Two-Bullet Russian Roulette

Let’s look at the arrangement of bullets. After each shot, the cylinder rotates clockwise by 1 chamber.

We know that the first shot is a dud, meaning that it came from chambers 1-4 with equal chance. The next chamber would then be 2, 3, 4, or 5 with equal chance, or a 1 in 4 chance of being a bullet. If we choose to spin the cylinder, we’ll end up with a 2 in 6 chance of landing on a bullet, or 1 in 3. Not spinning the cylinder gives us a better chance of survival.

Now for the other cases.

1 Bullet

If there is only 1 bullet, the first click would be in chambers 1-5, so the next chamber would be 2, 3, 4, 5, or 6: a 1 in 5 chance of being a bullet. Spinning again would give us a 1 in 6 chance of landing on a bullet. You should spin the cylinder.

2 Bullets on Opposite Sides

If there are 2 bullets on opposite sides, the first click would be in chambers 1, 2, 4, or 5; so the next chamber would be 2, 3, 5, or 6: a 1 in 4 chance of being a bullet. Spinning again would give us a 2 in 6 (or 1 in 3) chance of landing on a bullet. You shouldn’t spin the cylinder.

2 Bullets Placed Randomly

Notice that if we spin, we could land on any chamber with equal chance, leaving us a 1 in 3 chance of landing on a bullet no matter what the configuration of bullets are.

How about if we don’t spin? With 6 chambers and 2 bullets, there are $\frac{6\times 5}{2}=15$26×5​=15 different configurations of bullets that we could end up in.

Thankfully, some of them are rotations of each other, which we will use the earlier steps to compute the chance of landing on a bullet.

The first row of configurations are identical to the earlier example where the bullets were in chamber 5 and 6, which we know gives us a 1 in 4 chance of landing on a bullet. We can also see that there are 6 of 25 possible ways to land in this configuration in the first place = $\frac{2}{5}$52​ chance. This means that there is a $\frac{2}{5}$52​ chance of a $\frac{1}{4}$41​ chance of landing on the bullet.

The second row of configurations gives us the same result as the when the bullets were in chamber 4 and 6. By the same logic, there is a $\frac{2}{5}$52​ chance of a $\frac{1}{2}$21​ chance of landing on the bullet.

Finally, the last row of configurations gives us the same result as the when the bullets were in chamber 3 and 6. By the same logic, there is a $\frac{1}{5}$51​ chance of a $\frac{1}{2}$21​ chance of landing on the bullet.

Working our probabilities out, we get $(\frac{2}{5}\times \frac{1}{4})+(\frac{2}{5}\times \frac{1}{2})+(\frac{1}{5}\times \frac{1}{2})=0.4$(52​×41​)+(52​×21​)+(51​×21​)=0.4 or $40\%$40% chance of landing on the bullet. Remembering that by spinning the chamber, we’ll land on a bullet with a $\frac{1}{2}$21​ chance, or about $33.33\%$33.33% chance. Therefore, on average, we’re better off spinning the cylinder.