Survival May Not be for the Fittest (Lessons from some TV games)

In this paper we argue that biological fitness is a multi-objective concept hence the statement "fittest" is inappropriate. The following statement is proposed "Survival is mostly for those with non-dominated fitness". Also we use some TV games to show that under the following conditions: i) There are no dominant players. ii) At each time step successful players may eliminate some of their less successful competitors, Then the ultimate winner may not be the fittest (but close). 1. Multi-objective fitness: Definition (1): A measure of biological fitness of an animal is the number of its offspring that reaches maturity. An equivalent definition is that it is the number of its offspring that are in the best possible state (to ensure that they will have their offspring). It is clear that this definition is a multi-objective one since it can be decomposed into the following two objectives: i) Maximize the number of offspring. ii) Maximize their state (e.g. food, protection, training to hunt etc…). These objectives are contradictory hence biological fitness is a multi-objective concept. Now we recall some basics of multi-objective optimization [Ehrgott 2005]. Almost every real life problem is multi-objective (MOB). Methods for MOB optimization are mostly intuitive. Definition (2): A MOB problem is: Minimize (min) subject k 1,2,..., i), (= x Z i to 0) x (h , 0) x (g = ≤ (1) Definition (3): A vector * x dominates x ' if k i x Z x Z i i ,..., 2 , 1) ' () (= ∀ ≤ with strict inequality for at least one i, given that all constraints are satisfied for both vectors. A non-dominated solution * x is called Pareto optimal and the corresponding vector k 1,2,..., i *), (= x Z i is called efficient. The set of such solutions is called a Pareto set. Now we discuss some methods for solving MOB problems: The first method is the lexicographic method. In this method objectives are ordered according to their importance. Then a single objective problem is solved while completing the problem gradually with constraints i.e. 0) (h , 0) (g subject to Z min 1 = ≤ x x (2) then if ZMIN(1) is the solution the second step is (2) in s constraint the and), 1 (Z subject to Z min 1 2 ZMIN = and so on. A famous application is in university admittance where students with highest grades …