Russian Roulette

I came across an interesting probability problem.

In a game of Russian roulette, a revolver has six chambers and one bullet. Players A and B take turns firing at themselves.

1. A fires first, followed by B, without spinning again. What is the probability that A dies?
2. A fires first, followed by B, but the cylinder is spun before every shot. What is the probability that A dies?

Question 1

The first question is a straightforward calculation. A dies when the bullet begins in positions 0, 2, or 4, so $p=\frac{1}{6} \times 3 = \frac{1}{2}$.

Question 2

The second question is more interesting because the cylinder is shuffled before every shot, making the outcomes independent.

No matter how many shots have already been survived, A’s probability of being shot remains the same whenever it is A’s turn.

Thus, $P(\text{A dies, starting with A}) = P(\text{A dies, starting with A}\mid\text{no one died in the previous n rounds})$.

Applying the law of total probability to the first round gives:

$P(\text{A is shot, starting from A}) = P(\text{A is shot on round1}) + P(\text{A is not shot on round 1}) \cdot P (\text{A is shot after round 1} | \text{A is not shot on round 1})$

By independence:

$P (\text{A is shot after round 1} | \text{A is not shot on round 1}) = P (\text{A is shot after round 1})$

By the same independence, $P(\text{A is shot after round 1})$ is equivalent to $P(\text{B is not shot} \mid \text{an irrelevant shot occurs before B's turn}) = P(\text{B is not shot, starting from B}) = 1 - P(\text{A is shot, starting from A})$.

This gives the compact recurrence:

$P = 1/6 + 5/6(1-P)$

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If expanding a single turn is not intuitive enough, we can partition a complete round and apply the law of total probability again.

Let event $B$ mean “A eventually dies.” We divide the first round—one shot by A and one by B, after which the system returns to its original state—into three mutually exclusive and exhaustive cases.

1. $A_1$: A dies on the first shot. Probability: $P(A_1) = \frac{1}{6}$. Since A is already dead, $P(B|A_1) = 1$.

2. $A_2$: A survives the first shot, but B dies on B’s first shot. Probability: $P(A_2) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$. The game ends with B’s death and A survives, so $P(B|A_2) = 0$.

3. $A_3$: Both A and B survive their first shots. Probability: $P(A_3) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$. Because the cylinder is spun before every shot, the second round begins in exactly the same state as the first, again with A shooting first. Therefore, under this condition, A’s eventual probability of death remains $P(B)$: $P(B|A_3) = P(B)$.

Substitute the three cases into the law of total probability:

$$P(B) = P(A_1) \cdot P(B|A_1) + P(A_2) \cdot P(B|A_2) + P(A_3) \cdot P(B|A_3)$$

Substituting the values:

$$P(B) = \left(\frac{1}{6} \times 1\right) + \left(\frac{5}{36} \times 0\right) + \left(\frac{25}{36} \times P(B)\right)$$

Simplifying:

$$P(B) = \frac{1}{6} + \frac{25}{36}P(B)$$

Rearranging and solving:

$$P(B) - \frac{25}{36}P(B) = \frac{1}{6}$$ $$\frac{11}{36}P(B) = \frac{1}{6}$$ $$P(B) = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11}$$