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A, B and C are in a three way duel. Starting with A, they rotate in the order of A-B-C, each firing one shot at a time. They stand close to one another, so that each can kill one of the others or deliberately miss. They can also adopt probabilistic strategy, like shooting at an opponent with a certain miss probability, etc. A just referee will make sure the required probabilities are determined objectively by tossing dice.

If after many rounds there is still more than one players standing, the referee will randomly eliminate one player, until there's only one survivor.

Before the shooting begins, C can make a public statement, followed by B, and finally A. Specifically, a statement is a promise or threat which takes the form "if this happens, I will do this; if that happens, I will do that...". For example, C can state "if B makes no statement, I will kill them at my turn" or "if no one has being killed when it comes to my turn, I will shoot in the air" or "if A makes a statement to kill B, I will fire at A with a miss probability of 99%, else I will kill A at my turn", etc.

It is common knowledge that statements are binding and irrevocable. All players are selfish and intelligent, and will say and act to maximize their own surviving probability.

Question: What should C say to maximize their surviving probability?


More challenging questions: Suppose there're now 4 players dueling, firing in the order of ABCD and making statements in the order of DCBA. What should D say? What would E say if there're 5 players? Is there any way to tell how adding more players would affect the last person's surviving probability?

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  • $\begingroup$ What exactly is a statement? Which form it has to be of? $\endgroup$ – trolley813 Mar 28 at 16:20
  • $\begingroup$ Another question - do A and B always act in their own interests, i.e. trying to maximise their own survival probability? $\endgroup$ – trolley813 Mar 28 at 16:22
  • $\begingroup$ Does being eliminated count as survival, in this case? $\endgroup$ – AxiomaticSystem Mar 28 at 17:25
  • $\begingroup$ @AxiomaticSystem No. Being eliminated is being killed. $\endgroup$ – Eric Mar 29 at 0:56
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    $\begingroup$ I assume that specifying a probability that would require the referee to roll enough dice to allow all gunslingers to retire and raise great-grandkids before it was measured out precisely would be outside the spirit of the question? ^_^ $\endgroup$ – Zomulgustar Mar 29 at 4:41
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This puzzle is closely related to this one. However, an important difference is that the shooting order is fixed which makes the statement C and D have to make slightly different.

In the three person puzzle C says "If A makes any statement (or no statement) other than "My first shot will miss 1% of the time" I will shoot A with 100% accuracy and miss all other shots. If not, and if B makes any statement at all I will shoot B with 100% accuracy and miss all other shots. Otherwise I will always fire with 99% accuracy prioritizing A."

No statement B can make gives A a better deal than surviving 100% of the time, so if B makes any statement B will always die, either to A's first shot, or to whichever of A or C B does not kill. Only by making no statement does B have a non-zero chance of survival. If A does not commit to their first shot missing 1% of the time A also always dies for the same reason. If B is alive B will always shoot A first, since A's second shot won't miss while C's might so A will always target B with their first shot. Either way, when C's turn comes around only one of A and B are alive. We end up with C surviving 99% of the time, A .99% of the time and B .01% of the time.

I'm still working on the four person part, I think a statement similar to C's above might not work for D since I think C and B have a shared statement that coerces A into killing D instead.

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  • $\begingroup$ Nice! I believe your reasoning is correct. Incidentally, fixing the shooting order makes analysis and statements easier. That random order case in your link is still open. Statement can be a bit messy and complicated in that case. $\endgroup$ – Eric Mar 29 at 1:56
  • $\begingroup$ @Eric -- Doesn't seem ambiguous to me. $\endgroup$ – Voldemort's Wrath Mar 29 at 2:18
  • $\begingroup$ "I think C and B have a shared statement that coerces A into killing D instead". By this do you mean D is hopeless? $\endgroup$ – Eric Mar 30 at 2:08
  • $\begingroup$ Can't this be generalized/improved to C surviving 100-\epsilon percent of the time, by substituting \epsilon for 1%? $\endgroup$ – Oliphaunt - reinstate Monica Mar 30 at 20:23
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4 player answer

The best thing D can do is say:

If and only if C gives me a nonzero chance of survival, C will inherit my gun.

Why?

C can dictate the chances of survival exactly (assuming D does not doom their self), as long as he gives A and B a nonzero chance. It does not matter to Cs survival how he divides the chance of A,B and D surviving (as long as Cs own chance stays the same). Ds statement makes sure it is in Cs interest to give him some chance, which is the best D can hope for.

How?

