32
4
Let's play Russian Roulette!
Normally, this would be a race to write the shortest MOD 6 program, but that's not very realistic, as the chance of winning decreases with each click. Here are the rules:
- Emulate a real six-shooter:
- A single bullet is placed into one of the six chambers, and the barrel is spun once, only before playing.
- The chance of losing on the nth try is 1/6.
- The chance of losing after n tries is 1/(6-n)
- You are guaranteed to lose in, at most, 6 tries.
- Losing:
- Display the text
*BANG!*
- Terminate the program.
- Display the text
- Winning:
- "Winning" means the gun does not fire, but the bullet is one chamber closer to the hammer.
- Display the text
*click*
- Present a "trigger" to the user, along with the ability to terminate the program (e.g. "ctrl+c", see below).
- Program specific:
- Pulling the trigger is some form of user input, including the first try. (This can be a keystroke, a mouse click, whatever; text prompts are not required.)
- Only one instance of the program is allowed until it is terminated. (Running a new instance of the program is akin to giving the barrel a good spin, i.e. the probability of losing on the next click is reset to 1/6.)
Shortest code wins!
Leaderboard
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I'm unclear on the probabilities. Is the loss time uniformly distributed between 1 and 6? – xnor – 2015-12-16T09:39:13.410
@xnor The probability of losing on the nth try is 1/6 - however, the probability of losing after n tries is 1/(6-n), since the probability of firing a bullet reaches one as the number of remaining empty chambers reaches zero. – ayane – 2015-12-16T10:02:00.253
This is very hard to do nondeterministically, I find. – cat – 2015-12-16T14:01:11.737
3Your assumptions are wrong - the chance of losing on the n'th try is only ⅙ if you respin the bullets after each shot. Without respinning the chance of losing is ⅙ on the first shot, ⅕ on the second, ¼ on the third... ending in 1 on the 6th. You recognise this with "You are guaranteed to lose in, at most, 6 tries". – rhialto – 2015-12-16T21:52:15.090
2@user2956063 you are forgetting that there is a (n-1)/6 chance that you never reach the n:th try, and thus cannot lose. They balance out to a uniform 1/6. – Jacob Raihle – 2015-12-17T10:26:00.737
2maybe that's a difference in the way computer scientists and statisticians express probability then - to me "The chance of losing on the nth try is ⅙" say's it's constant - whether n is 1 or 100. – rhialto – 2015-12-17T11:53:28.270
3Why was this not titled "Russian Roulette, Reloaded"? – Hand-E-Food – 2015-12-18T04:51:35.023
@Hand-E-Food Because I'm not very creative, especially with titles :P – ayane – 2015-12-18T05:19:35.797
I posted my answer several hours ago and I only just got the title now. well done (or I might just be stupid—that's never outside of the realm of possibility) – Cyoce – 2015-12-18T08:38:32.377
1@user2956063: Your probabilities are conditional. P(lose on shot 2) = ⅙, but P(lose on shot 2 | didn't lose on shot 1) = ⅕. Also,
n
is (implicitly, I'll grant you) limited to [1,6], so 100 is out. – Tim Pederick – 2015-12-19T05:04:41.680