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That doesn't implement the algorithm nor does it meet the problem because you were able to evaluate all options and not one at a time (whereas you would need to

That doesn't implement the algorithm nor does it meet the problem because you were able to evaluate all options and not one at a time (whereas you would need to make a irreversible decision on the spot).

buro9 4y ago
Hot housing market with constant stream of options and where the good ones are barely on the market and require an instant decision and full commitment to a quick decision. Unsure what part doesn't meet the criteria, I consciously used that approach to make the decision.
- StevenWaterman 4y ago
  One of the criteria is a finite, pre-known number of options. In this scenario, you're never forced to buy a house simply because it's the last one and if you don't buy it there's never going to be more houses available.
  - buro9 4y ago
    Finite yes... As otherwise you wouldn't have a stopping problem as you could preselect from all options. What I didn't know is how good each option was, whether to wait for better or not. n is finite as time provides the limit to n. We were looking to move within 6 months (this was a hard requirement based on personal circumstances) and I'd already established the number of likely options according to the rate they appeared on the market.
    - sokoloff 4y ago
      If fewer than 8 places would meet your criteria from among the current inventory plus what comes onto the market in the next 6 months, then you may have used the algorithm. In all of the part of greater London you considered, 7 or fewer possibilities seems quite low, but I don’t know how tight your criteria were in practice relative to that market. If your criteria were that tight, I congratulate you on securing 1 of the 7 suitable properties.
  - Niksko 4y ago
    Set some timeframe you'd like to buy a house in, x months. For small enough x, you can choose some constant rate of houses coming on to the market per month y. xy is your finite number of options. The question then becomes: how much do the simplifications made affect the outcome of applying what we know about the secretaries problem to the real world situation? Someone could probably write a neat article about it.
    - wikfwikf 4y ago
      This is obviously artificial because after x months, you are not going to buy something much worse than previous things you've rejected just because it's the best remaining option, rather than wait one more month. But having to do that is the crux of the secretary problem. OP has not understood the significant differences between this clearly defined mathematical problem and his own experiences, not sure why people are indulging him.
      - buran77 4y ago
        It's perhaps important for people to understand that this type of problem is not necessarily a problem of hiring secretaries but rather a mathematical problem presented in less abstract form that people can easier relate to and digest. There could be dozens of variations, one could involve being able to go back one decision point (change your mind on the previous candidate), one could be to use parallelism and multiple evaluators, one could be to not know the total number so you might have seen the last secretary ever, etc. Most real life problems will never perfectly match to this kind of math problem simply because real life usually offers more flexibility than a strictly defined mathematical scenario. Reminds me of the joke with the mathematician and the engineer who can take as many steps as they want to get to the pot of gold in front of them with the condition that every step is maximum half of the distance remaining. The mathematician is absolutely livid because he knows he'll never reach it, while the engineer is happy because he'll be there for all practical purposes.
        aesthesia 4y ago
        Crucially, the optimal solution depends on the specific constraints. So you can’t just take a solution derived under one set of assumptions and use it in another situation without losing whatever guarantees it made in the original case.
- jamessb 4y ago
  The key part of the Secretary Problem is that you know the number of applicants, and use this to determine how many candidates to examine (1/e of the total under the optimal policy) to determine the selection threshold. Instead, it sounds like you arbitrarily decided to use 2 examples to pick a selection threshold for the "price to square foot" at which you'd accept a property.
  - deleted 4y ago
    [deleted]
- paulcole 4y ago
  1. You don’t know how many candidates there so you couldn’t apply the criteria. 2. You established a threshold of acceptability and took a candidate that met that threshold rather than saying, after X properties, I’ll take the next property that’s better than the best so far. Your approach is much closer to the standard method of buying a house - you have constraints, needs, and wants, and buy a house that fits those.
sytelus 4y ago
It doesn't implement the algorithm but not for the reasons you described. The fundamental issue that this algorithm solves is whether to keep waiting for something better to pop up or accept what you have now. The fundamental constraint is maximum number of trials you are willing to do. So, if you are looking for rental apartment, you might say that you may keep looking for each day for N days. On day 12 when you find the best apartment you have seen so far, the question is if you should commit to it or keep waiting for better in coming days. One of the conclusion of Secretory Problem is that you should not commit to anything until sqrt(N) and then you should pick the one that is the best you have seen so far. This maximizes the goodness of the apartment you might select. If you want to select the best possible apartment then reject first 37% (also called 1/e rule) then select the best you seen so far. This maximizes the probability that you will select the best of the best out of N but this probability is only 37%. The problem with this approach is that there is much higher chance that it wouldn't beat the quality from square root approach.
- pcthrowaway 4y ago
  What happens if the best apartment you could have looked at was the first one, which you rejected because you hadn't seen enough by that point. Following the above heuristic, isn't there a chance you never look at another apartment that would be "better than the others you've looked at" once you enter the decision window, and therefore keep looking until you reach N and have to take the last option? (the implications for dating should be obvious here, with it being fairly common for young lovers to break up so they can "see what's out there", and not uncommon for people to reach an age where they feel the need to settle for "what they can get" at that point)
  - deleted 4y ago
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  - deleted 4y ago
    [deleted]
  - eru 4y ago
    > Following the above heuristic, isn't there a chance you never look at another apartment that would be "better than the others you've looked at" once you enter the decision window, and therefore keep looking until you reach N and have to take the last option? Yes, obviously. As Wikipedia says: > The probability of selecting the best applicant in the classical secretary problem converges toward 1/e ~ 0.368. The classic solution to the classic secretary problem works in less than 37% of cases. But in reality, you tend to know a bit about the distribution beforehand, and you are also interested in getting eg the second best secretary (if the best is no longer available), instead of just going for best-or-bust.
nkmnz 4y ago
Think of the batch-search performed as „one secretary“. You either need to take what that search offered you, or you need to wait for the market to change enough in order to perform the next batch-search.
throw__away7391 4y ago
I'd say that's true of almost every real world example normally discussed. It is quite common for "buyers" to put multiple candidates on ice for as long they can possibly leverage to do so while they evaluate their options, be it in dating, hiring, or purchasing.

