Fun Russian Roulette probability question

Spoiler: If the image doesn't load:

We're playing Russian Roulette with a six-chambered revolver. I have loaded two rounds in adjacent chambers, and fired once. It was a blank, and the cylinder has advanced by one position.

Would you prefer to fire again straight away, or to spin the chamber before firing?

Of course, there are natural interesting generalisations which we can talk about once the starter question has been dispatched.

 

The likelihood of landing on a non-filled chamber on a random spin is 2/3. If it landed on a non-empty chamber, the probability of the second trigger pull being empty as well would be 3/4, since the loaded chambers are adjacent. Since the first attempt was not loaded, we can discount the outcome that we would land on the second loaded spot with the second trigger pull - we only have to worry about landing on the first loaded chamber.

If the gun had originally landed on a loaded chamber, we would choose to spin because our chances of getting the second bullet are 1/2 without spinning, where our chances of surviving would be a then 5/6 with spinning.

 

Let's say the chamber we fired is chamber 1. The bullets are in either chambers 2-3, 3-4, 4-5, or 5-6. There's a 1/4 chance that a bullet is in chamber 2. If we spin again, then there's a 2/5 chance we get a bullet in our chamber. So ez choice we don't spin again

 

For those of you who have answered the initial question, you may want to think about these two extensions:

• After a shot is fired, the revolver moves either one chamber forward, or three chambers forward, both with probability 1/2. Now what do you do?
• The two rounds are initially loaded at random, now which option is best?

 

Going to say spinning would be better for both options (assuming they are separate cases). 2/3 (spin) vs 5/8 (no spin) chance for survival for the first and 2/3 (spin) vs 3/5 (no spin) chance for survival for the second.

 

• The four possibilities are
  
  Empty - 100% safe
  Empty - 50% safe, 50% die
  Empty - 50% safe, 50% die
  Empty - 50% safe, 50% die
  Loaded - Not an outcome
  Loaded - Not an outcome
  
  So your likelihood of survival is (.75 * .5) + (.25 * 1) = 62.5%, vs another random spin that is 66.6% chance of survival.
  
  
• There are only 3 positions two bullets can hold in a six chambered revolver. If bullets were added at random, there would be a 40% chance of the bullets being adjacent (x), 40% chance of the bullets being separated by 1 chamber and 3 chambers (y), and 20% chance of the bullets being across from each other (2 chambers in between them) (z).
  
  2x + 2y + z are the outcomes for the bullet loads
  
  Now we have to weigh our survival outcomes
  
  50% chance of 1 chamber movement - x = .75, y = .5, z = .5
  50% chance of 3 chamber movement - x = .625, y = .5, z = .75
  
  Substitute and solve - both 1 and 3 movements result in a 60% chance of survival, averaging to 60% ofc. You should spin for a 66.6% to live.
  

 

oh man this is like those sf threads rich did a few months ago

rip jim

 

Mod squad Superfluous TTB and Entrr are about to show me up with actual equations for this shit lol.