Shooting Room Paradox and Time Travel
05 Nov 2026John Leslie, in his 1996 book,1 raises a curious puzzle. The setup is as follows.
A “game” is held against infinitely many participants. At stage 0, 1 person is randomly chosen to enter a room. In the room, a die is rolled, and if it lands on six, that person is killed and the game halts. Otherwise, the game proceeds to stage 1, whereby 9 people are randomly chosen from the remaining participants to enter the room. The die is rolled again, and if it lands on six, the 9 people are killed and the game halts. Otherwise, the game proceeds to stage 2, whereby 90 people are randomly chosen, and so on.
Let Jones be one of the participants. After a period in a coma, Jones finds himself in the room. How should Jones set his chance of being killed?
The puzzle is that there seem to be two different but equally justifiable answers.
- Jones knows that the die is fair, so he should put the chance at $1/6$.
- Jones knows that whenever the game halts, 90% of those ever chosen are killed.2 So, having been chosen, he should put the chance at $9/10$.
In this essay I approach the puzzle through several variants, first with a finite population and then with an infinite one. To put the conclusion bluntly, I will argue for the following.
The correct answer is 1/6. However, if a time traveler from the future testifies to Jones that the game eventually halts, then the correct answer is 9/10.
Throughout the essay, I will tacitly assume the following principle.
Epistemic Constraint (EC). At any moment, the agent should only consider epistemically possible worlds, i.e. worlds that are consistent with all the knowledge the agent has gained up to that moment.
Also, let $n_k$ stand for the number of people chosen at stage $k$. Thus,
\[n_k = \begin{cases} 1 & k = 0, \\ 9 \cdot 10^{k-1} & k \geq 1. \end{cases}\]Finite population
Let us first change the setting and assume that there are 100 participants. The game halts automatically once there are no remaining participants. Thus there is no game after stage 2.
Case 1
At the beginning of the game, Jones should put his chance of being killed at 71/600. This is a simple mathematical problem.
\[\begin{align} P(\text{dies}) &= \sum^{2}_{k = 0}P(\text{dies at stage $k$})\\ &= \left( \frac{1}{100}\cdot \frac{1}{6} \right) \\ &\quad + \left( \frac{99}{100}\cdot \frac{5}{6} \right)\left( \frac{9}{99}\cdot\frac{1}{6} \right) \\ &\quad + \left( \frac{99}{100}\cdot \frac{5}{6} \right)\left( \frac{90}{99}\cdot\frac{5}{6} \right)\left( \frac{90}{90}\cdot\frac{1}{6} \right) \\ &= \frac{71}{600} \approx 0.12 \end{align}\]Case 2
Right before the game begins, a time traveler from the future tells Jones that he gets selected at some stage. In this case, Jones should update his chance of being killed to 1/6:
\[\begin{align} &P(\text{chosen}) = \frac{1}{100} + \left( \frac{99}{100}\cdot \frac{5}{6} \right)\left( \frac{9}{99} \right) + \left( \frac{99}{100}\cdot \frac{5}{6} \right)\left( \frac{90}{99}\cdot\frac{5}{6} \right)\left( \frac{90}{90} \right) \\ &\phantom{P(\text{chosen})} = \frac{71}{100} \\\\ &P(\text{dies} \mid \text{chosen}) = \frac{P(\text{dies} \cap \text{chosen})}{P(\text{chosen})} = \frac{P(\text{dies})}{P(\text{chosen})} = \frac{71/600}{71/100} = \frac{1}{6} \end{align}\]Note that EC is implicitly used to justify using the time traveler’s testimony for the conditional update. Note also that the time traveler stands in for Leslie’s original setup, in which Jones finds himself in the room. The time traveler merely delivers this information as testimony rather than as self-locating evidence.
