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At 2:34 pm +0100 12/4/99, I wrote: >Suppose there are N applicants for a job, of which S are to be shortlisted >(S < N). A panel of P scrutineers decides on the shortlist. Each scrutineer >can vote for as many candidates as he/she wants, and the candidates >receiving the most votes are shortlisted. > >What is the optimal number of votes for each scrutineer to cast? If they >vote for all the candidates, or alternatively if they vote for none, they >make no difference to the process. Obviously some intermediate number of >votes is the most influential, and I suspect the optimal number is S. Am I >right? The answer to my question hinges on the precise definition of "optimal". Also I need to specify how to handle ties, and what to do if fewer than S candidates get any votes. If the question is viewed as a screening test, optimality could mean maximising the sum of sensitivity (voting for shortlisted candidates) and specificity (not voting for non-shortlisted candidates). Voting for all and voting for none then gives a sum (sensitivity+specificity) of 100% in each case. Some intermediate number of votes ought in general to increase this sum. In the absence of knowledge about the other scrutineers, one might assume as a first step that all candidates have the same expected number of votes (though this is a fairly unrealistic assumption). In this case the correct strategy is to vote just for the candidates you want shortlisted, which will involve no more than S votes, and possibly fewer. As Jim Slattery points out, any more than S votes are wasted as they cannot all be shortlisted. Anyway this analysis to the problem, though imperfect, was sufficient for my purposes, and I have now used it with fellow governors to select a shortlist for appointing a new head teacher. Very many thanks to John Kornak, Eryl Bassett, Dietrich Alte, Miland A I Joshi, Paul Seed, Jim Slattery, Mike Franklin, MM Peterson and Pat Gaffney for their comments. Their individual responses appear below. Tim Cole ------ From: John Kornak <[log in to unmask]> I suggest that you need to create some kind of cost/loss/quality of candidate function and some criterion for creating a cut off point. I would further suggest a Bayesian decision process (but I admit to being biased). Consider the case where there is one candidate you really want to get the job that far outshines everyone else. Also, it is difficult to choose anyone as being the second best i.e. they are all the same standard. Choosing 5 of these at random to vote for will decrease the chances of your preferred candidate getting shortlisted. ------ From: Eryl Bassett <[log in to unmask]> I strongly suspect that you need to specify optimality rather more closely, if you want to get an "optimum" strategy. Suppose, for example, you have one strongly favoured candidate, and your aim is to get that candidate onto the shortlist. I'm pretty sure (though I haven't got a proof on tap) that the best strategy is just to vote for that one candidate. It looks to me as if one might be able to prove that conjecture by some sort of proof by contradiction. If A is *your* candidate, and B and C are other candidates, then (for the case S=2) voting for A and B rather than just for A could result in B and C both beating A onto the shortlist. ------ From: dietrich alte <[log in to unmask]> this is an optimal voting problem. there (at least) are two good books on the subject which may answer your questions: * Mathematics and Politics : Strategy, Voting, Power and Proof (Textbooks inMathematical Sciences) by Alan D. Taylor; $47.95 * Basic Geometry of Voting, by Donald G. Saari; $39.00. i read both of them and like them very much. esp. taylors book is easy to read and gives fascinating insight in the topic. check amazon.com or other bookstores for more info. ------ From: Miland A I Joshi <[log in to unmask]> Your problem is very interesting. I would say that the minimum number is S, in that if it is any less, it is theoretically possible that there will be less than S candidates mentioned at all, hence eligible for the short list. However the problem also depends on the size of P, which needs to be large enough to guarantee that S candidates can be chosen out of the N. Even this will only work for sure if each panellist is compelled to cast S votes. So in practice we need to assume that the votes of the panellists will be correlated such that the short lists will not be very different, and thus the emergence of a short list is likely. Therefore it seems to me that the problem is best solved by practical experience rather than by mathematics. ------ From: Paul Seed <[log in to unmask]> Call the number of votes you cast V. I would have thought V=N/2 (or (N+1)/2 or (N-1)/2 if N is odd). The problem is similar to that of multiple voting in a multi-member