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Edit: This post is meant for fun: I was initially interested in seeing a wider range of mathematical techniques applied to computing the likelihood of a Deal or No-Deal outcome of the negotiations. Specifically, I was interested in how mathematics can be applied to a current real-world example, wanting to see how different fields of mathematics might approach the problem (i.e. Bayesian probabilists vs. Game Theorists).

Since there have been no answers so far, I have decided to narrow the "question" specifically to Game Theory, as I would be most interested in seeing how this particular discipline could be applied here. If any game theorists are interested in responding, I would be very interested in seeing an example.

Original post: This post is meant for fun: given that we have two days of negotiations left, use any mathematical technique or reasoning to compute the probability of a No-Deal. Be creative: game theory, probabilistic theory, Monte Carlo: it's all allowed.

I will post my own answer in a few hours.

As per the request in the comments below, here are some details:

  • Negotiations must end by Sunday, whether a deal has been reached or not (this post is written on Fri evening, i.e. two days before the deadline)
  • EU wants guarantees on aligned regulation ("level playing field"), to make sure that UK can't change it's regulatory environment in key areas to "undercut the EU" and gain a competitive advantage
  • UK rejects the above, because it argues the point of Brexit is to be able to pass its own laws (i.e doesn't wanna be kept in the EU's "regulatory orbit")
  • Both sides have much to loose: EU exports many goods to the UK (trade between Germany and UK alone was worth $139bn in 2019), whilst the goods exported by the UK to the EU amounted to £294bn in 2019 (43% of all UK exports): no deal means trade tariffs governed by WTO (mathematically, treat no deal as economically damaging to both sides)
  • A deal would be economically beneficial to both sides, but EU demands regulatory alignment, which is rejected on political grounds by the UK (and there is an argument for economic benefits in the longer run if own laws can be passed in the future)

Mathematically, you can look at current odds (and their evolution over the past week and months), current market moves in the GBP currency, or use other techniques: use anything you like to come up with a mathematically-backed answer.

Try to stay apolitical.

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    $\begingroup$ This is a very intriguing question. Anyway, I think that most of us (including me) are not really aware about all the major details of the discussion between UK and UE for the Brexit. If you post a list of the important details, I guess you will get more and more answers. $\endgroup$ Dec 11, 2020 at 15:15
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    $\begingroup$ Simple Bayesian estimate scheme: Start with whatever prior for a Bernoulli $p$ (uniform or Jeffreys, for example) and update according to count of how many times the Tories said good/bad things about hard Brexit vs soft Brexit. $\endgroup$ Dec 11, 2020 at 15:29
  • $\begingroup$ Here's a source for some probabilities: oddschecker.com/politics/brexit. They will have done the mathematics based on their book plus their own analysis. I don't know how bookies balance risk/return, but I would think that the odds offered are somewhat informative. $\endgroup$
    – Řídící
    Dec 12, 2020 at 21:40
  • $\begingroup$ Thanks @Řídící, I was more interested in seeing how real mathematics could be used and applied to the problem, but thank you for the resource. $\endgroup$ Dec 13, 2020 at 10:00

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Initial super-basic model to get the thread started (I hope to see some Game theorist & Bayesian probabilist contributions!)

  • Event $A$: EU willing to offer trade deal without UK fully committing to regulatory alignment

  • Event $B$: UK willing to accept a trade deal whilst agreeing to commit to regulatory alignment

  • Event $C$: No-Deal Brexit

  • Event $C^{'}$: Some sort of Deal

Now:

$$\mathbb{P}(C')=\mathbb{P}(A)+\mathbb{P}(A'\cap B)=\\=\mathbb{P}(A)+\mathbb{P}(A')\mathbb{P}(B)=\\=\mathbb{P}(A)+(1-\mathbb{P}(A))\mathbb{P}(B)$$

My subjective guesstimate is that $\mathbb{P}(A)=0.15$, whilst $\mathbb{P}(B)=0.35$. Then clearly:

$$\mathbb{P}(C')=0.15+0.85*0.35=0.4475$$

So that would mean:

$$\mathbb{P}(C)=1-0.4475=55.25\%$$

Latest odds of "no deal" are now $8/11$, which implies: $\mathbb{P}(C)=\frac{8}{8+11}=42.1\%$.

Clearly my very basic model is off. I might want to do some Bayesian refinements and assume distributions of probabilities, instead of constants, and regress these on latest events (as proposed in the comments in the original question: might have a shot 2mrw!).

I am sure a game theory answer can beat my initial super-basic model!

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There have been no answers from Game Theorists (at least not so far): so I have decided to implore google. A bit of searching has led me to the following two Game-Theoretic papers on Brexit (both in a PDF format), I share them here in case others are interested in this topic (both papers are relatively short and relatively easy to follow):

1) Brexit and Europe's future: A game theoretical approach

2) Brexit and game theory: A single-case analysis

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