That's what I thought. If each shot was a separate dice roll then getting two in a row would make a difference though.

Oh yes. In that case the sooner you shoot the better.

Whoa whoa whoa. I didn't say it was a statistical advantage (one moment while I double check...yep, didn't say statistical, since I haven't checked the statisticals on it), I said it gives you more chances. Specially, in ABBAAB, the first person gets initiative (which may work out, ala Agent vs Adalia), but going second gives you more chambers in the early slots.

It's (mostly) a fun random-shoot game for me too. And to be fair, you've had two chances to shoot at other people. ;)

Eh, okay, that's fair. I guess there is implicit probability involved.

-Running from a lack of initial knowledge, the chance of making the shot from the first chamber is 1/6. ~16% win rate?
-When you know the first chamber isn't loaded, then the naive chance of the second chamber holding the shot is 1/5, but using the second AND third chambers gives you 1/5 + 1/4. Rolling for that 20% + 25% win seems nicer to me than hoping your opponent fails the 25%. If you lose, then your opponent gets 33% + 50%.
-Of course, you could leave the third shot to your opponent, leaving you hoping the 25% fails so you can make the 33% + 50% rolls.

On the flip side, I also use the rule of thumb that "Opponents can't breathe bullets". The more shots I put in the air faster, the more likely I am to get a hit. :)

Aha, yes I understand now. I have no idea whether it's correct as statistics and probability aren't my thing but it seems to make sense. Thanks!

From a general point of view, if the chamber size is even, then:

If n shots have been fired and I have a choice of doing a single shot or double shot this means that:
_ n+1 shot is mine, no matter what I do.
_ If I do a double shot, I get n+2, opponent gets n+3. If I do a single shot, opponent gets n+2 and I get n+3.
_ P(n+2) is equal to P(n+3) since bullet was put randomly (Note to self: exception is if the mod (*cough* Bookwyrm *cough*) decides to disregard a draw and reroll).
_ Either choice recursively takes us to a state equivalent to one of (no shots fired OR one of us fired a single shot).

Ergo, statistically the choice is statistically irrelevant.

Hmmm...

Without prior knowledge, the first shot has a 1/6 chance of hitting. However, if I see that the first shot hasn't hit, then firing the second AND third chambers gives you a 3/6 chance of hitting. If neither of those hit, firing the fourth AND fifth shots gives a 5/6 chance of hitting. And obviously then you get a 6/6 chance of hitting if the first five shots were blanks.

Monty Hall Problem. I think that's a correct application.

I think wanting to fire first or second depends on what chances you want to take. Firing first give the other player a significant chance (1/2) of hitting before you get an even more significant chance (5/6) of hitting. Firing second gives you the significant chance first, but it's less significant.

EDIT:
You're right about the overall probability, but I do think there's a statistical difference between firing first and firing second.

... which adds up to 1/2 vs 1/2.

With an even sized chamber, no matter what choice you make (regarding hitting first or not), each choice ultimately ends with 3 (or half of whatever the chamber size is) possible shots each.

With an odd-sized chamber, there might be a situation where it's better not to do a double shot (i.e. not to go first). A zugzwang. e.g. if this was 5-chamber gun, then after agent went first, adalia is better off firing only a single shot. If he fires double, it goes MTTMM, so adalia should fire a single shot and hope agent blunders and fires a shot next, making it MTMTT. Of course with agent being smart enough, he'd not fire too, and it would end up MT... (deadline fires randomly).

Yes, I agree. But if I fire first, you get your 1/2 chance of hitting me BEFORE I get my 1/2 chance of hitting you.

No.

(Assuming all double shots):

If you fire first, I get a 2/5 of hitting you BEFORE you get another chance. If I miss you get a 2/3 chance of killing me.

So your probability is: first_chance + (assuming I miss)second_chance
= 1/6 + (5/6)*(3/5)*(2/3) = 1/2

Allowing me to fire first gives you, assuming I miss, a 2/5 chance of killing me straightaway and, assuming you miss and I miss again, a 100% chance of killing me with the last bullet.

= (5/6)*(2/5) + (5/6)*(3/5)*(1/3)*(1/1) = 1/2.

In the final round, it doesn't matter what choice you make. None of the choices made by you or adalia really mattered in the final round.
(except the trivial choice of firing if your opponent has done so already, lest deadline kills you).

OK, I've convinced myself that you're right.

Just read this post in detail.

No! You don't add here. You DON'T get 20%+25%.

Your probability is actually 20% + (assuming you miss first)25%, which is equal to 20% + 80%*25% = 2/5 or 40%.
(another way of arriving at the same result is looking at it as having 2 bullets out of 5).

So while 20% and 25% does seem nicer, in reality the combined effect is only 40%, and as shown before mathematically it amounts to the same thing.

Well no, I don't... because I didn't post that. :D

Sorry, wrong quotes.