Now, I’m not a game theorist—I’m just a guy who likes terrible reality TV. I have little expectation that this is groundbreaking work, but the problem struck me as a fun one.
For those less cultured than myself, Married at First Sight Australia is a programme that has been running for several years, loosely based on the Danish series Gift ved første blik—there’s also a UK version, but it’s much more wholesome and thus of far less interest. In the show, which consistently refers to itself as “The Experiment”, a team of people consistently referred to as “The Experts”, led by former New Zealand cricketer John Aiken, assign desperate singles to their ideal partner based on some loose concept of “science”. Based on their track record, I think it’s fair to say they may be more interested in making good telly than in lighting the fires of true love.
As the couples begin their married lives together—which, of course, consist of living in a big block of flats with ten or so other couples and plenty of scope for interpersonal conflict—they face a series of tests. The one I am most interested in for this post is the weekly “commitment ceremony”, in which the couples must decide whether to stay in The Experiment or cut their losses and leave. I’ll get to the details in a moment, but first, a small digression.
You may be familiar with the prisoner’s dilemma, the game theory classic in which two players are faced with a choice: to coöperate or to defect, with no knowledge of what the other player will choose. If both players coöperate, they receive a modest payoff; if both defect, they are worse off. But if Alice coöperates when Bob defects, Bob receives a substantial reward at Alice’s expense.
We can visualise this as a payoff matrix; for example:
B coöperates | B defects | |
---|---|---|
A coöperates | (2, 2) | (0, 3) |
A defects | (3, 0) | (1, 1) |
Here, the first value in each couplet is the payoff for Alice, and the second for Bob. If Alice coöperates, she will receive 2 points if Bob does the same, or zero otherwise. If Alice defects, she will receive 3 points if Bob coöperates, or one if Bob also defects. It is thus clearly in Alice’s interest to defect.
Except that if both players defect, they both end up with a suboptimal payout. The total return is actually greatest if both players coöperate—but this comes with substantial personal risk. Hence the dilemma.
In a more generalised form,
B coöperates | B defects | |
---|---|---|
A coöperates | (w, w) | (x, y) |
A defects | (y, x) | (z, z) |
For the classic prisoner’s dilemma, \(y > w > z > x\), and it’s at its most interesting when \(2w > x + y > 2z\) so that mutual coöperation maximises the overall returns and mutual defection minimises them.
There are lots of alternative versions with different payoff matrices. But what does this have to do with Australian people marrying strangers?
Once a week, the couples are split up, with the men (player A) and women (player B) sent into separate rooms. With them, they take a piece of paper and a pen, with which they are to write either “Stay” or “Leave”. So far, this setup sounds pretty prisoner-y.
So, if you’re happy with your relationship, you probably want to stay. If you’re unhappy, you probably want to leave. But here’s the catch: Because The Experts have a loose understanding of consent, a couple can only leave if both partners write “Leave” on their card.
So the matrix of outcomes looks a bit like this:
B stays | B leaves | |
---|---|---|
A stays | Stay | Stay |
A leaves | Stay | Leave |
Ostensibly, this gives couples a chance to really work on their relationship and turn things around. And that does happen. But in most cases, a Stay/Leave couple is just in for a really awkward week. Nobody wants that.
The exact payoff in each case depends on what it is you actually want.
First, let’s consider a happy couple, Arnold and Beatrice. They’re properly loved up and both want to stay in the experiment. Obviously, they should both write “Stay” on their cards and be done with it. Go home, open a bottle of wine, and have a lovely evening gossiping about all the other dysfunctional couples.
But what if Arnold and Beatrice are both just really insecure? They’re happy, but they’re just not sure their partner feels the same, y’know?
If they both stay, they’re as happy as can be. If they both leave, they’re very sad. If Arnold says leave (even though he’d rather stay) and Beatrice says stay, Arnold is perfectly happy—but Beatrice is a bit less happy than she could be because she’s worried she might have trapped Arnold in an unhappy marriage and he doesn’t really love her. That’s awkward—to account for it, let’s define a quantity, \(a\), that represents the desire of each party to avoid awkwardness as a value between 0 and 1. (Note that this does not need to have the same value to both participants, but we will see that any difference doesn’t actually affect anything.)
That would give us a payoff matrix something like this:
B stays | B leaves | |
---|---|---|
A stays | (1, 1) | (1−a, 1) |
A leaves | (1, 1−a) | (a, a) |
More generally, we have \(w = y\), \(y > x\), and \(y > z\); the sign of the difference between \(x\) and \(z\) depends on the value of \(a\). Realistically, this is a “Cake Eating” situation—everyone wants cake, and there’s enough for everyone, so the equilibrium solution maximises both overall and individual happiness for both players. (If \(a>0.5\), it becomes a Stag Hunt, but the optimal solution remains the same.)
Stay/Stay is the preferred option for everyone involved and maximises the overall return. There’s really no reason to pick anything else.
Now, consider Arthur and Becky. They’re miserable. Arthur’s obsession with model trains is driving Becky mad, and Becky has a really annoying habit of biting her toenails that Arthur just can’t stand. They’d really rather just go home and forget this whole debacle ever happened, thank you very much.
But once again, they’re wracked with doubt. “What if Becky still wants to try and make this work?” muses Arthur. “What if she makes me stay?” The only thing that could make their relationship worse would be the awkwardness of spending a week together after outright admitting that he hates her guts and breaking her annoying little heart. If Arthur has to stay anyway, he’d rather not have Becky spend another week going on and on about how she can’t believe he’d do that in front of The Experts and all the other couples. He should just grin and bear it.
