Probability in Two Deterministic Universes

Abstract

How can probabilities make sense in a deterministic many-worlds theory? We address two facets of this problem: why should rational agents assign subjective probabilities to branching events, and why should branching events happen with relative frequencies matching their objective probabilities. To address the first question, we generalise the Deutsch–Wallace theorem to a wide class of many-world theories, and show that the subjective probabilities are given by a norm that depends on the dynamics of the theory: the 2-norm in the usual Many-Worlds interpretation of quantum mechanics, and the 1-norm in a classical many-worlds theory known as Kent’s universe. To address the second question, we show that if one takes the objective probability of an event to be the proportion of worlds in which this event is realised, then in most worlds the relative frequencies will approximate well the objective probabilities. This suggests that the task of determining the objective probabilities in a many-worlds theory reduces to the task of determining how to assign a measure to the worlds.

Appendices

Consistent Fine-Graining

For which norms can the fine-graining argument work?

There are multiple ways to fine-grain a game with unequal coefficients into a symmetric game. For example, if $\mu = \big \Vert \big (\Vert {\mathbf {1}}^{(n)}\Vert ,\Vert {\mathbf {1}}^{(m)}\Vert \big )\big \Vert$, where ${\mathbf {1}}^{(n)}$ is the vector with n ones, one can fine-grain the game

$$\begin{aligned} G = \begin{pmatrix} 1 &{}\quad r_1 \\ \mu &{}\quad r_2 \end{pmatrix} \end{aligned}$$

(73)

either by first taking it to

$$\begin{aligned} G' = \begin{pmatrix} 1 &{}\quad r_1 \\ \Vert {\mathbf {1}}^{(n)}\Vert &{}\quad r_2 \\ \Vert {\mathbf {1}}^{(m)}\Vert &{}\quad r_2 \end{pmatrix}, \end{aligned}$$

(74)

and then applying two more fine-grainings to take it to

$$\begin{aligned} G'' = \begin{pmatrix} 1 &{}\quad r_1 \\ {\mathbf {1}}^{(n)} &{}\quad {\mathbf {r}}_2^{(n)} \\ {\mathbf {1}}^{(m)} &{}\quad {\mathbf {r}}_2^{(m)} \end{pmatrix}, \end{aligned}$$

(75)

or by taking G directly to $G''$, which will be possible only if $\mu = \Vert {\mathbf {1}}^{(n+m)}\Vert$. We want all possible ways of fine-graining a game to give same result, so we demand the norm to be such that for all vectors ${\mathbf {v}}$ and ${\mathbf {w}}$ with disjoint support

$$\begin{aligned} \Vert {\mathbf {v}} + {\mathbf {w}}\Vert = \big \Vert \big (\Vert {\mathbf {v}}\Vert ,\Vert {\mathbf {w}}\Vert \big )\big \Vert . \end{aligned}$$

(76)

We also demand the norm to be permutation-invariant, as it seems unphysical to attribute meaning to the labelling of the vectors, and that $\Vert (1,1)\Vert \ne 1$, because otherwise $\Vert {\mathbf {1}}^{(n)}\Vert =1$ for all n, and it is therefore impossible to fine-grain any non-trivial game.

These conditions are enough to show that these norms must be equivalent to p-norms when restricted to vectors of rational numbers, as can be seen by adapting an argument by Bohnenblust [48]. We have then

Theorem 6

Let $\Vert \cdot \Vert :{\mathbb {C}}^N \rightarrow {\mathbb {R}}$ be a permutation-invariant norm for $N\ge 3$ such that $\Vert (1,1)\Vert \ne 1$ and

$$\begin{aligned} \Vert {\mathbf {v}} + {\mathbf {w}}\Vert = \big \Vert \big (\Vert {\mathbf {v}}\Vert ,\Vert {\mathbf {w}}\Vert \big )\big \Vert \end{aligned}$$

(77)

for all vectors ${\mathbf {v}},{\mathbf {w}}$ with disjoint support. Then for any vector ${\mathbf {c}}$ such that the absolute value of its components is rational,

$$\begin{aligned} \Vert {\mathbf {c}}\Vert = \left( \sum _i |c_i|^p\right) ^{\frac{1}{p}}, \end{aligned}$$

(78)

for some real number $p \ge 1$.

