# The dating math problem

The last assumption, for example, that \(n\) is known, is more appropriate for the secretary interpretation than for the marriage interpretation. What is the probability of success with this strategy?What happens to the strategy and the probability of success as \(n\) increases? It is simple to state and not difficult to solve, but the solution is interesting and a bit surprising.Also, the problem serves as a nice introduction to the general area of statistical decision making.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

The last assumption, for example, that \(n\) is known, is more appropriate for the secretary interpretation than for the marriage interpretation. What is the probability of success with this strategy?What happens to the strategy and the probability of success as \(n\) increases? It is simple to state and not difficult to solve, but the solution is interesting and a bit surprising.Also, the problem serves as a nice introduction to the general area of statistical decision making.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.As always, we must start with a clear statement of the problem.

||The last assumption, for example, that \(n\) is known, is more appropriate for the secretary interpretation than for the marriage interpretation. What is the probability of success with this strategy?

What happens to the strategy and the probability of success as \(n\) increases?

It is simple to state and not difficult to solve, but the solution is interesting and a bit surprising.

Also, the problem serves as a nice introduction to the general area of statistical decision making.

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.

As always, we must start with a clear statement of the problem.

The assumptions, of course, are not entirely reasonable in real applications.

$ 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.As always, we must start with a clear statement of the problem.The two solutions provided differ slightly in their approach in this regard.

(Since this is a decay problem, I expect the constant to be negative.

If I end up with a positive value, I'll know that I should go back and check my work.) In Its radiation is extremely low-energy, so the chance of mutation is very low.

However, I note that there is no beginning or ending amount given.

How am I supposed to figure out what the decay constant is?