Choosing the best or second best secretary
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We have a hall with $N$ secretaries. We choose randomly one secretary at a time and interview them. After the interview either we hire the secretary (and all others go home) or move to the next one (we can't go back and hire them later).
The interviewer wants to hire either the best or second best secretary with the highest probability
My approach:
I know we should use this strategy: after we interview the first $aN$ candidates (which we don't hire), then we choose the first one that's better than the previous $aN$ candidates. I am just stuck knowing what is the value of $a$. What is the probability of this value? and how we get to it?
I have seen the "Secretary Problem" where we want to choose the best one, the difference here is that we are satisfied with the best or second best one equally
probability
add a comment |
up vote
5
down vote
favorite
We have a hall with $N$ secretaries. We choose randomly one secretary at a time and interview them. After the interview either we hire the secretary (and all others go home) or move to the next one (we can't go back and hire them later).
The interviewer wants to hire either the best or second best secretary with the highest probability
My approach:
I know we should use this strategy: after we interview the first $aN$ candidates (which we don't hire), then we choose the first one that's better than the previous $aN$ candidates. I am just stuck knowing what is the value of $a$. What is the probability of this value? and how we get to it?
I have seen the "Secretary Problem" where we want to choose the best one, the difference here is that we are satisfied with the best or second best one equally
probability
I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath
– Adddison
Nov 25 at 17:36
I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version.
– hardmath
Nov 25 at 18:38
1
The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath
– Adddison
Nov 25 at 21:00
Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries.
– hardmath
Nov 27 at 3:40
Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire).
– hardmath
Nov 27 at 19:22
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
We have a hall with $N$ secretaries. We choose randomly one secretary at a time and interview them. After the interview either we hire the secretary (and all others go home) or move to the next one (we can't go back and hire them later).
The interviewer wants to hire either the best or second best secretary with the highest probability
My approach:
I know we should use this strategy: after we interview the first $aN$ candidates (which we don't hire), then we choose the first one that's better than the previous $aN$ candidates. I am just stuck knowing what is the value of $a$. What is the probability of this value? and how we get to it?
I have seen the "Secretary Problem" where we want to choose the best one, the difference here is that we are satisfied with the best or second best one equally
probability
We have a hall with $N$ secretaries. We choose randomly one secretary at a time and interview them. After the interview either we hire the secretary (and all others go home) or move to the next one (we can't go back and hire them later).
The interviewer wants to hire either the best or second best secretary with the highest probability
My approach:
I know we should use this strategy: after we interview the first $aN$ candidates (which we don't hire), then we choose the first one that's better than the previous $aN$ candidates. I am just stuck knowing what is the value of $a$. What is the probability of this value? and how we get to it?
I have seen the "Secretary Problem" where we want to choose the best one, the difference here is that we are satisfied with the best or second best one equally
probability
probability
edited Nov 27 at 2:23
Xander Henderson
14.1k103553
14.1k103553
asked Nov 25 at 17:21
Adddison
826
826
I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath
– Adddison
Nov 25 at 17:36
I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version.
– hardmath
Nov 25 at 18:38
1
The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath
– Adddison
Nov 25 at 21:00
Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries.
– hardmath
Nov 27 at 3:40
Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire).
– hardmath
Nov 27 at 19:22
add a comment |
I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath
– Adddison
Nov 25 at 17:36
I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version.
– hardmath
Nov 25 at 18:38
1
The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath
– Adddison
Nov 25 at 21:00
Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries.
– hardmath
Nov 27 at 3:40
Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire).
– hardmath
Nov 27 at 19:22
I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath
– Adddison
Nov 25 at 17:36
I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath
– Adddison
Nov 25 at 17:36
I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version.
– hardmath
Nov 25 at 18:38
I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version.
– hardmath
Nov 25 at 18:38
1
1
The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath
– Adddison
Nov 25 at 21:00
The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath
– Adddison
Nov 25 at 21:00
Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries.
– hardmath
Nov 27 at 3:40
Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries.
– hardmath
Nov 27 at 3:40
Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire).
– hardmath
Nov 27 at 19:22
Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire).
– hardmath
Nov 27 at 19:22
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.
The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:
- Gusein-Zade, S. M. "The problem of choice and the optimal stopping rule for a sequence of random trials," Teor. Veroyatnost. i Primenen., 1966, [Vol. 11,Issue 3 (1966), pp. 534-537] (English version: Theory of Probability and its Applications, 1966, 11:3, 472–476)
- Gilbert, John P. and Mosteller, Frederick "Recognizing the Maximum of a Sequence," Journal of the American Statistical Association Vol. 61, No. 313 (Mar., 1966), pp. 35-73
Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N to infty$, the asymptotic chance of success for the classic problem $1/e approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.
The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,ldots,N$.
In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 le i le Ncdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $Ncdot m_1(N) lt i le Ncdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $Ncdot m_2(N) lt i le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).
It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $Nto infty$, namely as Gusein-Zade showed:
$$ lim_{Nto infty} m_2(N) = frac{2}{3} $$
and:
$$ lim_{Nto infty} m_1(N) = varphi $$
where $varphi$ is the unique solution in $(0,1)$ of:
$$ varphi - ln varphi = 1 - ln(2/3) $$
Now $varphi approx 0.3470$, and the exact limiting probability of success for large $N$ is $varphi (2-varphi)$, whose approximate value $0.5736$ was mentioned before.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.