Proof

Let f be such that $f(1):=\Vert 1\Vert$ and $f(n+1) := \Vert (1,f(n))\Vert$. First note that $f(1) = 1$, as $\Vert 1\Vert = \Vert (1,0)\Vert = \Vert (\Vert (1,0)\Vert ,\Vert (0,0)\Vert )\Vert = \Vert (1,0)\Vert ^2$.

We need to show that f(n) is monotonous, and that $f(n^k) = f^k(n)$. For the former, consider the identity $2(f(n),0) = (f(n),1) + (f(n),-1)$ and take the norm of both sides. By the triangle inequality

$$\begin{aligned} 2f(n) \le \Vert (f(n),1)\Vert + \Vert (f(n),-1)\Vert = 2f(n+1). \end{aligned}$$

(79)

For the latter, first we show that $f(n+m) = \Vert (f(m),f(n)\Vert$. Assume that it holds for some m. Then

$$\begin{aligned} f(n+1+m) = \Vert (f(m),f(n+1))\Vert = \Vert (f(m),1,f(n))\Vert = \Vert (f(m+1),f(n))\Vert .\nonumber \\ \end{aligned}$$

(80)

Since it holds for $m=1$, by induction it holds for all m. Now assume that $f(nm) = f(n)f(m)$ holds for some m. Then

$$\begin{aligned} f(n(m+1)) = f(nm+n) = \Vert (f(nm),f(n))\Vert = f(n)\Vert (f(m),1)\Vert = f(n)f(m+1),\nonumber \\ \end{aligned}$$

(81)

and therefore by induction this is true for all m, as it obviously holds for $m=1$. This implies that $f(n^k) = f^k(n)$.

Now let $m,n\ge 2$ be some fixed integers, and h the integer such that for any positive integer k

$$\begin{aligned} m^h \le n^k < m^{h+1}. \end{aligned}$$

(82)

Applying $f(\cdot )$ to these numbers, it follows that

$$\begin{aligned} h\log f(m) \le k \log f(n) \le (h+1) \log f(m), \end{aligned}$$

(83)

and using the fact that $h \le k \frac{\log n}{\log m}$ and $h > k \frac{\log n}{\log m}-1$ we have that

$$\begin{aligned} \frac{\log f(m)}{\log m}-\frac{\log f(m)}{k\log n} < \frac{\log f(n)}{\log n} \le \frac{\log f(m)}{\log m}+\frac{\log f(m)}{k\log n}. \end{aligned}$$

(84)

Taking the limit of k going to infinity lets us conclude that

$$\begin{aligned} \frac{\log f(m)}{\log m} = \frac{\log f(n)}{\log n}, \end{aligned}$$

(85)

which means that this fraction is a constant independent of n (and different than 0 as $f(2)>1)$. Calling this constant 1 / p, we conclude that

$$\begin{aligned} f(n) = n^\frac{1}{p}. \end{aligned}$$

Now for any rational number m / n we have that

$$\begin{aligned} \Vert (1,m/n)\Vert = \frac{1}{n}\Vert (n,m)\Vert = \frac{1}{n}\Vert (f(n^p),f(m^p)\Vert = \frac{1}{n}f(n^p+m^p) = (1+(m/n)^p)^\frac{1}{p}, \end{aligned}$$

so by homogeneity $\Vert (a,b)\Vert = (|a|^p + |b|^p)^{\frac{1}{p}}$ for any rationals |a| and |b|, and by induction for any vector ${\mathbf {c}}$ such that the absolute values of the components are rational numbers

$$\begin{aligned} \Vert {\mathbf {c}}\Vert = \left( \sum _i |c_i|^p\right) ^{\frac{1}{p}}. \end{aligned}$$

(86)

$\square$

If one furthermore assumes some regularity condition, then the result is valid for any complex vector.

