We play Russian Roulette. There are 4 blanks 2 bullets (assuming you randomly pick 2 places the bullets can go in the cycle, then spin it randomizing the how the rotation ends up). If someone shoots a blank next to you, would you take another shot or spin.

How I approached the problem:

Ok so the chance of surviving if you spin is 2/3. The chance of you surviving if you don't spin depends on the layout of the bullets and blanks. If the two bullets are adjacent then you have a 3/4 chance of surviving, otherwise you have a 1/2 chance of surviving. Where I start to get confused is when I have to find out the chance of getting the case of 2 adjacent bullets versus the 2 bullets separated.

SIDE QUESTION: This might not make sense as a natural progression to the problem but I first tried to think about how many different unique arrangements of bullets you can have (not factoring in perspective differences from rotation). Intuitively it seems like there are only 3: the bullets are adjacent, the bullets are separated by 1 (separated by 3), the bullets are separated by 2 (separated by 2 the other way too). Mathematically though I don't understand why this method doesn't work:

I first tried to think about putting 2 unique people in a line of 6. Obviously the different amount of combinations would be $6\cdot5=30$. Now if this were a circular permutation as in now 2 people sit in a circle of 6 chairs, each unique layout can be put into 6 different perspectives (because you can rotate), giving $\frac{6\cdot5}{6}$. Now each bullet is not unique unlike the persons in chairs, so I would do $\frac{6\cdot5}{6\cdot2!} = 2.5$ which intuitively makes no sense...

Anyways, I assumed there were 3 cases as outlined above that the bullets could be placed, and each would happen with the same probability so the probability of you living given you didn't spin is $$\frac{2}{3}\frac{1}{2} + \frac{1}{3}\frac{3}{4} = \frac{7}{12}$$

I was wondering if this approach was reasonable and how exactly to mathematically deal with the different probabilities of arrangements of the bullets.


  • $\begingroup$ Your $3$ cases are not equally likely. $\endgroup$ May 22, 2012 at 21:18

2 Answers 2


Given the first shot is blank, if you don't spin then the probability that the next shot is a bullet is $\frac{2}{5}$ with two bullets in five places. If you do it is $\frac26=\frac13$. So you should spin.

The possible patterns are



If the two bullets are adjacent then there is a $\frac14$ chance of of the next shot being a bullet and if they are not adjacent there is a $\frac36=\frac12$ chance, as you say. But given a blank first shot, there is a probability of $\frac{4}{10}=\frac25$ of them being adjacent and $\frac35$ not. So the overall probability of the next shot being a bullet is $$\frac25 \times \frac14 + \frac35 \times \frac12 =\frac25.$$

As for your seating people on a round table, the gaps of zero and four and the gaps of one and three are twice as likely as the gaps of two and two. If the individuals and seats can be identified, there are $30$ ways of placing the two individuals: $12 + 12 + 6$; if the seats can be identified but the individuals cannot then there are $15$ ways: $6+6+3$.


Your basic idea was right, but you overlooked one important detail in counting the number of distinguishable arrangements of the bullets. Label the chambers $0$ through $5$ in order. If the bullets are adjacent, they can be in chambers $0$ and $1$, $1$ and $2$, $2$ and $3$, $3$ and $4$, $4$ and $5$, or $5$ and $0$, a total of $6$ possibilities, There are also $6$ possibilities if they are separated by one chamber on one side and three on the other. But there are only three possibilities if they are diametrically opposite each other: $0$ and $3$, $1$ and $4$, and $2$ and $5$. Thus, the probability of adjacent bullets is $6/15=2/5$, as is the probability of bullets separated by one chamber, while the probability of bullets separated by two chambers is only $1/5$: the three cases aren’t equally likely.

Suppose that you don’t spin. As you said, if the bullets are adjacent, your probability of living is $3/4$, and if they’re not, it’s $3/4$, so your overall probability of living is $$\frac25\cdot\frac34+\frac35\cdot\frac12=\frac35\;.$$

Now suppose that you do spin: you’ll live with probability $5/6$, since only one of the six chambers is now loaded. Clearly, then, you should spin.


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