October | 2025 | Martynas M.

In this article I want to argue that the Lottery paradox and the Penrose stairs paradox belong to the same category of paradoxes because they both have the same pattern:

There is a set of points;
there is a point s that is self-evidently P(s);
there is a point e that is self-evidently ~P(e);
there is an induction rule that transfers property P from one point to the other;
in reality¹ there is a point (guard) between (2) and (3) which is ~P and prevents (2) and (4) from proving P(e);
because of (5) the contradiction P(e) and ~P(e) can’t be derived;
there is a lossy compression of reality which moves the guard point from ~P to P.
because of (7) the contradiction P(e) and ~P(e) can be derived.

Penrose Stairs

Let’s start with the Penrose stairs first. It is a paradox because after taking the last step you are both at the end and at the beginning. However, there is no paradox in the 3D version (see Figure 1) of the staircase — you just reach the end as expected. Only when we project 3D to 2D (see Figure 2) at a specific angle we get the paradox.

This situation can be depicted formally by first noting two self-evident truths about staircases:

A1: If step B is higher than step A, and step C is higher than step B, then step C is higher than step A: $H(s_b, s_a) \land H(s_c, s_b) \Rightarrow H(s_c, s_a)$ .

A2: No step is higher than itself, $\neg H(s_i,s_i)$ .

where step B is “higher” than step A can be defined as the bottom side of step B is higher than the top side of step A. Whilst your staircase model might be different than mine, it would nonetheless include the axioms above.

From Figure 1 we can tell that in the actual 3D structure, step 2 is above step 1, step 3 is above step 2, …, step $n$ is above step $n-1$ , and step 1 is not above step $n$ . Formally, this can be expressed as:

D3: $H(s^{3D}_2, s^{3D}_1) \land H(s^{3D}_3, s^{3D}_2) \land \dots \land H(s^{3D}_n, s^{3D}_{n-1}) \land \neg H(s^{3D}_1, s^{3D}_n)$

Because the first step in D3 is not above the last one, we cannot derive a contradiction $H(s^{3D}_i, s^{3D}_i) \land \neg H(s^{3D}_i, s^{3D}_i)$ for some $i$ .

Meanwhile, the specific 2D configuration shown in Figure 2 can be described as:

D2: $H(s^{2D}_2, s^{2D}_1) \land H(s^{2D}_3, s^{2D}_2) \land \dots \land H(s^{2D}_n, s^{2D}_{n-1}) \land H(s^{2D}_1, s^{2D}_n)$

The only difference between D3 and D2 is that $\neg H(s^{3D}_1, s^{3D}_n)$ has been projected to $H(s^{2D}_1, s^{2D}_n)$ . From this description, we can derive a contradiction:

$H(s_n, s_1) \qquad\qquad\qquad\qquad\qquad\qquad\ \text{(by repeated use of A1 on D2)}$
$H(s_1, s_n) \land H(s_n, s_1) \Rightarrow H(s_1, s_1) \quad\quad \text{(A1)}$
$H(s_1, s_1) \qquad\qquad\qquad\qquad \text{(from (1), (2), and the assumption } H(s^{2D}_1, s^{2D}_n)\text{)}$
$H(s_1, s_1) \land \neg H(s_1, s_1) \qquad\qquad\qquad\quad \text{(from (3) and A2)}$

Given the initial list that describes the general pattern and the specific case of the Penrose stairs paradox we can make the following table:

Having outlined how the Penrose stairs embody this pattern, we can now turn to a paradox of belief that exhibits the same structure: the Lottery paradox.

Lottery Paradox

Given that it is highly likely that it will rain tomorrow, it seems rational to believe that it will. Similarly, if you have a lottery with 1000 tickets and only one winner, it appears rational to believe that some arbitrary ticket i will be a loser once the winner is announced. Here, “highly likely” means that the probability of an event is $\ge t$ (acceptance rule), where $t$ is a threshold close to 1. We can state the latter formally as:

$B(p_i)$ where $p_i \Leftrightarrow$ “ticket $i$ is a loser”, and $B \Leftrightarrow$ “I rationally believe that”.

Additionally, if you happened to rationally believe that it will be cloudy tomorrow, and rationally believe that it will rain tomorrow, then it seems rational to believe that tomorrow it will be both cloudy and rainy. Generally, it sounds rational that if one rationally believes A and rationally believes B, one also rationally believes A and B:

P.AGGR: $B(p) \land B(q) \Rightarrow B(p \land q)$

However, there is a problem. If we were to accept the ideas above, then this would lead to paradoxical situations. For example, let’s say that for us it is rational to believe in the events with probability $\ge 0.95$ . It then would be rational to believe in every proposition $p_i$ in the lottery case we saw above or $B(p_1) \land B(p_2) \land \dots \land B(p_{1000})$ (S1). It would also be rational to believe that $\neg B(p_1 \land p_2 \land ... \land p_{1000})$ (S2) since it is certain there will be a winner. Using P.AGGR and S1, we would then conclude $B(p_1 \land p_2 \land ... \land p_{1000})$ (S3), producing a contradiction. Thus, rational belief in highly likely events, combined with P.AGGR, leads to a contradiction known as the Lottery paradox.

