ML Interview Q Series: Solving Russian Roulette Player Probabilities with Geometric Series

Browse all the Probability Interview Questions here.

Three desperados A, B, and C play Russian roulette in which they take turns pulling the trigger of a six-cylinder revolver loaded with one bullet. Each time the magazine is spun to randomly select a new cylinder to fire as long as the deadly shot has not fallen. The desperados shoot in the order A, B, C, A, B, C, and so on. Determine for each of the three desperados the probability that this desperado will be the one to shoot himself dead.

Short Compact solution

The probabilities that each desperado will shoot himself dead are given by summing the relevant geometric series for the turns in which they actually pull the trigger. The results are:

• For desperado A:

  

  For desperado B:

• For desperado C:

Comprehensive Explanation

Overview of the Problem Setup

We have a six-chamber revolver with one live bullet. Before each trigger pull, the cylinder is spun so that one of the six chambers is chosen uniformly at random. The question asks: given that A shoots first, B shoots second, C shoots third, and then the cycle repeats in the order A, B, C, …, what is the probability that A (respectively B or C) ends up shooting himself?

Key Insight: Geometric Series in the Probabilities

Each time the gun is fired, the probability of that shot being deadly is 1/6. The probability the gun does not go off in one particular attempt is 5/6. Because the cylinder is re-spun each time, each shot is independent with probability 1/6 of being fatal and 5/6 of being safe.

For desperado A, he pulls the trigger on shots numbered 1, 4, 7, 10, and so on. In general, these are the shots 1 + 3k for k = 0, 1, 2, 3, … . The probability that the revolver actually kills someone on the 1 + 3k-th shot (and that “someone” is specifically A in that shot) is:

(5/6)^(i-1) * (1/6)

where i = 1 + 3k is the shot index in the entire sequence of all shots. However, the sum can be more directly handled by focusing on the event that A shoots on the 1 + 3k-th trial for any k. Since these events (i.e., the gun goes off on the 1st, 4th, 7th shot, and so on) are mutually exclusive, we sum the probabilities.

Concretely, the probability that A is the one who ends up shooting himself can be expressed as:

Here is why:

• 1/6 is the probability that the bullet fires on a given shot (the deadly event).

• (5/6)^(3k) is the probability that in each of the previous 3k shots (which correspond to one full cycle of A, B, C repeated k times), the gun did not go off.

Because the sum of a geometric series sum_{k=0 to infinity} r^k = 1 / (1 - r) for |r| < 1, we have:

(1/6) / [1 - (5/6)^3]

which numerically is approximately 0.3956.

Probabilities for B and C

Desperado B shoots on shots numbered 2, 5, 8, 11, and so forth. This leads to a factor (5/6) in front because we need to survive the first shot (A’s shot) for B to get to shoot. Thus, B’s probability is:

Similarly, desperado C shoots on shots numbered 3, 6, 9, 12, and so on. To get to the third shot in each cycle, the first two must have been safe. Hence there is a factor (5/6)^2 in front. This gives:

Why the Three Probabilities Sum to 1

If exactly one bullet is loaded and is guaranteed to fire at some point, exactly one of A, B, or C must be the one who gets that fatal shot (there is no scenario where two people die). Hence, these probabilities must sum up to 1. Checking numerically:

0.3956 + 0.3297 + 0.2747 = 1.0000 (approximately)

which is a consistency check on the correctness of the calculations.

Potential Follow-up Question 1

What if the bullet is never re-spun, i.e., the cylinder is not rotated randomly for each shot but only once at the beginning?

In that scenario, the probability analysis changes drastically because the chambers are no longer reset to a uniform random position each time. The bullet might be in a certain position, and each pull moves the cylinder to the next chamber. We then have to reason about the random initial alignment of the cylinder and how it shifts with each shot. The independence is lost because each shot’s outcome depends on whether the bullet has already been fired and on how the cylinder advances. A completely different approach, typically enumerating all initial bullet placements, is needed.

Potential Follow-up Question 2

What if there are two bullets in the cylinder instead of one?

With two bullets loaded and the cylinder re-spun each time, the probability that the revolver fires on a single pull becomes 2/6 = 1/3. The survival probability of any single pull becomes 2/3. Then we would set up a similar geometric sum approach. For A, you would have:

1/3 * sum_{k=0 to infinity} (2/3)^(3k)

and similarly for B and C but with the appropriate powers of 2/3 factored in for each to reflect surviving the shots before them in each cycle.

Potential Follow-up Question 3

Why are these events (A_{1+3k}, B_{2+3k}, and C_{3+3k}) mutually exclusive?

They are mutually exclusive because exactly one shot can be the first deadly shot. If the first deadly shot occurs on the 4th trigger pull, it cannot simultaneously occur on the 1st or 2nd or 3rd pull. Thus, for instance, A_{1} means the first shot was deadly and that excludes all other events from happening for that round. A_{4} means the first shot was safe, the second was safe, the third was safe, and the fourth was deadly, which excludes the scenario that it was deadly at the 1st, 2nd, or 3rd shot, etc. Hence each event occurs in disjoint circumstances, allowing us to sum their probabilities directly.

Below are additional follow-up questions

What if the number of desperados changes?