Where, then, does the 90%-intuition go wrong? For every participant, we have
\[\begin{gather} P(\text{dies}) = \frac{\mathbb{E}[\text{#dead}]}{100}, \quad P(\text{chosen}) = \frac{\mathbb{E}[\text{#chosen}]}{100}, \\\\ P(\text{dies} \mid \text{chosen}) = \frac{\mathbb{E}[\text{#dead}]}{\mathbb{E}[\text{#chosen}]}. \end{gather}\]Also, for each possible world $W$,
\[P(W \cap \text{chosen}) = P(W) \cdot \frac{\text{#chosen}(W)}{100}.\]Hence,
\[P(W \mid \text{chosen}) = \frac{P(W)\cdot \text{#chosen}(W)}{\mathbb{E}[\text{#chosen}]}.\]That is, from a chosen participant’s point of view, each world is weighted by its probability times its chosen population. There are four possible worlds:
| World | Outcome | $P(W)$ | $\text{#chosen}(W)$ | $\text{#dead}(W)$ |
|---|---|---|---|---|
| $W_0$ | six at stage 0 | $\frac{1}{6}$ | 1 | 1 |
| $W_1$ | six at stage 1 | $\frac{5}{36}$ | 10 | 9 |
| $W_2$ | six at stage 2 | $\frac{25}{216}$ | 100 | 90 |
| $W_{\text{none}}$ | no six at all | $\frac{125}{216}$ | 100 | 0 |
This gives $\mathbb{E}[\text{#dead}] = \frac{2556}{216}$ and $\mathbb{E}[\text{#chosen}] = \frac{15336}{216}$, and everything checks out:
\[P(\text{dies} \mid \text{chosen}) = \frac{\mathbb{E}[\text{#dead}]}{\mathbb{E}[\text{#chosen}]} = \frac{1}{6}.\]Note that the 90%-intuition is incorrect in this case, because the 90%-intuition refers only to worlds in which the game halts by a six, i.e. $W_1$ and $W_2$. What it overlooks is $W_{\text{none}}$, the world in which the game runs out and nobody dies. Yet for a chosen participant, $W_{\text{none}}$ is the most probable world. In general, with a population of $10^N$, we have
\[\begin{align} P_N(\text{dies} \mid \text{chosen}) &= \frac{\sum_{k=0}^{N} \left(\frac{5}{6}\right)^{k} \cdot \frac{1}{6} \cdot n_k}{\sum_{k=0}^{N} \left(\frac{5}{6}\right)^{k} \cdot n_k} = \frac{1}{6}, \\\\ P_N(W_{\text{none}} \mid \text{chosen}) &= \frac{\left(\frac{5}{6}\right)^{N+1} \cdot 10^N}{\sum_{k=0}^{N}\left(\frac{5}{6}\right)^{k} n_k} \xrightarrow{N \to \infty} \frac{22}{27} \approx 0.81. \end{align}\]Case 3
Right before the game begins, a time traveler from the future tells Jones that he gets selected at some stage, and also that the game halts by killing. In this case, the time traveler eliminates $W_{\text{none}}$ from Jones’s consideration (note again the use of EC here), which makes the 90%-intuition valid. Jones should thus update his chance of being killed to approximately 9/10. It is only “approximately” because of the $W_0$ case; the actual calculation yields
\[P(\text{dies} \mid \text{chosen} \cap \lnot W_{\text{none}}) = \frac{2556}{36 + 300 + 2500} = 0.9012\ldots\]Infinite population
Now there are infinitely many participants, and the game does not halt unless a die lands on six. So it is almost certain that the game halts.
Case 4
At the beginning of the game, Jones should put his chance of being killed at 0, because it is almost certain that he will not be chosen. Indeed, with a population of $10^N$,
\[P(\text{chosen}) = \frac{1}{10^N}\sum_{k=0}^{N}\left(\frac{5}{6}\right)^{k} n_k \xrightarrow{N \to \infty} 0.\]Case 5
Right before the game begins, a time traveler from the future tells Jones that he gets selected at some stage. In this case, Jones should update his chance of being killed to 1/6. This is easier to explain in contrast with Case 6, so let me proceed.
Case 6
Right before the game begins, a time traveler from the future tells Jones that he gets selected at some stage, and that the game eventually halts. In this case, Jones should update his chance of being killed to 9/10.
Why do Case 5 and Case 6 differ? The difference lies in whether the time traveler eliminates $W_{\text{none}}$ for Jones. In the infinite case, $W_{\text{none}}$ is a world in which the game proceeds for eternity. Of course, the probability of such a case is $\lim_{k \to \infty}(5/6)^k = 0$. This may tempt one to ignore $W_{\text{none}}$ altogether in the infinite case.
But recall the table in Case 2. In its infinite limit, although $P(W_{\text{none}})$ would be $0$, we also would have $\text{#chosen}(W_{\text{none}}) = \infty$. Thus we have a $0 \cdot \infty$ situation. In such a case, we cannot simply ignore the term.
In Case 2, the probability of Jones’s dying was $1/6$ because $W_{\text{none}}$ was a live option. When $W_{\text{none}}$ is removed as a live option by EC, the probability increases to $9/10$. The case where $W_{\text{none}}$ is live corresponds to Leslie’s original setup. Jones finds himself in the room but does not know whether he is located in a world in which the game would continue eternally. This mirrors Case 2, and thus it is rational for Jones to set his chance at $1/6$. However, if a time traveler tells Jones that he is not in an eternally running game, then he should set his chance at $9/10$, mirroring Case 3.
The force of this conclusion can be strengthened by considering supertasks. Assume that this game takes place in a suitable Malament–Hogarth spacetime, so that the game is always finished by time $t$. Suppose that Jones finds himself in the room, and beside him is a time traveler from a time later than $t$. Three scenarios are possible, each of which updates Jones’s chance of being killed differently.
- The time traveler testifies nothing ⇒ $1/6$.
- The time traveler testifies that the game has resulted in a killing ⇒ $9/10$.
- The time traveler testifies that the game has resulted in no killing ⇒ $0$.
The striking fact is that, once Jones finds himself in the room, the probability of scenario 3 is no longer zero. Recall that the limit of $P_N(W_{\text{none}} \mid \text{chosen})$ was $22/27$. This limit carries real weight in the infinite case. That is, once Jones finds himself in the room, the probability that the game will run eternally is 22/27.
This may seem strikingly counterintuitive. But because the prior probability of Jones being selected from an infinite population is $0$, the occurrence of this event can turn a zero prior into a nonzero posterior. Indeed, this is consistent with the Malament–Hogarth tripartition, as follows:
\[P(\text{dies} \mid \text{chosen}) = \frac{5}{27}\cdot\frac{9}{10} + \frac{22}{27}\cdot 0 = \frac{1}{6}.\]I hope that this provides a satisfactory resolution to the puzzle.