Just as there is a 3D model of Penrose stairs there is also a 3D model equivalent for the Lottery paradox, namely the multiverse model. We can treat propositions about the future as being true in their own separate universes. For example, given a 1000-ticket-1-winner lottery, there would be a universe, say #10, in which the 10^th ticket is a winner, the rest are losers. Generally, we could just say that in the i^th universe it is true that: $p_1 \land p_2 \land \dots \land \neg p_i \land \dots \land p_{1000}$ .

In each future universe, everything is certain since it already happened. Therefore, if you rationally believe A and rationally believe B in some universe $i$ , then it is also rational to believe A and B in the universe $i$ . Additionally, in the multiverse model of the lottery there will be no universe in which all of the tickets are losers. All of this can be formally stated as follows:

A1 (or P.AGGR): $B(p, u_i) \land B(q, u_i) \Rightarrow B(p \land q, u_i)$

A2: $\neg B (p_1 \land p_2 \land \dots p_n, u_i)$ .

For simplicity of further analysis, let’s change the number of tickets from 1000 to 3.

By using the multiverse model, we can describe a 3-ticket-1-winner lottery as follows:

M1: $\neg B (p_1, u_1) \land B(p_2, u_1) \land B(p_3, u_1)$

M2: $B(p_1, u_2) \land \neg B (p_2, u_2) \land B(p_3, u_2)$

M3: $B(p_1, u_3) \land B(p_2, u_3) \land \neg B (p_3, u_3)$

where $u_i$ stands for some universe $i$ .

In each universe $u_i$ there is a proposition $\neg B(p_i, u_i)$ that prevents the rule A1 to derive $B(p_1 \land p_2 \land p_3, u_i)$ . Therefore, a contradiction $B(p_1 \land p_2 \land p_3, u_i) \land \neg B (p_1 \land p_2 \land p_3, u_i)$ can’t be derived either.

A good candidate for a 2D version of the multiverse model is a universe model. Just like we can project 3D onto 2D, we can project multiverse onto universe. One way to do it is by using the acceptance rule:

$\frac{\mid B(y, u_j) \mid}{\mid B(y, u_j) \mid + \mid \neg B (y, u_j) \mid} \ge t \Rightarrow B(y, u)$

where $\mid x \mid$ stands for count of formulas of form $x$ .

Because of the acceptance rule, we get the following description of the 3-ticket-1-winner lottery in the universe:

S1: $B (p_1, u) \land B(p_2, u) \land B(p_3, u)$

Guards $(p_1, u_1) \in \neg B, (p_2, u_2) \in \neg B, (p_3, u_3) \in \neg B$ that existed in M1, M2 and M3 respectively and prevented the contradiction got compressed to $(p_1, u) \in B, (p_2, u) \in B, (p_3, u) \in B$ . This in turn allows derivation of a rational belief in a conjunction of all $p_i$ propositions from S1 and A1. Therefore, in the universe model we get a contradiction $B (p_1 \land p_2 \land p_3, u) \land \neg B (p_1 \land p_2 \land p_3, u)$ .

Based on this analysis we can fit the Lottery paradox to the general pattern:

reality here means some bigger space. ↩︎

If you hear your friend Moore say, “It is raining, but I don’t believe it,” you might say that it is absurd, ridiculous or sounds illogical. It would also be absurd if he stated, “The stock market is down, but I don’t believe it.” He would seem to assert that the market is down, but he is not convinced. It would equally sound illogical if he stated, “The market is down, but I believe it is not.” Here, Moore seems to say that the market is down, but he is convinced otherwise, e.g., that it is up. These sentences are called Moorean sentences, and they come in two forms: (O) “X but I don’t believe it” and (C) “X but I believe not-X”. O sentences are called omissive, while C sentences are commissive.

Even though asserting Moorean sentences sound illogical, there is no clear logical contradiction. If Moore only stated that “the stock market is down” (D), there would be no problem, nor if he only stated “I don’t believe that the stock market is down” (~B(D)). In logic a proposition of the form D & ~B(D) is not a contradiction either. Yet, when Moore asserts D & ~B(D), it sounds illogical. This might seem paradoxical, and it is, hence the name Moore’s paradox.

Moorean sentences are problematic in the future tense too, e.g., “The stock market will go down, but I won’t believe it.” However, the past tense is fine: “The stock market was down, but I didn’t believe it.”