If there were only two desperados, say A and B, then the turns would cycle A, B, A, B, …, and we would look at each shot’s index mod 2. Desperado A would shoot first (shot index 1), then B would get shot index 2, and so on. Formally, you would again sum geometric series but with a “period” of length 2 instead of 3. Specifically, for A, you would consider i = 1 + 2k, and for B, i = 2 + 2k. Each term’s probability would be (5/6)^(i-1) * (1/6), and you would sum over k = 0, 1, 2, … . The resulting probabilities would still sum to 1 because exactly one of A or B must eventually fire the deadly shot.

If there were four or more desperados, you would extend the same reasoning to a longer cycle. For instance, with four desperados A, B, C, D, the shot sequence would go A, B, C, D, A, B, C, D, and so on. Then each person’s probability is found by considering the sum of probabilities that the bullet fires for that person’s turn in positions i = p + 4k (for p being 1, 2, 3, or 4, depending on the person). The only subtlety is keeping track of each distinct position mod 4 and multiplying by (5/6)^(i-1) for all preceding shots that must have been safe.

A potential pitfall is forgetting that each new player only gets a turn after all previous turns in that cycle fail to fire the bullet. One must carefully align the indices to the correct formula for each person.

How do we handle conditional probabilities once we know a certain number of shots were safe?

Suppose you learn that the first n shots do not fire the bullet. You might ask: “After those n safe shots, what is the conditional probability that A will eventually be the one who dies?” Once you know n shots were safe, you effectively start the scenario again at shot n+1, which is determined by (n+1) mod 3 in the original problem. For example:

• If n ≡ 0 (mod 3), the next shot belongs to A.

• If n ≡ 1 (mod 3), the next shot belongs to B.

• If n ≡ 2 (mod 3), the next shot belongs to C.

From that point on, the probability analysis resets, but with the shooting order starting at the person whose turn is shot n+1. You can then apply the same geometric-series approach, just shifting who acts as the “first” shooter in the new cycle. The main pitfall is to correctly handle the cyclical nature of who gets to shoot next and not assume that it always starts with A after the conditioning.

What happens if we slightly change the spinning mechanism so it is not perfectly uniform?

If the revolver spin is biased (for example, some chambers are more likely to come up than others), the probability of firing on a single pull is no longer exactly 1/6. Instead, let p be the chance that the loaded chamber is selected on a single spin, where p might be different from 1/6. Then each safe shot has probability 1 - p. The formulas become:

• Probability that A kills himself: p * ∑ (1 - p)^(3k), summing from k = 0 to infinity.

• Probability that B kills himself: p * (1 - p) * ∑ (1 - p)^(3k), and so on.

But if the bias is more complicated (e.g., certain chambers are more likely, but not identically each time), you’d need to model that distribution carefully. A significant subtlety arises if the probability distribution changes over time or depends on which chamber was chosen in the previous spins. In that case, you cannot simply raise (1 - p) to a power for each cycle. You might need a Markov chain or other more advanced probabilistic framework. The pitfall is assuming that a “simple multiplication” approach remains valid when the spins are not identically and independently distributed.

How do we interpret results if the bullet is known to have a nonzero chance of misfire?

Sometimes a bullet can fail to fire (a “dud round”) even if it is aligned under the hammer. If the probability of a misfire is q (where 0 <= q <= 1), then the probability that the gun “goes off” on any given pull is (1 - q) * (1/6) if the spin is still uniform. One might reframe the question: “What is the probability that each desperado is the one to get a successful fire (not a misfire) that also ends the game?” That would modify each safe-shot probability. Specifically, being safe is now: either the chamber is empty or the chamber is loaded but the bullet misfires. The overall chance of safe outcome on a single pull = (5/6) + (1/6) * q = 5/6 + q/6. Then all the geometric arguments proceed with that new safe probability. A hidden subtlety is that the bullet can keep “failing” to fire even when it is in the correct position, which prolongs the game. A pitfall is forgetting to account for the new safe probability vs. the bullet “being in place” probability.

What if each desperado has their own “escape mechanism,” i.e., they do not always pull the trigger?

Imagine each desperado might “chicken out” with some probability r_A, r_B, and r_C on their turn (they skip their turn or pass the revolver). If someone skips, the next player gets to shoot next in the order. That changes the distribution of which shot index belongs to which desperado. The game can become more complicated, because the skipping probabilities can chain in ways that are not simply cyclical. You might need a state-based approach, enumerating sequences of skip/no-skip until the bullet fires. The pitfall here is to incorrectly treat the scenario as if the turns remain strictly in order A, B, C when in fact some players may pass, effectively reshuffling the order of who gets the next attempt.

Does it matter if the revolver keeps spinning after the bullet has fired?

This might sound trivial at first, but consider a hypothetical scenario: once a bullet is fired, we might wonder about subsequent events. However, in traditional Russian roulette rules, the game ends immediately after a fatal shot. There is no “post-shot spinning.” So the probabilities we compute assume the game ends as soon as the bullet fires. If, hypothetically, the game did not end (say the question was about a non-lethal event but still a shot, or if we had multiple rounds loaded and the game continues), you must reevaluate the entire stopping criterion. The major pitfall is forgetting that the process is supposed to stop at the first fatal shot and accidentally summing probabilities for subsequent events that should never come into play.