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The dating math problem

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However, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them (

However, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one .The $1/e$ number presumably arises as the limit as $N \to \infty$.Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.As always, we must start with a clear statement of the problem.The assumptions, of course, are not entirely reasonable in real applications.(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.

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However, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.

Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one .

The $1/e$ number presumably arises as the limit as $N \to \infty$.

Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.

/e$, to be precise), and then marry the next one .The

However, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one .The $1/e$ number presumably arises as the limit as $N \to \infty$.Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.As always, we must start with a clear statement of the problem.The assumptions, of course, are not entirely reasonable in real applications.(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, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.

Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one .

The $1/e$ number presumably arises as the limit as $N \to \infty$.

Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.

/e$ number presumably arises as the limit as $N \to \infty$.Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are

The task requires the student to use logarithms to solve an exponential equation in the realistic context of carbon dating, important in archaeology and geology, among other places.

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

However, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one .The $1/e$ number presumably arises as the limit as $N \to \infty$.Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.As always, we must start with a clear statement of the problem.The assumptions, of course, are not entirely reasonable in real applications.(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, you can only tell how good the mates are , so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.

Also, for each potential mate, you can either stay with them forever or leave forever, i.e. The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one .

The $1/e$ number presumably arises as the limit as $N \to \infty$.

Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.

$ 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.

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.

$.As always, we must start with a clear statement of the problem.The assumptions, of course, are not entirely reasonable in real applications.(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.

The task requires the student to use logarithms to solve an exponential equation in the realistic context of carbon dating, important in archaeology and geology, among other places.

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 task requires the student to use logarithms to solve an exponential equation in the realistic context of carbon dating, important in archaeology and geology, among other places.

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.

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.

||

The task requires the student to use logarithms to solve an exponential equation in the realistic context of carbon dating, important in archaeology and geology, among other places.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. 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.In particular, when \(n\) is large, is there any reasonable hope of finding the best candidate?

$ 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.

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.