However, not all present-tense cases are problematic. For example: ‘He’s coming, but I can’t believe it!’¹ is acceptable, since ‘I can’t believe it’ expresses surprise. Similarly, ‘The disaster is contained, but I don’t believe it’ may be reasonable if the speaker is referring to a propaganda announcement over the radio. Or consider: ‘Train No. … will arrive at … o’clock. Personally, I don’t believe it,’² said by an announcer at the station.

We can turn both the omissive and commissive versions of Moorean sentences into a clear contradiction by adding “I believe” in front of X. For example:

“I believe it is raining, but I don’t believe it” or $B(R) \land \neg B(R)$
“I believe it is raining, but I believe it is not raining” or $B(R) \land B(\neg R) \to B(R) \land \neg B(R)$ ³

By contrast, by providing a source of assertion that is not a personal belief, we can defuse the absurdity. For instance:

“My eyes say it is raining, but I don’t believe it” (skepticism of one’s sensory impulses is not absurd)
“Science says that the universe started with the Big Bang, but I don’t believe it”
“The government says that unemployment is down, but I believe it is not”

In general, Moorean sentences appear incomplete, as they leave unspecified who is asserting X. This missing part is left for us, the interpreters, to fill in. If we choose personal belief as the source of assertion, we get a contradiction. If we choose something else, we get a coherent sentence.

We can analyze the situation by mapping all the different languages that are used when talking about the Moorean sentences. First, there are Moorean sentences which belong to the subset of English, this is level 0, the object-language. These are then talked about in a bigger language (the meta-language) which embeds the object-language and also has propositional logic, this is level 1. Level 1 only talks about the logical form. Finally, we extend the analysis in an even bigger language (the meta-meta-language) which is complete English. This is level 2. The meta-meta-language allows exploring what stating a proposition X might actually mean. What are problematic and non-problematic cases? Is it equivalent to believing it or not? What if the source of assertion is not one’s personal belief?

Level 1 shows no contradiction. However, level 2 provides two contradictory interpretations of what it means to state X: illogical and logical. The illogical interpretation comes from interpreting X as equivalent to B(X), and not-X as equivalent to B(not-X). The logical interpretation comes from interpreting X as equivalent to A(X), and not-X as equivalent to A(not-X), where A is some source of assertion that is not a belief. We can visualize this as follows:

The classical Moore’s paradox arises if one ignores the logical interpretation (green box) in level 2. This creates a situation where the logical interpretation in level 1 (blue box) sounds illogical, because of the highly likely illogical interpretation (red box) in level 2. However, the actual paradox is due to the contradiction in level 2, where a problematic Moorean sentence is both logical and illogical.

Ludwig Wittgenstein, Remarks on the Philosophy of Psychology, Volume 1 (ed. G. E. M. Anscombe & G. H. von Wright, trans. G. E. M. Anscombe, 1980), §485. ↩︎
Ludwig Wittgenstein, Remarks on the Philosophy of Psychology, Volume 1 (ed. G. E. M. Anscombe & G. H. von Wright, trans. G. E. M. Anscombe, 1980), §486. ↩︎
Wikipedia, Doxastic logic – Consistent reasoner: https://en.wikipedia.org/wiki/Doxastic_logic#Types_of_reasoners ↩︎

General pattern	Penrose stairs
A set of points	Pairs of steps $(s_i, s_j)$
P(x)	$H(x)$
S	$(s^{3D}_2, s^{3D}_1), (s^{3D}_3, s^{3D}_2), (s^{3D}_3, s^{3D}_1) \dots$
E	$(s^{3D}_1, s^{3D}_1), (s^{3D}_2, s^{3D}_2), (s^{3D}_3, s^{3D}_3), \dots$
Induction rule	$H(s_b, s_a) \land H(s_c, s_b) \Rightarrow H(s_c, s_a)$
Guard point	$(s^{3D}_1, s^{3D}_n) \in \neg H$ .
Lossy compression	$(s^{2D}_1, s^{2D}_n) \in H$
Contradiction in 2D	$H(s_1, s_1) \land \neg H(s_1, s_1)$

General pattern	Penrose stairs
A set of points	$(p_i, u_j)$
P(x)	$B(x)$
S	$(p_i, u_j)$ where $i \neq j$
E	$(p_1 \land p_2 \land \dots \land p_n, u_i)$
Induction rule	$B(p, u_i) \land B(q, u_i) \Rightarrow B(p \land q, u_i)$
Guard points	$(p_i, u_j)$ where $i = j$
Lossy compression	$(p_i, u)$
Contradiction in 2D	$B(p_1 \land p_2 \land \dots \land p_n, u) \land \neg B(p_1 \land p_2 \land \dots \land p_n, u)$

Martynas M.

Monthly Archives: October 2025

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