The trouble with Everettian accounts of probability

[Bruce Kellett]

In a series of papers, Simon Saunders (arxiv:2103.01366 and arxiv:2103.03966) offers an extensive argument for the Everettian interpretation  of quantum mechanics. The paper on structure (arxiv:2103.01366) contains the following paragraph:

"Everett called them branches. It is not hard to see that the connection between amplitudes and Born-rule probability is retained for multiple experiments. The amplitude of each branch, at the end of N experiments, as determined by the unitary evolution alone (together with the initial state), equals the square root of the Born-rule probability for that sequence of outcomes )just multiply together the probabilities for the results taken sequentially). Now consider the superposition of all those branches with the same relative frequency for the "+" outcome; not quite so obviously, the amplitude of this superposition is highly sensitive to the discrepancy, if any, between that relative frequency and the Born rule quantity for the "+" outcomes, the quantity p. Let the discrepancy be eps; then the amplitude falls off exponentially as exp( - N eps^2/kappa), where kappa = 4p(1 - p) and N is, as before, the number of trials. It is the first of a number of quantum Bernoulli theorems, the quantum analogues of the laws of large numbers: the amplitudes of branches with the "wrong" relative frequencies fall off exponentially quickly in the number of trials, in comparison with the amplitudes of Born-rule compliant branches."

This argument was criticized by Adrian Kent (arxiv:0905.0624), but the argument persists, largely because it is central to the Everettians' case that their theory is consistent with the Born rule.

While Kent's criticism still stands, he has made much the same mistake as Saunders. This is to assume that in Everett's picture, each trial is effectively a Bernoulli trial, with a probability of success p. This is not the case. Since every outcome occurs on every Evettian trial, the process cannot be seen as a Bernoulli trial. The crucial point of Saunders' argument hinges on the normal approximation to the Binomial (Bernoulli) distribution as the number of trials becomes large. Since the distribution from Everettian trials is not binomial, this approximation does not hold, and the whole argument collapses.

The essence of a Bernoulli trial is that one outcome occurs with probability p (a 'success'), and other outcomes do not occur. This is in obvious conflict with Everett's approach in which every outcome occurs on every trial.

Bruce

[John Clark]

On Sat, Sep 13, 2025 at 7:49 PM Bruce Kellett <bhkel...@gmail.com> wrote:

> This is to assume that in Everett's picture, each trial is effectively a Bernoulli trial, with a probability of success p. This is not the case. Since every outcome occurs on every Evettian trial, the process cannot be seen as a Bernoulli trial.

You're making an implicit assumption that is usually harmless but not when you're talking about the nature of quantum reality. You are forgetting that a Bernoulli trial is not a Bernoulli trial unless the Bernoulli trial has been OBSERVED.  And every observer observes that only two outcomes are possible, the coin lands heads or the coin lands tails. And after many trials all the observers in the Multiverse deduce that the outcome of one trial does not influence the outcome of another trial. And all the observers in the Multiverse also deduce that the probability of heads and the probability of tails remains the same for every trial. And that is exactly what is required for something to be a Bernoulli trial.

John K Clark    See what's on my new list at  Extropolis

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[Bruce Kellett]

Which probability textbook did you get that out of?

Bruce

[John Clark]

On Sun, Sep 14, 2025 at 7:20 AM Bruce Kellett <bhkel...@gmail.com> wrote:

>> You're making an implicit assumption that is usually harmless but not when you're talking about the nature of quantum reality. You are forgetting that a Bernoulli trial is not a Bernoulli trial unless the Bernoulli trial has been OBSERVED.  And every observer observes that only two outcomes are possible, the coin lands heads or the coin lands tails. And after many trials all the observers in the Multiverse deduce that the outcome of one trial does not influence the outcome of another trial. And all the observers in the Multiverse also deduce that the probability of heads and the probability of tails remains the same for every trial. And that is exactly what is required for something to be a Bernoulli trial.

> Which probability textbook did you get that out of?

 

Any probability textbook will tell you that a "trial" is a formal statistical procedure in which a hypothesis is tested by analyzing data from trials or experiments. But something cannot be analyzed unless there is an analyzer. Or do you think I'm wrong and Bernoulli trials existed during the Jurassic age?

John K Clark    See what's on my new list at  Extropolis

vws

[Brent Meeker]

A Bernoulli trial does not require that the probability of heads and tails be equal.  But Everett entails that on every trial there will be an equal number of heads and tails across all worlds.

Brent

[John Clark]

On Sun, Sep 14, 2025 at 6:51 PM Brent Meeker <meeke...@gmail.com> wrote:

>>> This is to assume that in Everett's picture, each trial is effectively a Bernoulli trial, with a probability of success p. This is not the case. Since every outcome occurs on every Evettian trial, the process cannot be seen as a Bernoulli trial.

 

>> You're making an implicit assumption that is usually harmless but not when you're talking about the nature of quantum reality. You are forgetting that a Bernoulli trial is not a Bernoulli trial unless the Bernoulli trial has been OBSERVED.  And every observer observes that only two outcomes are possible, the coin lands heads or the coin lands tails. And after many trials all the observers in the Multiverse deduce that the outcome of one trial does not influence the outcome of another trial. And all the observers in the Multiverse also deduce that the probability of heads and the probability of tails remains the same for every trial. And that is exactly what is required for something to be a Bernoulli trial.

 

> A Bernoulli trial does not require that the probability of heads and tails be equal. 

The probability of heads or tails does not have to be equal, it might be a weighted coin where the probability of heads on each coin flip is 75% and tails only 25%, but the probability of each throw of the coin must be constant for all coin flips.  

> But Everett entails that on every trial there will be an equal number of heads and tails across all worlds.

