Credits: This way of looking at probability is due to Bruno de Finetti; this particular framing was taught to me by Andrew Critch.

Out of the blue, you get the following email from me:

Dear You:

I extend to you, and you alone, a chance to take part in my free lottery. Please choose at most one of the following options:

- Option A: On November 9th, I’ll roll two standard six-sided dice, and if I roll a double one (“snake-eyes”), I’ll send you $200
- Option B: On November 9th, if Washington DC has been declared for Clinton, I’ll send you $200

If I don’t hear from you within 24 hours, or if your answer isn’t a clear preference for one of these, I won’t do either. Thanks!

If you know me, you know that I’d straightforwardly honour the promise in the email; for this thought experiment set aside all questions about that. It may help you to know that Obama got over 90% of the vote in DC in 2012. There seems zero benefit in refusing the offer or not replying – it’s totally free, and the worst that can happen if you lose is that I don’t send you $200. Would you reply to this email, choosing one of the options? If so, which one? I think it’s obvious what the right choice is, but stop for a moment and decide what you’d do.

What about if I’d offered you the opposite choice: A’ means $200 if I don’t roll snake-eyes, while B’ means $200 if Clinton doesn’t win DC? Does that change your choice?

I hope you chose B in the first case, and A’ in the second. That’s because it seems very clear that Clinton’s chances of winning in DC are very high, and certainly higher than the chances of snake-eyes on a single roll of two dice, which is 1/36 or less than 3%.

However, for many people, this statement contradicts their understanding of what probability is. The most common and widely taught view of probability is strictly frequentist; it makes sense to say that a pair of dice have a less than 3% chance of snake-eyes only because you can roll the dice many times and in the long run they will land snake-eyes one time in 36. You cannot rerun the 2016 Presidential election in DC many times, so it means nothing to say that Clinton has a greater than 3% chance of winning.

If you’re prepared to choose between the options above – if you agree that a single roll of a pair of dice producing snake-eyes is less likely than a Democratic victory in DC in 2016 – then there’s an important sense in which you already reject this view and accept a subjective view of probability.

Interesting post, thanks.

“If you’re prepared to choose between the options above – if you agree that a single roll of a pair of dice producing snake-eyes is less likely than a Democratic victory in DC in 2016”

Those aren’t quite the same things, I don’t think. If I were a hardline frequentist I might be prepared to choose between the bets, and even favour the Clinton-DC option, but I might still bridle at using the term ‘likely’ here since it implies the strong statistical sense of likelihood that doesn’t apply in this context.

There’s one way in which this is hairsplitting nonsense and it doesn’t really matter. But there’s another in which it isn’t and it does.

I think it’s terribly important to keep clear the distinction between a number you’ve generated from a very powerful and accurate model of the world and one you have made up as a wild guess. In this instance, I’m really very confident in the accuracy of my model of the probability of two fair dice coming up snake eyes – particularly two fair dice in a thought experiment run by you. It’s a very powerful and accurate model. I’m pretty unconfident in the accuracy of my model of the chances that Clinton will win DC. My knowledge of US politics is somewhat limited – as I was reading your post, I was half expecting there to be a surprise reveal at some point where it turned out that DC isn’t part of the electoral college at all so the chances of Clinton getting its electoral college votes were nil. (For my and anyone else’s reference: DC does have 3 electoral college votes, following the 23rd Amendment, but no Senator, and they have a single delegate in the House who isn’t allowed to vote substantively.) The information that Obama took over 90% of vote in 2012 certainly helps a lot, but Clinton and Obama have different voter profiles and I know DC is an outlier in population profile, so I wouldn’t be surprised to learn that for some specific reason Obama was loved by the DC electorate but Clinton was loathed. So the model in my head is a very rubbishy one, and being implicitly forced to put a number on the average value feels like pushing it beyond its design parameters. Much better to bet where you are more confident of your numbers.

As time passes, I’m increasingly convinced of the veracity of your aphorism that – sometimes at least – “it is better to pull numbers out of your arse and use those to make a decision than to pull a decision out of your arse”. But numbers you’ve pulled out of your arse are not the same as numbers you’ve generated from a robust and well-tested model. The danger with the rectal digits strategy is you forget where they came from and treat them as cleaner than they are.

Of course, if it became important I could build a better model of the election in DC, and people have. (For reference, as of now, 538 has >99.9% chance of Clinton winning DC, and it looks like Betfair doesn’t even run a market on it. Turns out it is an outlier, and in that direction. I should have guessed from the fact that you chose it in this context.)

