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This is the same problem, but your algorithm was not n/e. If there are 100 homes on the market, doesn't the optimal algorithm suggest 100/2.718 => 37 homes mu

This is the same problem, but your algorithm was not n/e. If there are 100 homes on the market, doesn't the optimal algorithm suggest 100/2.718 => 37 homes must be assessed after which you commit hard to the first home that would be above all of those top 37? I mean: you can't argue with the results. You underpaid and are delighted with your home! But you share the problem space without having implemented anything close to the ideal solution. In this case your intuition and sophistication resoundingly /trounced/ the ideal solution in terms of efficiency. So you won!

shepherdjerred 4y ago
> If there are 100 homes on the market, doesn't the optimal algorithm suggest It's not about the number of homes on the market, it's about how many homes you want to look at total (which is capped by the number of homes on the market). So, you say that you want to look at 10 homes max. You take 37% of either quantity as the evaluation period. After the evaluation period (3.7 houses) you commit to the first house that's better than the best you saw in the evaluation set.
- teawrecks 4y ago
  People in this thread also seem to not be aware that the evaluation period can be measured in time as well (as long as the candidates come in uniformly at random, which depending on the time of year, doesn't often happen with houses).