buro9 4y ago

Hot housing market with constant stream of options and where the good ones are barely on the market and require an instant decision and full commitment to a quick decision. Unsure what part doesn't meet the criteria, I consciously used that approach to make the decision.

StevenWaterman 4y ago
One of the criteria is a finite, pre-known number of options. In this scenario, you're never forced to buy a house simply because it's the last one and if you don't buy it there's never going to be more houses available.
- buro9 4y ago
  Finite yes... As otherwise you wouldn't have a stopping problem as you could preselect from all options. What I didn't know is how good each option was, whether to wait for better or not. n is finite as time provides the limit to n. We were looking to move within 6 months (this was a hard requirement based on personal circumstances) and I'd already established the number of likely options according to the rate they appeared on the market.
  - sokoloff 4y ago
    If fewer than 8 places would meet your criteria from among the current inventory plus what comes onto the market in the next 6 months, then you may have used the algorithm. In all of the part of greater London you considered, 7 or fewer possibilities seems quite low, but I don’t know how tight your criteria were in practice relative to that market. If your criteria were that tight, I congratulate you on securing 1 of the 7 suitable properties.
- Niksko 4y ago
  Set some timeframe you'd like to buy a house in, x months. For small enough x, you can choose some constant rate of houses coming on to the market per month y. xy is your finite number of options. The question then becomes: how much do the simplifications made affect the outcome of applying what we know about the secretaries problem to the real world situation? Someone could probably write a neat article about it.
  - wikfwikf 4y ago
    This is obviously artificial because after x months, you are not going to buy something much worse than previous things you've rejected just because it's the best remaining option, rather than wait one more month. But having to do that is the crux of the secretary problem. OP has not understood the significant differences between this clearly defined mathematical problem and his own experiences, not sure why people are indulging him.
    - buran77 4y ago
      It's perhaps important for people to understand that this type of problem is not necessarily a problem of hiring secretaries but rather a mathematical problem presented in less abstract form that people can easier relate to and digest. There could be dozens of variations, one could involve being able to go back one decision point (change your mind on the previous candidate), one could be to use parallelism and multiple evaluators, one could be to not know the total number so you might have seen the last secretary ever, etc. Most real life problems will never perfectly match to this kind of math problem simply because real life usually offers more flexibility than a strictly defined mathematical scenario. Reminds me of the joke with the mathematician and the engineer who can take as many steps as they want to get to the pot of gold in front of them with the condition that every step is maximum half of the distance remaining. The mathematician is absolutely livid because he knows he'll never reach it, while the engineer is happy because he'll be there for all practical purposes.
      - aesthesia 4y ago
        Crucially, the optimal solution depends on the specific constraints. So you can’t just take a solution derived under one set of assumptions and use it in another situation without losing whatever guarantees it made in the original case.
jamessb 4y ago
The key part of the Secretary Problem is that you know the number of applicants, and use this to determine how many candidates to examine (1/e of the total under the optimal policy) to determine the selection threshold. Instead, it sounds like you arbitrarily decided to use 2 examples to pick a selection threshold for the "price to square foot" at which you'd accept a property.
- deleted 4y ago
  [deleted]
paulcole 4y ago
1. You don’t know how many candidates there so you couldn’t apply the criteria. 2. You established a threshold of acceptability and took a candidate that met that threshold rather than saying, after X properties, I’ll take the next property that’s better than the best so far. Your approach is much closer to the standard method of buying a house - you have constraints, needs, and wants, and buy a house that fits those.

