# Dating problem probability

As always, we must start with a clear statement of the problem.The assumptions, of course, are not entirely reasonable in real applications.It hits all my requirements for a practice activity: it's self-checking, promotes dialog, allows for some differentiation, requires a little movement, and the kids are doing all the work.To prepare, you will need cards or slips of paper with problems on one side and the answers on the other.

If you are breaking the class into two distinct groups, you need half as many problems as students, but two copies.Arrange your desks in two rows facing each other, like this: Each student gets a problem.Here's where you can differentiate, by giving quick workers more difficult problems.Also, given a strategy that you wait through the first $K$ partners and then marry the next one that is the best so far, I calculate your expected chances of succeeding as \begin \frac \end (Let $M$ be the maximum value among the first $K$ people you meet, where

If you are breaking the class into two distinct groups, you need half as many problems as students, but two copies.

Arrange your desks in two rows facing each other, like this: Each student gets a problem.

Here's where you can differentiate, by giving quick workers more difficult problems.

Also, given a strategy that you wait through the first $K$ partners and then marry the next one that is the best so far, I calculate your expected chances of succeeding as \begin \frac \end (Let $M$ be the maximum value among the first $K$ people you meet, where $1$ is the value of the worst person, $2$ is the next, and so on, with your desired mate having value $N$.

Given $M$, your chances of winning are $\frac$, because the value $N$ must be the first to appear out of the highest $N - M$ values.

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If you are breaking the class into two distinct groups, you need half as many problems as students, but two copies.Arrange your desks in two rows facing each other, like this: Each student gets a problem.Here's where you can differentiate, by giving quick workers more difficult problems.Also, given a strategy that you wait through the first $K$ partners and then marry the next one that is the best so far, I calculate your expected chances of succeeding as \begin \frac \end (Let $M$ be the maximum value among the first $K$ people you meet, where $1$ is the value of the worst person, $2$ is the next, and so on, with your desired mate having value $N$.Given $M$, your chances of winning are $\frac$, because the value $N$ must be the first to appear out of the highest $N - M$ values.

$is the value of the worst person,$ is the next, and so on, with your desired mate having value $N$.Given $M$, your chances of winning are $\frac$, because the value $N$ must be the first to appear out of the highest $N - M$ values.

This sometimes makes first dates a daunting proposition.