Best Bets vs. Russian Roulette (April '23 Challenge Question)

First of all, the host can't pull the trigger, the contestant must(haven't you ever seen the Deer Hunter?)...assuming this revolver has 6 shots, the first spin would have a 1 in 3 chance of ending badly, if you spin again after the first try, your odds improve to 1 in 6 chance of blowing your brains out....How do you ever get yourself into these situations?

Assuming that the host holds the gun with the barrel vertical (either up or down) when spinning, so that the weight of the bullets has no effect, and that the chamber spins enough times so that the original position of the bullets on each spin has no effect.

If you choose before starting to have two spins you stand 2/3 X 2/3 = 4/9 chance of surviving. Whereas if you choose to have only one spin you stand 1/2 chance of surviving.

However if as in the question you have already survived one "pull" then if you choose to spin again you stand 2/3 chance, but if you choose not to spin again you stand 3/4 chance.

So don' t spin again.

it sounds like the the game show where you get to choose what’s behind door number one, two or three… and you odds changes, (variable change)

Nope,… I still don’t want to play…

In my scenario your opponent was not so lucky, so on your second turn there is only one bullet remaining, but you are unsure whether the gun has been tampered with, as the gun was wiped off after your opponents unfortunate demise...A spin in full view will be your best chance for survival...

Opponent?

Bingo, bango, bongo - first to get it right!

Not playing. Thanks for the invitation, though.

I'm with you. It's a stupid "game"

The Deer Hunter. That movie still gives me the creeps. Not one I'd care to see again.

Correct me if I'm wrong but as I know the barrel of the gun will move to the next position based on the energy of the previous shot. So, without a shot the barrel will remain in the same position and the second pull of the trigger will be harmless.

So, obviously, the answer is "don't spin".

You're thinking of a semi-automatic, in the six shooter, or revolver, the cylinder is advanced by cocking the trigger...

https://en.wikipedia.org/wiki/Revolver

The assumption is that the revolver has six shots but what if it is one of the old cowboy movie revolvers where 10 to 12 shots can be fired off at the baddies?

Beginning cylinder positions: bullets loaded in adjacent chambers: cylinder position 1 is the chambered bullet.

1 2 3 4 5 6

B B 0 0 0 0 Odds surviving 1st round: 4/6 or .67

0 B B 0 0 0

0 0 B B 0 0

0 0 0 B B 0

0 0 0 0 B B

B 0 0 0 0 B

New Orientation after 1st round: The 1st and last positions don't count because you died.

B B 0 0 0 0

0 B B 0 0 0

0 0 B B 0 0

0 0 0 B B 0

The last pull of the trigger has the odds of 3/4 or 0.75 so don't spin the cylinder or your odds go back down to 0.67...

I think people are coming up with the correct answer of not spinning the chambers but with the wrong reasoning. After surviving the first trigger pull, one survival position has been lost, and so has one lethal position. Unless the first bullet misfired, the next chamber cannot be the second bullet. Thus, given n chambers in the gun immediately after any spin, death should occur 2/n of the time. Death should occur 1/(n-1) of the time on the second trigger pull after one spin.

Sorry, but your second formula is wrong. Since "After surviving the first trigger pull, one survival position has been lost, and so has one lethal position", it's not 1/(n-1) but 1/(n-2).

I agree with Alex @ #16. Your own answer @ #3 was correct. If you have already survived the first "pull", and, choose not to spin again you stand 1/4 chance of dying, or, 3/4 chance of surviving.

Huh?

Sorry: Rixter's answer at #3.