sytelus 4y ago

It doesn't implement the algorithm but not for the reasons you described. The fundamental issue that this algorithm solves is whether to keep waiting for something better to pop up or accept what you have now. The fundamental constraint is maximum number of trials you are willing to do. So, if you are looking for rental apartment, you might say that you may keep looking for each day for N days. On day 12 when you find the best apartment you have seen so far, the question is if you should commit to it or keep waiting for better in coming days. One of the conclusion of Secretory Problem is that you should not commit to anything until sqrt(N) and then you should pick the one that is the best you have seen so far. This maximizes the goodness of the apartment you might select. If you want to select the best possible apartment then reject first 37% (also called 1/e rule) then select the best you seen so far. This maximizes the probability that you will select the best of the best out of N but this probability is only 37%. The problem with this approach is that there is much higher chance that it wouldn't beat the quality from square root approach.

pcthrowaway 4y ago
What happens if the best apartment you could have looked at was the first one, which you rejected because you hadn't seen enough by that point. Following the above heuristic, isn't there a chance you never look at another apartment that would be "better than the others you've looked at" once you enter the decision window, and therefore keep looking until you reach N and have to take the last option? (the implications for dating should be obvious here, with it being fairly common for young lovers to break up so they can "see what's out there", and not uncommon for people to reach an age where they feel the need to settle for "what they can get" at that point)
- deleted 4y ago
  [deleted]
- deleted 4y ago
  [deleted]
- eru 4y ago
  > Following the above heuristic, isn't there a chance you never look at another apartment that would be "better than the others you've looked at" once you enter the decision window, and therefore keep looking until you reach N and have to take the last option? Yes, obviously. As Wikipedia says: > The probability of selecting the best applicant in the classical secretary problem converges toward 1/e ~ 0.368. The classic solution to the classic secretary problem works in less than 37% of cases. But in reality, you tend to know a bit about the distribution beforehand, and you are also interested in getting eg the second best secretary (if the best is no longer available), instead of just going for best-or-bust.

nkmnz 4y ago

Think of the batch-search performed as „one secretary“. You either need to take what that search offered you, or you need to wait for the market to change enough in order to perform the next batch-search.

throw__away7391 4y ago

I'd say that's true of almost every real world example normally discussed. It is quite common for "buyers" to put multiple candidates on ice for as long they can possibly leverage to do so while they evaluate their options, be it in dating, hiring, or purchasing.