NEWRY, Maine — A Finnish couple has added to their victories by taking first place in the North American Wife Carrying Championship at Maine’s Sunday River ski resort.

Taisto Miettinen and Kristina Haapanen traveled from Helsinki, Finland – where they won the World Wife Carrying Championship – for Saturday’s contest. The Sun Journal (bit.ly/Q30QWq) reports that the couple finished with a time of 52.58 seconds on a course that includes hurdles, sand traps and a water hole.

The winners receive the woman’s weight in beer and five times her weight in cash.

The model: At date 0 each of N husbands decides how fat his wife should be. At date 1 they run a wife-carrying race, where the husband’s speed is given by some function f(s,w) where s is the strength of the husband, and w is the weight of his wife. The function f is increasing in its first argument and decreasing in the second. The winner gets K times his wife’s weight in cash and beer. Questions

If the husbands are symmetric what is the equilibrium distribution of wife weights?

Under what conditions on f does a stronger husband have a fatter wife?

RT @jacobgrier: If people aren't seeing your tweets, try locking your account, committing your tweets to memory, and whispering them to pas… 15 hours ago

RT @retsoor: [pounding podium] the right word is worth 10k pictures 2 days ago

1. 0 is the equilibrium. Any husband choosing a wife with a nonzero weight will not be better off since he will lose. This equilibrium appears to be unique, including mixed strategies.

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.

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.

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.

We could make race times stochastic, or we could make strength s private information. Then we choose weights simultaneously prior to seeing others’ strengths. That turns the game into a first-price auction, with lower wife weights corresponding to higher bids. The private information is on the probability of winning conditional on a bid, though, not value conditional on winning. Are there any auction papers with that form of private info?

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.

[…] so much outrage about Steven Landsburg’s thought experiment, but so little about Jeff Ely’s excellent prelim question? [note, I'm not at all outraged] Jeff puts up a video from the wife-carrying competition, […]

[…] so much outrage about Steven Landsburg’s thought experiment, but so little about Jeff Ely’s excellent prelim question? [note, I'm not at all outraged] Jeff puts up a video from the wife-carrying competition, […]

## 11 comments

Comments feed for this article

April 3, 2013 at 2:44 am

Will Dearden1. 0 is the equilibrium. Any husband choosing a wife with a nonzero weight will not be better off since he will lose. This equilibrium appears to be unique, including mixed strategies.

April 3, 2013 at 3:07 am

TMWhat 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.

April 3, 2013 at 3:41 am

Will Dearden1. 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.

April 3, 2013 at 12:02 pm

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

April 3, 2013 at 8:52 am

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

Click to access 65.pdf

April 3, 2013 at 12:21 pm

Will DeardenThank 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.

April 3, 2013 at 12:52 pm

Alex FWe could make race times stochastic, or we could make strength s private information. Then we choose weights simultaneously prior to seeing others’ strengths. That turns the game into a first-price auction, with lower wife weights corresponding to higher bids. The private information is on the probability of winning conditional on a bid, though, not value conditional on winning. Are there any auction papers with that form of private info?

April 3, 2013 at 6:51 pm

jeffThis is an interesting observation and offhand I cannot think of other examples

April 4, 2013 at 8:13 pm

WatchmakerThat 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.

September 3, 2014 at 10:23 pm

…and while we’re taking offence | News and Features[…] so much outrage about Steven Landsburg’s thought experiment, but so little about Jeff Ely’s excellent prelim question? [note, I'm not at all outraged] Jeff puts up a video from the wife-carrying competition, […]

September 5, 2014 at 2:49 pm

…and while we’re taking offence | weychi.com[…] so much outrage about Steven Landsburg’s thought experiment, but so little about Jeff Ely’s excellent prelim question? [note, I'm not at all outraged] Jeff puts up a video from the wife-carrying competition, […]