Well that is certainly wrong, and it's wrong on several levels. First of all, contrary to what you seem to be implying, Everett never claimed that probability can be determined by counting branches (a.k.a. worlds) nor has anybody in his school of thought claimed that. I've pointed out on this list before exactly why counting branches can't produce a probability. 

And on every trial there will be an equal probability BUT that doesn't mean each trial will produce an equal number of heads. If I was blindfolded and flipped a fair coin 100 times and just let the coins fall to the ground, and then before the blindfold was removed, had to make a bet on what sort of world I was in I would bet I was in the sort of world where there were 50 heads and 50 tails on the ground, but I would prefer not to bet on it at all because the probability of getting EXACTLY 50 heads is only about 8%. So if I'm not in a 50 heads world I'm probably in a world where the number of heads is close to 50, but there's a finite chance I'm in a 100 heads world, although the probability of that is very low, and so is its quantum wave amplitude. 

Until I take the blindfold off and look at the hundred coins on the ground I will always have some uncertainty about what sort of world I'm in. And even in the worlds where they are exactly 50 heads there would be slight differences on exactly where the coins landed.  

 John K Clark    See what's on my new list at  Extropolis

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[Bruce Kellett]

On Mon, Sep 15, 2025 at 11:13 PM John Clark <johnk...@gmail.com> wrote:

On Sun, Sep 14, 2025 at 6:51 PM Brent Meeker <meeke...@gmail.com> wrote:

> But Everett entails that on every trial there will be an equal number of heads and tails across all worlds.

Where did you get that quote from? It seems that you are indulging in your standard pastime of putting words into people's mouths. What I actually said was:

"This is in obvious conflict with Everett's approach in which every outcome occurs on every trial."

Which is not the same as your suggestion.

Bruce

[Brent Meeker]

It's not a quote.  I made it up (when have I ever put words in your mouth?).  If there are two possible outcomes Heads and Tails then according to Everett on each trial they both occur, albeit in different "worlds", and hence there will be equal numbers of H's and T's regardless of whether the coin is fair or loaded.

Brent

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[Bruce Kellett]

On Tue, Sep 16, 2025 at 2:23 PM Brent Meeker <meeke...@gmail.com> wrote:

It's not a quote.  I made it up (when have I ever put words in your mouth?).  If there are two possible outcomes Heads and Tails then according to Everett on each trial they both occur, albeit in different "worlds", and hence there will be equal numbers of H's and T's regardless of whether the coin is fair or loaded.

I am sorry. I thought I was responding to Clark's comment on my position. You are right, according to Everett's theory there will be equal numbers of heads and tails. across the whole 2^N sequences. And this occurs regardless of whether the coin is fair or not (regardless of the 'probability of success'). My point was rather that there is a head and tail outcome on every trial -- which probably amounts to the same thing, though with different emphasis.

Bruce

[Brent Meeker]

John correctly  observes that he hasn't proposed branch counting, but requires that the Born rule attach probability weights to branches.  But he's wrong that nobody ever proposed branch counting.  Even Sean Carroll, an Everettian, discusses differential branching as a way to realize the Born rule.

Brent

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[John Clark]

On Tue, Sep 16, 2025 at 1:05 AM Brent Meeker <meeke...@gmail.com> wrote:

> John correctly  observes that he hasn't proposed branch counting, but requires that the Born rule attach probability weights to branches.  But he's wrong that nobody ever proposed branch counting.  Even Sean Carroll, an Everettian, discusses differential branching as a way to realize the Born rule.

I've read Sean Carroll's books and he does indeed discuss branch counting, but then he goes on to explain why it WILL NOT WORK. See for yourself in a paper that he wrote with Charles T. Sebens:

Self-Locating Uncertainty and the Origin of Probability in Everettian Quantum Mechanics

Right there in the abstract they say "it is tempting to regard each branch as equiprobable, but we give new reasons why that would be inadvisable".

And on page 14 of the article they say: 

"It is tempting to think that the number of copies of Alice cannot change without her physical state changing—this is the way things work in classical physics. But, in Everettian quantum mechanics, changes that purely affect her environment can change the number of copies of Alice in existence"

And on page 17 they say: 

"In this section we will derive the Born rule probabilities as the rational assignment of credences post-measurement pre-observation. We will first derive the rule in a case with two branches that have equal amplitudes, then use similar techniques to treat a case with two branches of unequal amplitude. It is straightforward to extend these methods to more general cases  (see appendix C)"

And finally, although they have two words in common, "differential branch counting" is very different from naïve "branch counting". Instead of discrete branches, in

differential branch counting the wave function of the Multiverse splits in a way that is similar to a continuum. If it really is a continuum then it's impossible to count them because there is literally an uncountable number of universes. However you can group similar universes together in a countable, and possibly finite, number of sets; in a previous post I made the analogy of cutting a rope, which contains an infinite number of points, into a finite number of pieces of rope of various lengths, putting the pieces into a hat and then picking one out of that hat at random. If you knew the number of cuts that were made and their length then you could assign a probability of picking any particular piece out of that hat.   

If you want something that is never negative and always adds up to exactly 1, and is

additive (if you group together multiple branches then the total weight is the sum of the individual weights) then ∣ψ∣^2 is the only way to obtain a probability. Gleason’s theorem proves that in any dimension greater than or equal to 3, the only probability measure that is compatible with the structure of Hilbert space is the Born rule.

You can thus assign a measure over the branches, where the density of branches corresponds to ∣ψ∣^2, you weigh them according to their “thickness” or “differential weight” in Hilbert space. As I've mentioned before, when I see a map of branching universes drawn on a 2D surface, I imagine the line having a thickness determined by ∣ψ∣^2 because that is the natural measure of Hilbert space; it's what makes Hilbert space be Hilbert space. 

John K Clark    See what's on my new list at  Extropolis

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