Of course,
Becky is having the exact same thoughts
(modulo s/Becky/Arthur/g
).
That is,
B stays | B leaves | |
---|---|---|
A stays | (a, a) | (0, 0) |
A leaves | (0, 0) | (1, 1) |
or \(z > w > x = y\). This is a rotated version of the “Let’s Party” game. If \(a=1\), this becomes a pure coördination game with no obvious solution for noncommunicating participants,^{1} but for any other value of \(a\) the optimal solution is relatively obvious: If you both want out, just get out. Write “Leave” on your little cards, go back to your separate homes on opposite sides of Australia and never speak to each other again.
This is where it gets interesting. So far, the partners’ incentives have aligned. Arnold and Beatrice both wanted to stay; Arthur and Becky both wanted to leave. But what of Archie and Belinda?
You see, Archie and Belinda see things differently. Archie thinks everything’s going swimmingly: They haven’t had a single argument, everything is very chill, and, as in any healthy marriage, they’ve been getting plenty of alone time to focus on their individual interests. What he doesn’t realise is that he hasn’t actually said a single word to Belinda, or even really acknowledged her presence, since a polite nod on their wedding day and to be honest it’s sort of freaking her out now. Archie wants to stay; Belinda wants to leave.
Let’s work through our familiar little table one cell at a time.
Say they both write “Stay” on their little cards. Archie will be thrilled: He gets to stay with Belinda, and she wants to stay with him too! Belinda will be less chuffed about it, but at least things won’t be any more awkward than they were before, which she values at \(a\).
B stays | B leaves | |
---|---|---|
A stays | (1, a) | ? |
A leaves | ? | ? |
What if they both write “Leave”? Belinda will be thrilled. Archie will be very sad, but at least he avoids some awkwardness and knows it’s probably for the best, so the payoff looks something like this:
B stays | B leaves | |
---|---|---|
A stays | (1, a) | ? |
A leaves | ? | (a, 1) |
We can also fill in the upper right quadrant. Archie gets to make Belinda stay, albeit with some amount of awkwardness, which we already know he values at \(1-a\). Belinda is forced to stay against her will, making her very unhappy:
B stays | B leaves | |
---|---|---|
A stays | (1, a) | (1−a, 0) |
A leaves | ? | (a, 1) |
The final quadrant is a weird one. This is what would happen if both players tried reverse psychology: Archie, who wants to stay, votes “Leave” and Belinda, who wants to leave, votes “Stay”. Archie wouldn’t actually mind this outcome. He really wanted to stay anyway, and Belinda has just told him she wants to stay too—score! Belinda would hate it. She could be home and far away from Archie if only she hadn’t tried to be so bloody tactical. I’d guess they’d value it something like this:
B stays | B leaves | |
---|---|---|
A stays | (1, a) | (1−a, 0) |
A leaves | (1, 0) | (a, 1) |
The symmetry we saw when incentives aligned is gone now. To Archie, this looks like a Cake Eating game; to Belinda, it looks like “Let’s Party”. Since their perceived optimal strategies are opposed, there’s no way to really satisfy everyone here. Archie would like Stay/Stay, but Belinda would prefer Leave/Leave. Either of these options would maximise the overall return, but it’s likely they’d find themselves stuck in a suboptimal Stay/Leave situation, which is the worst of all worlds.
Addendum (2021-06-03): If both players had sufficient understanding of game theory, they might decide to look for Nash equilibria, the points at which neither player can gain anything by changing their strategy. There is a Nash equilibrium at Stay/Stay: Belinda would lose \(a\) points of utility by switching to Leave, while Archie gets a full point whatever he picks. At Stay/Leave, Belinda can gain \(a\) by changing tack; at Leave/Stay, she can gain 1. At Leave/Leave, Belinda has nothing to gain but Archie would see his reward change from \(a\) to \(1-a\) if he were to switch to Stay, so this is also a Nash equilibrium if (and only if) \(a \ge 0.5\). If this doesn’t make much sense to you, feel free to ignore it—it’s not really central to the thesis of this post.
Of course, you might not have thought to draw out these tables with your partner in advance of the ceremony, and now you find yourself sat in a room with a little card and a pen and no way of getting in touch. You don’t know what your partner is thinking, but you could make a guess at the probability that they’ll write “Stay”. What’s the optimal strategy?
If you want to stay, your expected returns are \(P(\mathrm{s}) + (1 - a) (1 - P(\mathrm{s}))\) if you write “Stay”, and \(P(\mathrm{s}) + a (1 - P(\mathrm{s}))\) if you write “Leave”, where \(P(\mathrm{s})\) is the probability that your partner will also write “Stay”. The difference between these outcomes depends entirely on the value you assign to \(a\): How much do you care about not making things awkward? If you really couldn’t care less and just want to spend more time with your assigned partner, \(a = 0\) and you should write “Stay”; if you can’t bear the idea of making them feel trapped and having a bit of an awkward week, \(a = 1\) and you should write “Leave”. The general solution is to write “Stay” as long as
\[a < \frac{1}{2}.\]If you want to leave, your expected returns are \(1 - P(\mathrm{s})\) if you write “Leave”, and \(a P(\mathrm{s})\) if you write “Stay”. You now have two things to consider: The value of avoiding awkwardness, and the probability that your partner has written “Stay”. You should write “Leave” if \(1 - P(\mathrm{s}) > a P(\mathrm{s})\); that is,
\[P(\mathrm{s}) < \frac{1}{a + 1}.\]That looks something like this:
Hopefully this has cleared things up for next year’s participants.
Correction (2021-03-22): An earlier version of this article suggested that no obvious solution existed for any value of a; this is of course false and has now been corrected. ↩