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

\[ \text{Probability} = \f{\text{Matches}}{\text{Total}} = \f{n}{n^2} = \f{1}{n} \]... 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

\[ \text{Probability}_\text{odd} = \f{n+(n-1)}{n^2} = \f{2n-1}{n^2} = \f{2}{n}-\f{1}{n^2} \]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

\[ \text{Probability}_\text{even} = \f{n+(n-1)-1}{n^2} = \f{2n-2}{n^2} = \f{2}{n}-\f{2}{n^2} \]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!