(Daughter gets kidnapped…whose dad was working on Riemann Hypothesis) ]]>

Charlie: “Well, I think there’s a few problems with that, and thanks for the gesture of friendship by the way. First, obviously you were speaking colloquially when you said Colby was a dead shot. He’s really alive, and deadly to boot. And second of all, you’re breaking a rule on how the problem is supposed to be set up. We take turns – you don’t get to keep going again and again until you win – you get ONE shot and then it’s my turn if I should be so fortunate as to be alive. Finally, I DO think you’re OK in how you phrased your question. Anyone who speaks English as a first language would be able to figure out it was implied the ONE shot was the ONE shot of THIS turn. I understand we need to be precise, but my brother Don and all the non-mathematicians for whom we are expounding already get the rules of the game.”

Colby: “Yeah, I’m not an idiot. And besides, I have a better solution. While Marshall is trying to aim the gun, I wrestle him to the ground, take his gun away, and scare the bejesus out of Charlie, causing him to pause while I shoot my gun.. It a 100% chance of me winning.”

Marshall: “Uh…yeahhhhh……”

]]>The short version of the analysis goes something like this: C, if he’s alive and has a choice, will always shoot B, because doing so maximizes his chances of survival – his chance of surviving is 2/3rds if A gets a shot at him, and 1/3rd if B gets a shot at him (since C is guaranteed to kill the other survivor next turn if he’s still alive), so ‘eliminate the biggest threat’ is correct for C. Because of this, B (if he’s alive and has a choice) must shoot C to maximize *his* chances of survival. These are pure, optimum strategies for B and C; B will never shoot A unless C is already dead.

Now, the only potentially-infinite situation is the one in which only A and B are left alive, and continually shoot at each other. It’s easy to see that if A shoots first, his chance of survival P is 1/3 (A hits) + 2/3*1/3*P (A misses, B misses, and we’re back to square one); P = 1/3 + (2/9)*P, (7/9)*P = 1/3, P = 3/7. If B shoots first, then A’s chance of survival is 1/3*P (B misses, and we’re back to the A-shoots-first case) = 1/7. We’ll hang on to these for later.

Now, suppose A misses with his first shot (whoever he aims at). Then with probability 2/3, B will kill C, and we’ll be at our A vs. B case with A shooting first. With probability 1/3, B will miss C; C will then kill B, and A has a 1/3 chance of killing C and surviving. So if A misses, the chances of his surviving are (2/3)*(3/7) + (1/3)*(1/3) = 2/7 + 1/9 = 25/63.

So suppose A chooses to shoot B. Then with probability 1/3 he succeeds, and immediately dies (because C shoots A as the only other survivor). With probability 2/3 he misses; this puts us in the situation that we just described, and his odds of survival there are 25/63. So his overall odds of survival if he aims at B are 2/3*25/63 = 50/189.

If A aims at C instead, then with probability 1/3 he succeeds, and we’re in the ‘A vs. B’ case with B having first shot (so A has a 1/7 chance of survival); with probability 2/3 he misses and we go back to the ‘after A misses’ case. So A’s total chances of survival here are 1/3*1/7 + 2/3*25/63 = 9/189 + 50/189 = 59/189.

But if A aims into the air and deliberately misses, then we know he has a 25/63 = 75/189 chance of surviving. So A’s best bet here, similar to the other outlined case, is to not shoot anyone! This gives him roughly a 8.5% better chance of surviving than if he shoots C.

]]>Andrew shoots at Bob.

2/3 chance – Bob lives

Bob shoots at Andrew (but both end in him dying)

1/3 chance – Andrew lives

Con shoots Bob. Bob dies. Andrew shoots at Con.

2/3 chance – Con lives

Con shoots Andrew. Game Over.

1/3 chance – Con dies. Game Over.

2/3 chance – Andrew dies

Con shoots Bob. Game Over

Bob shoots at Con

1/3 chance – Con lives

Con shoots Bob. Bob dies. Andrew shoots at Con.

2/3 chance – Con lives

Con shoots Andrew. Game Over.

1/3 chance – Con dies. Game Over.

2/3 chance – Con dies. Andrew shoots at Bob and they play each other. (4/9)*(1/3) + (4/9)(1/3)(1/3)(1/3) + … = 1/6

1/3 chance – Bob dies

Con shoots Andrew. Game Over

Case 2 – Total chance Andrew wins is 2/9 + 1/24 = 0.26

Andrew shoots at Con.

2/3 chance – Con lives

Con shoots Bob. Bob dies. Andrew shoots at Con.

2/3 chance – Con lives

Con shoots Andrew. Game Over.

1/3 chance – Con dies. Game Over.

1/3 chance – Con dies.

Bob shoots Andrew.

1/3 chance – Andrew lives

Andrew shoots at Bob and they play each other. (1/9)*(1/3) + (1/9)(1/3)(1/3)(1/3) + … = 1/24

2/3 chance – Andrew dies. Game Over.

So Andrew should shoot Conner. Hopefully I didn’t make any mistake.

]]>