual binomial probabilities, when we find that one is the most likely outcome by a tiny margin.

RonL