Reducing Sequential Games to Simple Games

Consider the sequential game

$$\begin{aligned} G = \begin{pmatrix} c_1 &{}\quad r_1 \\ c_2 &{}\quad d_1 &{}\quad s_1 \\ &{}\quad d_2 &{}\quad s_2 \end{pmatrix}, \end{aligned}$$

(87)

where in the worlds with outcome 1 reward $r_1$ is given, but in the worlds with outcome 2 the subgame $H = \begin{pmatrix} d_1 &{} s_1 \\ d_2 &{} s_2 \end{pmatrix}$ is played. We want to reduce it to a simple game in both Kent’s universes and Many-Worlds.

In the regular Kent’s universe $c_1$ worlds are created with reward $r_1$, $c_2d_1$ worlds are created with reward $s_1$, and $c_2d_2$ worlds are created with reward $s_2$, so it seems natural to postulate that G is equivalent to the simple game

$$\begin{aligned} G' = \begin{pmatrix} c_1 &{}\quad r_1 \\ c_2d_1 &{}\quad s_1 \\ c_2d_2 &{}\quad s_2 \end{pmatrix}. \end{aligned}$$

(88)

In the reverse Kent’s universe, there are $M(c_1+c_2)(d_1+d_2)$ worlds in the beginning. After the first branching, $Mc_1(d_1+d_2)$ worlds are imprinted with outcome 1, and the remaining $Mc_2(d_1+d_2)$ worlds are split again, with $Mc_2d_1$ being imprinted with a further outcome 1, and $Mc_2d_2$ with outcome 2. In the end there are $Mc_1(d_1+d_2)$ with reward $r_1$, $Mc_2d_1$ worlds with reward $s_1$, and $Mc_2d_2$ worlds with reward $s_2$, so it seems natural to postulate that G is equivalent to the simple game

$$\begin{aligned} G'' = \begin{pmatrix} c_1(d_1+d_2) &{}\quad r_1 \\ c_2d_1 &{}\quad s_1 \\ c_2d_2 &{}\quad s_2 \end{pmatrix}. \end{aligned}$$

(89)

Note that in the reverse Kent’s universe Substitution is satisfied: the value of the subgame H is

$$\begin{aligned} V(H) = \frac{1}{d_1+d_2}(d_1s_1+d_2s_2), \end{aligned}$$

(90)

and the value of G there is

$$\begin{aligned} V(G)&= \frac{1}{(c_1+c_2)(d_1+d_2)}\Big (c_1(d_1+d_2)r_1 + c_2(d_1s_1 + d_2s_2) \Big ) \end{aligned}$$

(91)

$$\begin{aligned}&= \frac{1}{c_1+c_2}(c_1r_1+c_2V(H)), \end{aligned}$$

(92)

which is equal to the value of $\begin{pmatrix} c_1 &{} r_1 \\ c_2 &{} V(H) \end{pmatrix}$, as required. We shall not prove the general case, as that is quite straightforward.

In Many-Worlds, the game G is instantiated by making a measurement on the state $| \psi \rangle = c_1| 1 \rangle + c_2| 2 \rangle ,$ giving reward $r_1$ in the worlds with outcome 1, and in the worlds with outcome 2 doing a measurement on the state $| \varphi \rangle = d_1| 1 \rangle + d_2| 2 \rangle ,$ finally giving rewards $s_1$ and $s_2$ in the worlds with the second outcomes 1 and 2. These measurements take the initial state $| \psi \rangle | M_? \rangle | \varphi \rangle | D_? \rangle$ to the final state

$$\begin{aligned} | G \rangle = c_1| 1 \rangle | M_1 \rangle | \varphi \rangle | D_? \rangle | r_1 \rangle + c_2d_1| 2 \rangle | M_2 \rangle | 1 \rangle | D_1 \rangle | r_2 \rangle + c_2d_2| 2 \rangle | M_2 \rangle | 2 \rangle | D_2 \rangle | r_3 \rangle .\nonumber \\ \end{aligned}$$

(93)

