What are the odds?

Pub: Oct. 11, 2014 | Wri: Oct. 11, 2014

There is a game called "What are the odds?" which goes like this: One person dares another person to do something, that person picks a probability $1$ in $n$ they will do it, and they both say a random number between $1$ and $n$ inclusive at the same time. If the two numbers are equal, then the person has to do the dare.

For example:

Mark: What are the odds you drink this disgusting mixture of milk and apple juice?
John: One in thirty.
Mark: 1... 2... 3...
*both shout 15*
*John drinks the mixture*

What I want to find out is, is the game mathematically correct? What is the probability that the two numbers are equal?

The first person has $n$ choices and the second person has $n$ choices, giving $n^2$ possible pairings. The pairs where the two numbers are equal are $(1, 1), (2, 2), ..., (n, n)$ giving $n$ pairs.

That makes the probability of picking two equal numbers

... which is indeed the correct probability.

Now if you play a variation where the two numbers can also add up to $n$ as well as be the same, the probabilities are different:

If $n$ is odd, the pairs adding to $n$ are $(1, n-1), (2, n-2), ..., (n-1, 1)$, giving $n-1$ pairs, none of which have both numbers the same. That makes the number of matches, $n+(n-1)$, the number of equal number pairs, plus the number of pairs adding to $n$.

The probability is then

If $n$ is even, there are also $n-1$ pairs adding to $n$, but this time there is one pair in that set where both numbers are the same, $(\f{n}{2}, \f{n}{2})$. That makes the number of matches, $n+(n-1)-1$, the number of equal number pairs, plus the number of pairs adding to $n$, minus the one pair counted twice because it is in both groups.

The probability is then

This means that, with this variation, the probability of the person having to do the dare is almost twice as high as they intended it to be!