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.
Thanks