I tend to be fairly pragmatic about Bayesian vs frequentist approaches, I think. I’m happy treating probabilities as either subjective beliefs or the limit of the value from a repeated series of trials, or even a mixture of both, depending on context – but keeping track of which is which is important. And one needs to be careful to account for both the accuracy of the model and for its applicability in a particular context.

With the dice example, the model is really pretty accurate. Its applicability in this context is also really pretty high, given that it’s Paul Crowley (who is good at sums) talking about hypothetical fair dice. But if it were a real-life Fat Tony (the streetwise Jersey trader from Nicholas Taleb’s Dr John vs Fat Tony idea) offering me a bet, I would walk away: them dice gotta be loaded, or there’s some other scam. If you want to be Bayesian about it, you say that your prior that it’s a scam is high; if you want to be frequentist about it, you say that the dice roll here is not being drawn from the parent distribution you’ve modelled.

In case this post really is an indirect offer – and to make it clear I agree with you in broad terms – I totally choose Option A (out of A and B) and Option B’ (out of A’ and B’). If we start getting in to me having to pay to play, it’s a very different question, and I suspect our answers might be different, since I’m on principle very reluctant to pay to play that sort of game. But that’s a different question that you have wisely bracketed off in this post.

I more or less believe in subjective probabilities, but I think this argument proves less than you think it does.

Consider the general problem where you have a probability 0 <= p <= 1 and are offering the lottery of $200 with probability p vs Clinton winning DC. Your argument successfully shows that for p <= 1/36 I will pick the Clinton lottery instead of the purely random one. (Technically it only shows that for p=1/36 but I'm going to assume monotonicity here because non-monotonic decisions on this bet are obviously silly).

Given the 2012 result you mentioned, I expect the same argument would more or less work for almost any p < 1.

But now consider the event of Clinton getting a swing state. Say Florida.

If you offered me $200 with p=1/36 vs that I would definitely take Clinton getting Florida. If you offered me $200 with probability 1 – 1 / 36 vs Clinton getting Florida I would definitely take that.

But I don't think you can conclude from this that there's some probability 1 / 36 <= p = 1 – 1/36 in the middle where below that I'd pick Clinton and above that I'd pick the lottery, and at that exact point I'm indifferent between the two.

An alternate scenario which I think is equally plausible is that there are two probabilities 1 / 36 <= p <= q <= 1 – 1/36 where below p I would definitely pick Clinton and above q I would definitely pick Florida, and in between I'd probably just shrug and flip a coin or something.

This feels like a better model of my behaviour in most scenarios, and I also think it captures something interesting about the situation that the idea that I have a single subjective probability doesn't: I think the gap q – p is highly likely to depend on the stake. If you're offering me $20 it's probably going to be quite large because unless I'm intrinsically interested in the question I won't put much effort into determining what I think are reasonable odds. If you're offering me $200 then I'd likely put in the effort to narrow the odds a bit. If you're offering me $2000 then I'd *definitely* put in the effort, and I'd expect q – p to be quite small indeed (say, well in the region of 0.01, which makes the expected difference between the two to be in the region of that $20 I didn't care that much about).

(All this assumes I don't just go to an odds checker and use whatever numbers they use, which is what I'd do in practice, but I think for the purpose of this thought experiment offloading to an external source of probability and claiming that as a subjective probability would be cheating. If necessary we can replace the whole "Clinton wins Florida" thing with something more obscure that the bookies aren't running bets on)

This model doesn't of course prove that I don't *have* a subjective probability, and I'm not intending it to reject the concept, but I think it's just a good a fit to your argument as a purer subjective probability would, and highlights an important point – having a subjective probability does not mean that it's accessible to me, and the process of trying to access it may require changing its value.

Oh, and as a completely off-the-side pointless nitpick, I’m not sure “However, for many people, this statement contradicts their understanding of what probability is.” is entirely the case, depending on what one means by ‘many’. I agree that frequentist probability is taught more often, but it’s not taught particularly clearly to most people, and most people who have been taught it don’t remember it particularly clearly. The group of people who know and care enough about statistics to form a view here is very small as a proportion of humanity. Sadly.

In conclusion, 7,000 children under 5 died of malnutrition today.

I do in fact believe that a great many lives depend on people’s ability to think clearly on matters of probability.