An equivalent way to play this game is to make a joint measurement on the state

$$\begin{aligned} | \psi \rangle | \varphi \rangle = c_1d_1| 1 \rangle | 1 \rangle + c_1d_2| 1 \rangle | 2 \rangle + c_2d_1| 2 \rangle | 1 \rangle + c_2d_2| 2 \rangle | 2 \rangle , \end{aligned}$$

(94)

but in the worlds where the measurement on $| \psi \rangle$ resulted in 1 apply the label ? to both outcomes of the measurement on $| \varphi \rangle$, leading to the final state

$$\begin{aligned} | G' \rangle= & {} c_1d_1| 1 \rangle | M_1 \rangle | 1 \rangle | D_?' \rangle | r_1 \rangle + c_1d_2| 1 \rangle | M_1 \rangle | 2 \rangle | D_?'' \rangle | r_1 \rangle \nonumber \\&+ \, c_2d_1| 2 \rangle | M_2 \rangle | 1 \rangle | D_1 \rangle | s_1 \rangle + c_2d_2| 2 \rangle | M_2 \rangle | 2 \rangle | D_2 \rangle | s_2 \rangle \end{aligned}$$

(95)

where $| D_?' \rangle$ and $| D_?'' \rangle$ are measurements results physically distinct from $| D_? \rangle$, but with the same label.

If one does not, however, coarse-grain the results (1, 1) and (1, 2) together, then this measurement procedure can be regarded as playing the simple game

$$\begin{aligned} G''' = \begin{pmatrix} c_1d_1 &{}\quad r_1 \\ c_1d_2 &{}\quad r_1 \\ c_2d_1 &{}\quad s_1 \\ c_2d_2 &{}\quad s_2 \end{pmatrix} \end{aligned}$$

(96)

instead. We postulate therefore that a rational agent in Many-Worlds should regard G and $G'''$ as equivalent, or more formally that:

• ReductionThe sequential game

  $$\begin{aligned} G = \begin{pmatrix} \alpha _1 &{}\quad r_1 \\ \vdots &{}\quad \vdots \\ \alpha _n &{}\quad \beta _1 &{}\quad s_1 \\ &{}\quad \vdots &{}\quad \vdots \\ &{}\quad \beta _m &{}\quad s_m \end{pmatrix} \end{aligned}$$

  (97)

  where subgame$(\varvec{\beta },{\mathbf {s}})$is played in the worlds with outcomen, has the same value as the simple game

  $$\begin{aligned} G' = \begin{pmatrix} \alpha _1\varvec{\beta } &{}\quad r_1 \\ \vdots &{}\quad \vdots \\ \alpha _{n-1}\varvec{\beta } &{} r_{n-1} \\ \alpha _n\beta _1 &{}\quad s_1 \\ \vdots &{}\quad \vdots \\ \alpha _n\beta _{m-1} &{}\quad s_{m-1} \\ \alpha _n\beta _m &{}\quad s_m \end{pmatrix}. \end{aligned}$$

  (98)

This Reduction postulate suffices to prove Substitution as a theorem, as the value of the subgame $(\varvec{\beta },{\mathbf {s}})$ is

$$\begin{aligned} V(\varvec{\beta },{\mathbf {s}}) = \frac{1}{\Vert \varvec{\beta }\Vert ^p_p}\sum _{j=1}^m|\beta _j|^p, \end{aligned}$$

(99)

and the value of G is

$$\begin{aligned} V(G)= & {} \frac{1}{\Vert \varvec{\alpha }\Vert ^p_p\Vert \varvec{\beta }\Vert ^p_p}\left( \Vert \varvec{\beta }\Vert ^p_p\sum _{i=1}^{n-1}|\alpha _i|^pr_i + |\alpha _n|^p\sum _{j=1}^m|\beta _j|^p s_j\right) \end{aligned}$$

(100)

$$\begin{aligned}= & {} \frac{1}{\Vert \varvec{\alpha }\Vert ^p_p}\left( \sum _{i=1}^{n-1}|\alpha _i|^pr_i + |\alpha _n|^p V(\varvec{\beta },{\mathbf {s}})\right) , \end{aligned}$$

(101)

as required.