The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:
- Gusein-Zade, S. M. "The problem of choice and the optimal stopping rule for a sequence of random trials," Teor. Veroyatnost. i Primenen., 1966, [Vol. 11,Issue 3 (1966), pp. 534-537] (English version: Theory of Probability and its Applications, 1966, 11:3, 472–476)
- Gilbert, John P. and Mosteller, Frederick "Recognizing the Maximum of a Sequence," Journal of the American Statistical Association Vol. 61, No. 313 (Mar., 1966), pp. 35-73
Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N to infty$, the asymptotic chance of success for the classic problem $1/e approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.
The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,ldots,N$.
In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 le i le Ncdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $Ncdot m_1(N) lt i le Ncdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $Ncdot m_2(N) lt i le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).
It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $Nto infty$, namely as Gusein-Zade showed:
$$ lim_{Nto infty} m_2(N) = frac{2}{3} $$
and:
$$ lim_{Nto infty} m_1(N) = varphi $$
where $varphi$ is the unique solution in $(0,1)$ of:
$$ varphi - ln varphi = 1 - ln(2/3) $$
Now $varphi approx 0.3470$, and the exact limiting probability of success for large $N$ is $varphi (2-varphi)$, whose approximate value $0.5736$ was mentioned before.
add a comment |
up vote
1
down vote
accepted
I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.
The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:
- Gusein-Zade, S. M. "The problem of choice and the optimal stopping rule for a sequence of random trials," Teor. Veroyatnost. i Primenen., 1966, [Vol. 11,Issue 3 (1966), pp. 534-537] (English version: Theory of Probability and its Applications, 1966, 11:3, 472–476)
- Gilbert, John P. and Mosteller, Frederick "Recognizing the Maximum of a Sequence," Journal of the American Statistical Association Vol. 61, No. 313 (Mar., 1966), pp. 35-73
Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N to infty$, the asymptotic chance of success for the classic problem $1/e approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.
The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,ldots,N$.
In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 le i le Ncdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $Ncdot m_1(N) lt i le Ncdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $Ncdot m_2(N) lt i le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).
It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $Nto infty$, namely as Gusein-Zade showed:
$$ lim_{Nto infty} m_2(N) = frac{2}{3} $$
and:
$$ lim_{Nto infty} m_1(N) = varphi $$
where $varphi$ is the unique solution in $(0,1)$ of:
$$ varphi - ln varphi = 1 - ln(2/3) $$
Now $varphi approx 0.3470$, and the exact limiting probability of success for large $N$ is $varphi (2-varphi)$, whose approximate value $0.5736$ was mentioned before.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.
The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:
- Gusein-Zade, S. M. "The problem of choice and the optimal stopping rule for a sequence of random trials," Teor. Veroyatnost. i Primenen., 1966, [Vol. 11,Issue 3 (1966), pp. 534-537] (English version: Theory of Probability and its Applications, 1966, 11:3, 472–476)
- Gilbert, John P. and Mosteller, Frederick "Recognizing the Maximum of a Sequence," Journal of the American Statistical Association Vol. 61, No. 313 (Mar., 1966), pp. 35-73
Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N to infty$, the asymptotic chance of success for the classic problem $1/e approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.
The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,ldots,N$.
In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 le i le Ncdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $Ncdot m_1(N) lt i le Ncdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $Ncdot m_2(N) lt i le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).
It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $Nto infty$, namely as Gusein-Zade showed:
$$ lim_{Nto infty} m_2(N) = frac{2}{3} $$
and:
$$ lim_{Nto infty} m_1(N) = varphi $$
where $varphi$ is the unique solution in $(0,1)$ of:
$$ varphi - ln varphi = 1 - ln(2/3) $$
Now $varphi approx 0.3470$, and the exact limiting probability of success for large $N$ is $varphi (2-varphi)$, whose approximate value $0.5736$ was mentioned before.
I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.
The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:
- Gusein-Zade, S. M. "The problem of choice and the optimal stopping rule for a sequence of random trials," Teor. Veroyatnost. i Primenen., 1966, [Vol. 11,Issue 3 (1966), pp. 534-537] (English version: Theory of Probability and its Applications, 1966, 11:3, 472–476)
- Gilbert, John P. and Mosteller, Frederick "Recognizing the Maximum of a Sequence," Journal of the American Statistical Association Vol. 61, No. 313 (Mar., 1966), pp. 35-73
Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N to infty$, the asymptotic chance of success for the classic problem $1/e approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.
The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,ldots,N$.
In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 le i le Ncdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $Ncdot m_1(N) lt i le Ncdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $Ncdot m_2(N) lt i le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).
It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $Nto infty$, namely as Gusein-Zade showed:
$$ lim_{Nto infty} m_2(N) = frac{2}{3} $$
and:
$$ lim_{Nto infty} m_1(N) = varphi $$
where $varphi$ is the unique solution in $(0,1)$ of:
$$ varphi - ln varphi = 1 - ln(2/3) $$
Now $varphi approx 0.3470$, and the exact limiting probability of success for large $N$ is $varphi (2-varphi)$, whose approximate value $0.5736$ was mentioned before.
answered Dec 4 at 16:11
hardmath
28.6k94994
28.6k94994
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I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath
– Adddison
Nov 25 at 17:36
I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version.
– hardmath
Nov 25 at 18:38
1
The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath
– Adddison
Nov 25 at 21:00
Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries.
– hardmath
Nov 27 at 3:40
Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire).
– hardmath
Nov 27 at 19:22