That sounds like a good model for contractors bidding on projects where they have a certain efficiency. A lower bid yields a higher chance of winning but a lower profit if they win.

]]>This is an interesting observation and offhand I cannot think of other examples

]]>Thank you. My mistake was only considering where the support is all positive numbers for the weaker player. In this case, the weaker player mixes in a certain way but the stronger husband chooses a weight which would make him tie the weaker player carrying zero weight. There is a more general case where the weaker player mixes over all positive weights greater than some threshold. I describe it here: https://docs.google.com/a/lehigh.edu/file/d/0BzIVoInXfu-CeXBQMlRoUzlLMEE/edit .

I am having trouble saying anything general about who has a heavier wife. In some cases the expected value is infinite. If the weaker player mixes over all positive weights, then neither player stochastically dominates the other.

]]>Also if we make the speed stochastic (beware the mud pit!) with mean f(s,w) all equilibria will have positive payoffs.

]]>there are conditions under which a mixed equilibrium with positive payoffs exist. see for example here:

]]>1. At least two husbands choose weight 0, the rest choose some nonnegative weight.

2. Suppose the tiebreaker is that the heavier wife wins, then one equilibrium is the weaker husband chooses weight 0 and the stronger husband chooses the weight which will produce a tie. This holds under all conditions on f.

Under a couple of functional forms (f(s,w)=(w+1)/s and f(s,w)=(w+2)^(1/s)), there appear to be no mixed strategy equilibria for two players.

]]>What happens in case of ties? If couples still get the same prize, then the distribution will be degenerate on any weight, not just zero. Suppose all N men pick weight w. Then, by symmetry, they all win so they all get kw.If any deviate, say by e, then you win and get k(w-e) but this is worse so no deviations below w.

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