Algorithms & Data Structures
HW# The Hiring Problem Mathematically
The problem. Searching for a new hire and interviewing potential candidates. When is the candidate good enough? What’s the stopping criteria?
Formalize an abstract problem. Let us consider each candidate as an integer, the integer representing a ranking criterion. For example: nine candidates whose rank = {1,3,7,5,8,3,1,9,4}.
This problem would be trivial, just pick the element with maximum value, if it weren’t for two properties.
There is no look-ahead. When I’m selecting any one candidate, you are unable to look forward into the future and consider who you will select in the future. No crystal ball.
There is no undo. If you select a candidate and after a while decide to fire them in a misguided attempt to find someone better, there’s a good chance this person will be unavailable in the future gone working for a competitor.
We can think of it visually as a machine which is fed a tape of integers. It has two actions:
it can either stop; or
it can consider the next integer.
The machine’s objective is to stop on the highest integer.
Real world problem. At the heart of the hiring problem is conflict. Do I reject the current possibility in hopes of landing something better if I keep looking, or do I stick with what I have? Can be applied to almost anything a selection choice decision is required. Take it or leave it?
Solving the hiring problem analytically.
Random selection. Choose the 7th element in the list. No reason just 7th element in the list.
The probability, then, of picking the best element from an integer sequence of length N with this random pick rule is (1 divided by N).
To improve on this random selection strategy, search for a while, gain some insight, determine your options, and then choose the next best element that presents itself. In terms of the hiring problem, such a strategy would be to scan through the first r integers and then choose the first option that is greater than any of the integers in [1,r].
How does this new strategy compare to random selection? The above image is a prop to help understand the discussion that follows. Assume that i, is the greatest integer, occurs at n+1.
In order for this strategy to return the maximum integer, two conditions must hold:
- The maximum integer cannot be contained in [1,r]. Our strategy is to scan through [1,r], so if the solution is in [1,r], we necessarily lose. This can also be stated as n≥r.
- Our strategy is going to select the first integer, i, in [r,N] that’s greater than max([1,r]) Given this, there cannot be any integers greater than i that come after i, otherwise the strategy will lose. Alternatively put, the condition max([1,r])==max([1,n]) must be true.
Thus, to calculate the effectiveness of our strategy, we need to know the probability that both of these will hold. For some given n, this is: (r divided by n) multiplied by (1 divided by N). (1 divided by N) is the probability that i occurs at n+1 (remember, this is the probability for some n, not the n), while (r divided by n) is a consequence of the second condition, the probability that the condition max([1,r])==max([1,n]) is true.
To calculate the probability for some r, P®, not for arbitrary n, but for everything, we need to sum over n≥r:
This is a Riemann1 approximation of an integral so we can rewrite it. By letting
Now, we can find the optimal r by solving for P′®=0.
By plugging roptimal back into P® we will find the probability of success.
What The Math Says
Well, the optimal solution is for us to estimate how many people we believe we might reasonably interview in the future, say 20. We plug this into the equation (N divided by e), where N=20, (20 divided by e) ≈7.
This result says that, if we want to maximize our probability of ending up with the best possible candite, we should interview 7 candidates and then, choose the next candidate who is better than all of those candidates.
1 Gary L. Miller. Riemann’s hypothesis and tests for primality. Journal of Computer and System Sciences, 13(3):300–317, 1976.
However, the typical hiring problem maximizes the chances of landing the best candidate and considers all other outcomes equally bad. Most on the choosers are not thinking this way, they want to maximize the probability that they end up with a pretty good candidate. It is not all or nothing.
Maximizing the Probability of a Good Outcome
Fear not, there’s a modification of the hiring problem that maximizes the probability of finding a high-value candidate. Suffice it to say, the strategy is the same except we use a cutoff of √N rather than (N divided by e).
What Sort of Optimal?
At the end of the day, the secretary problem is a mathematical abstraction and fails to take into account much of complexity of, you know, reality.
The solution to the secretary problem suggests that the optimal hiring strategy is to estimate the maximum number of people you are willing to interview, N, and then interview √N people and hire the next person who is better than all of those. In laboratory experiments, people often stop searching too soon when solving searching problems. This suggests that the average person doesn’t search enough candidates prior to choosing.
At the end of the day, the hiring problem is a mathematical abstraction and there is more to finding the “right” candidate than interviewing a certain number of people.
Question: Existing assistant ranks 4. Quantity of thirty candidate elements each with a low- high ranking criterion 0-9. A Reject/Accept decision is to be made after the interview and cannot go back to a rejected candidate. Each scenario below, explain your method and who would you hire?
Scenario #1 Candidate Elements
3 8 3 9 2 3 8 5 1 4 2 7 9 2 4 4 9 7 1 0 1 7 4 2 1 8 1 9 5 1
Scenario #2 Candidate Elements
3 8 3 9 2 3 8 5 1 4 2 6 8 2 4 3 8 6 1 0 1 6 4 2 1 7 1 8 5 1