C can state: "If B does not state to shoot at A with 100% unless A states to shoot only if B and D are eliminated and then with 2a(100-d)%, I will shoot D with 100%.
If B does not also state to otherwise shoot only if A and D are eliminated and then with 2b(100-d)%, I will shoot D with 100%.
If B makes those statements and A does not state to shoot at most once and only at me with 2a(100-d)%, I will shoot A with 100%" If A and B make the requested statements, I will shoot D with 100-d% chance and if successful A and B with 100% in random order in later rounds."

Why does it work:

If B does not comply:
A can kill B, C is forced to kill D and A can then kill C: 100% survival for A, thus 0% survival for B

If B complies, but A does not:
Either B or C is forced to kill A, thus 0% survival for A

Thus both comply, and then:
round 1: A and B must pass, C shoots at D, and D shoots C if not killed
if D survives round 1: D survives since A and B cannot shoot at D (d% survival)
if C survives round 1: A cannot only shoot C if B is killed before A (50% times Cs round 1 survival chance times the chance that A hits) -> A has a% chance of survival. Similarly: B has b% chance of survival.

Conclusion

C can get e.g. choose a=b=c=0.0001% for 99.9997% survival rate.

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  • $\begingroup$ But players can state anything, not just in the form of "if who and who is killed, I will do thus and such". I've edited the question to clarify this (and other issues in the comments). $\endgroup$ – Eric Mar 29 at 1:40
  • $\begingroup$ True, but see extended answer $\endgroup$ – Retudin Mar 29 at 6:38
  • $\begingroup$ Hm, ignore that, the other answer is correct $\endgroup$ – Retudin Mar 29 at 6:51
  • $\begingroup$ D can't require C to inherit D's gun, whatever that means. Statements are only promises or threats that the speaker themself must fulfill. It takes the form "if this happens, I will do this", not "if this happens, you will do this". $\endgroup$ – Eric Mar 30 at 7:59
  • $\begingroup$ It was meant as shorthand for "then I will make a will, giving you the ownership of my gun when I die" It is (intended as) a promise to do something during/just before the duelling.(promises to do things are allowed) I used it to emphasize that D cannot say anything that really influences the outcome. If not allowed, whatever D states does not matter at all (according to my logic) $\endgroup$ – Retudin Mar 30 at 8:26
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1. Four players

In the four players case, regardless of what D says, C can make a statement to secure arbitrarily high surviving probability for themself. There're many ways C can do that, one of which is like the following: enter image description here where enter image description here


2. Five players

The above analysis for 4 players gives us a leverage to handle the 5 players case. Notice that if E makes the statement "if D does $x$, I will do $y$; otherwise I will miss all my shots", D will comply as long as doing $x$ gives them positive surviving probability. Why? Because if E misses all their shots, D is in the exact situation as if there were now just four players, and we know D has arbitrarily low chance in that case.

Now E has D on their side. If the duo can coerce C into compliance by promising them an arbitrarily small surviving probability, then E's goal is achieved. That's because the trio can then coerce A and B into compliance by threatening to kill whoever first disobeys their orders. Because A and B can kill at most 2 of the trio, whoever defies first will face certain death.

The only thing E and D have to worry when coercing C, is that C may try to form alliances with A and/or B. They can thwart that attempt if

  1. E and D can deliver a certain surviving probability $P_c\gt 0$ for C, if C complies (yellow branch in the diagram below); and

  2. If C doesn't comply, E and D can deliver a certain surviving probability $1-\frac{1}{2}P_c$ for B (red branch in the diagram below)

by stating "if C complies we do 1; otherwise we do 2".

So we have the following statement for E (all the "shoot", "kill" and "miss" in the diagrams below refers to players' actions in their first shots, unless otherwise specified): enter image description here

where $P_c$ is surviving probability for C down the yellow branch, and

enter image description here enter image description here

By using higher probabilities than 99% in the above diagrams, E can achieve a surviving probability arbitrarily close to 100%.


3. More than 5 players

I have no idea (for now!). When thinking about 6 players, a couple of meta questions came to my mind:

  1. On the surface, it seems the statements are just a bunch of interacting "if...otherwise..." whose end nodes are always "shoot somebody with certain probability". Is it possible to write a computer program of reasonable complexity that solves the puzzle (at least for small or medium number of players)?
  2. Is it a general feature that for any number of players, the last player either can achieve arbitrarily high surviving probability, or is doomed? What're some reason to suspect whether this feature is true or false?
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  • $\begingroup$ 2 No, see e.g. 4 player. I believe the last odd numbered player can always reach arbitrarily high surviving probability (using the threat of throwing the game to the previous odd player). I did try to write it down in an understandable way, but that is quite hard. Maybe I'll try again using your notation style $\endgroup$ – Retudin Mar 31 at 16:15
  • $\begingroup$ correction: I Missed the 'or doomed' earlier, I think you are right $\endgroup$ – Retudin Mar 31 at 20:37

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