The Spouse vs. Spouse Payout Theorem

[This is an article that I thought was funny, in a math-nerd way.]

My wife, Kathleen Davis Barr, and I used to daydream a lot about playing together as partners
in a Doubles tournament and winning. We tried several times over the years. We entered some
Limited Doubles events for Intermediate/Advanced players and made it to the finals once, but
lost. In Open Doubles events, I don’t think we ever even got as far as the semifinals.
After some first round losses at a couple tournaments in a row, we put our Doubles dream on hold
until we improved.

In the meantime, we also had a chance to restrategize. Maybe it’s better if we don’t put all our
eggs in one basket in the Doubles tournaments. Maybe entering separately with other partners
is a better idea. Assuming that we split the entry fees with our partners, we would then each be
only paying half an entry fee. Of course, that’s the same amount as paying a full entry fee if we
teamed together, but separately we have two chances to win! As a married couple that shares
all their finances, it doesn’t matter which one of us wins because we’ll share the prize money.
(And when I say it doesn’t matter which one of us wins, I mean that strictly financially).
However, we also would only win half as much prize money compared to if we played together
since whoever wins would have to split the winnings with their partner. So really this is a
lower-risk, lower-reward strategy.

In this scenario, our daydream Doubles tournament would be if our teams played against each
other in the Doubles final. Not only would it be fun, but then we would both be winning prize
money! After thinking about this some more, it’s not hard to realize that if a spouse plays
another spouse in a Doubles final, and those spouses combine their winnings, then no matter
what the prize distribution percentage is between 1st and 2nd place, the two spouses will go
home with exactly half of the total prize money available to 1st and 2nd place. One way to think
about that is if we played each other one-on-one in a normal final, then we’d definitely take
home 100% of 1st and 2nd place money, but in a doubles final we would each share with our
partners, so we would take home 50% of the total pot. That’s also assuming that each pair of
doubles partners is splitting their winnings equally.

This “Spouse vs. Spouse Payout Theorem” turns out to be easy to prove mathematically and
can be generalized for any number of players per team, from 1 to n. Assume:
P = total prize pool for 1st and 2nd place, and
x = amount paid to 1st place, and
(P - x) = amount paid to 2nd place.
n = number of players on each team

Again, assuming that each team is splitting winnings equally, then each spouse gets 1/n of the
prize money that the team wins. So the spouse on the winning team wins (1/n) * x dollars and the
other spouse on the losing team wins (1/n) * (P - x) dollars. When the spouses combine those
winnings, together they have:
(1/n)x + (1/n)(P - x)
= (1/n)x + (1/n)P - (1/n)x
= (1/n)P
So in a Doubles tournament, the couple always takes home ½ of P.

There are a couple interesting corollaries to this proof too.

1) There is no point in either spouse being involved in any hedging decisions because they
will win the same amount regardless of the distribution percentages. Whether the split is
60/40, 80/20, etc., it won’t matter. That means any hedging decisions can be left entirely
up to the other partners.

2) In Doubles, if each spouse chooses for their partner one spouse of another married
couple (e.g, Barr/Rockwell vs. Barr/Rockwell) then no matter what the prize distribution,
each couple will win exactly half the pot.

So if David Rockwell and I are ever playing against Linda Rockwell and Kathleen in a Doubles
final, we’ll have to play for something else since the money won’t matter, like the losing team
buys the winning team dinner. Oh wait...never mind