constituency. Usually, the maximum number of votes is S, but less are permitted. Your vote is only decisive if it makes or breaks a tie between two candidates. This occurs if (without your votes) a candidate you voted for has the same number of votes or one less vote than a candidate that you did not vote for. Without information about what the other voters will do, you cannot say what ties or near-ties are more likely. To maximise you chance of being effective, you need to have as many candidate pairs with one vote between them as possible. The number of such pairs is V*(N-V) This is 0 for V=0 or V=N , as you noted, and maximum for V=N/2 (or as near as possible). If S<N/2, as it usually is in major elections, and you have only S votes the best option is V=S. A problem will arise if V> N/2. Most people think that more votes means more influence, and try to use their full allocation. This is a mistake. If you have one candidate to whom you are strongly opposed, the solution is to vote for every candidate except this one. If votes are restricted, you should vote for every _marginal_ candidate except this one, assuming that you can identify them. The strong candidates will get in without your suport. ------ From: JIM SLATTERY <[log in to unmask]> S makes a lot of sense. If they vote for more than S they will each vote for candidates they would not shortlist if left to themselves. If they vote for less they will fail to support candidates who they would shortlist. It seems reasonable to choose a number which would make sense no matter how many scrutineers you have. Again, S is the only possibility since you might have only one scrutineer. I suspect that what you really want is a natural definition of 'influential' which makes the intuitive solution the right solution. No doubt someone will supply one, something like maximising the expected satisfaction of the scrutineers. ------ From: Michael Franklin <[log in to unmask]> Not an answer but I think you may need to tighten up on what you mean by 'optimal'. Optimality criteria tend to be very specific and different criteria lead to different results. (I guessed you mean which scoring will lead on average to the highest number of the 'top' s candidates being interviewed - though 'top' is still undefined.) I'm not sure the answer is S I suspect it may be nearer N/2. Think about 'extreme' case of shortlist of S=1 which in effect is very similar to our electoral voting system but would it be better for each elector to vote for, say, 2 candidates as distinguishing between the top two candidates would be based on roughly twice as many 'successes'? ------ From: MM Peterson <[log in to unmask]> A difficulty with the problem as posed is that no rule is given for resolving cases in which after the voting S-x applicants (0<x<=S) receive more votes than any of the remainder, and a further x+y applicants (0<y) each receives the SAME number v of votes, where v is less than the number received by any applicant in the first set but greater than that received by any other applicant. For example, suppose there are 10 applicants, A, B, C, D, E, F, G, H, I and J, the short list is to consist of 3, and that a panel of four scrutineers casts votes as follows Votes for A B C D E F G H I J Scrutineer 1 1 1 1 Scrutineer 2 1 1 1 Scrutineer 3 1 1 1 Scrutineer 4 1 1 1 Totals 4 2 2 2 2 0 0 0 0 0 Clearly applicant A with 4 votes is to be on the short list, but which of B, C, D, E each with 2 votes is to be short-listed? I shall call applicants in this second category MARGINAL, and those who are in that category when all votes except those of scrutineer i are counted i-MARGINAL. Since the number r of i-marginal applicants can by the rules proposed be anything from 0 up to N, and since voting by scrutineer i for any positive number of i-marginal applicants less than r will alter the number of marginal applicants, the solution to the problem of influence for the i-th scrutineer is in my view bound to depend on the rule devised for resolving the problem of the marginal applicants. I don't see how it is possible to proceed further without some information on this. ------ From: Patrick Joseph Gaffney <[log in to unmask]> I don't know if statistics can answer the question without asking more questions (e.g. modelling how people cast their votes -- some interviewers may be highly correlated and vote similarly, other panels might have more independent folks). My suspicion would be to have more than S votes (2S perhaps), if S is small. But I'd think of S as the number of candidates required for 2nd interview in order to get 1 qualified candidate. Again, nothing statistical about this. Discupleme. [log in to unmask] Phone +44 171 242 9789 x 0740 Fax +44 171 242 2723 Epidemiology & Public Health, Institute of Child Health, London WC1N 1EH, UK %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%