probability question

If you spin again, your chances of surviving are 2/3. If he pulls again without spinning they are 3/4.

How on Earth do you come up with that?

If he spins again…the chances of being shot are 1 in 3.. (There are six chambers…twois loaded. A spin will land anywhere…1/3…or 3 to 1 against being shot.)

If he fires again without spinning….ahhh…there’s the rub. You now know that one of the empty chambers is lost to the equation. There are five more chambers…two of which are loaded. The chances of being shot appear to be 2 in 5…or 5 to 2 against being shot. Lower odds of not being shot; higher odds of being shot.)

The fact that both loads are contiguous may impact…but I cannot figure how. I’d love to hear some reasoning from someone skilled in probability.

If you spin, there are four empty chambers, two filled ones so the chance of surviving is 4 in 6.

For not spinning, it is a given that the first spin selected one of the contiguous four empty chambers. Three of those chambers have an empty chamber beside them, one does not so the chance of not getting shot is 3 in 4. This problem is small enough that you can draw out the problem. You have six chambers and let's say 1-4 are empty and 5 and 6 are full. On the first spin you selected 1 ,2 ,3 or 4. On 1-3 you survive a second pull, on 4 you don't.

Okay...thanks.

I notice that I screwed my post up royally. I had corrected it during composition...and then apparently selected the old text to submit.

Sorry for that.

P(1) (no spin) = # of ways to roll on loaded chambers = 2
over total # of (remaining) chambers = 5
The probability (P) that you will be shot is at 2/5.


P(2) (spin) = # of ways to roll on a loaded chamber = 2
total # of chambers = 6
The probability (P) that you will be shot is at 1/3.

So rationally, I suggest you should choose not to spin the chamber since the probability that you'll get shot is lesser than spinning it again.