-1

There is such a thing as a math course for relatively non-mathematically inclined people that is intended to challenge students' intelligence more than to teach them some mathematics. (It is true that the first-year calculus course is used for that purpose by medical schools, etc., and if you don't know that that works very very very badly then you haven't paid attention. Here I have in mind courses that could work well.) And there is such a thing as textbooks designed only for such courses. In such a textbook one may find this:

You are guarding 100 murderers in a field, and you have a gun with a single bullet. If any one of the murderers has a non-zero probability of surviving, he will attempt to escape. If a murderer is certain of death, he will not attempt to escape. How do you stop them from escaping?

The well-ordering of the natural numbers solves this.

But now suppose infinitely many murderers. If infinitely many conspire to escape simultaneously, knowing exactly one would be shot, then each one's probability of survival exceeds your chance of surviving the hazards of getting out of bed in the morning, but if David Hilbert can talk about his "hotel", then we may suppose that no one will try to escape if it is certain that they will be shot. Here the well-orderability of all sets, equivalent to the axiom of choice, solves the problem.

My question is whether there is a converse: If you have a strategy for preventing all escapes, does the axiom of choice follow?

Michael Hardy
  • 11,922
  • 11
  • 81
  • 119
  • 5
    I think there's an interesting question here, but a lot of the flavor text could be trimmed (I don't see what the first paragraph, or the bit about how long you personally took on the original version, or the bit about getting out of bed in the morning, have to do with the actual math here). And as usual with such common/shared-knowledge problems, the most obvious difficulty is even posing the problem precisely in the first place (which is why I think it's a good idea to avoid unnecessary flavor-text in this case). – Noah Schweber Nov 25 '23 at 19:41
  • @NoahSchweber : Perhaps . . . . but I thought some here might benefit from knowing of that book. I don't have it in front of me now and don't remember the specifics, but I should be able to dig it up. – Michael Hardy Nov 25 '23 at 19:46
  • @NoahSchweber : ok, I've deleted part of this. – Michael Hardy Nov 25 '23 at 19:47
  • @NoahSchweber : The bit about getting out of bed in the morning does have something to do with the actual math, if not with the question at the end. – Michael Hardy Nov 25 '23 at 19:48
  • Define "infinitely many". – Asaf Karagila Nov 25 '23 at 19:55
  • @AsafKaragila : Now I suspect that you want me to say more than any finite cardinality, so you can point out something wrong with that. – Michael Hardy Nov 25 '23 at 20:20
  • No, I am just curious as to the parameters of the question. Sometimes people think about this in very "bounded" terms, i.e. countably infinite. Or something like that. – Asaf Karagila Nov 25 '23 at 22:53
  • I found the 100 murderers question on some websites about quant job interviews. – Gerry Myerson Nov 26 '23 at 00:01
  • How does a well-ordering solve the problem? – bof Nov 26 '23 at 12:06
  • 1
    @bof: You declare that any set of prisoners that escape, the one with the least number is gonna get shot. This guarantees that 0 is not escaping, and more generally, if nobody escapes with index less than $n$, then $n$ is not escaping either. So by (transfinite) induction, no well-orderable set of prisoners is going to escape. – Asaf Karagila Nov 26 '23 at 16:32
  • @bof : The prisoners are assigned (ordinal) numbers. A group may conspire to run away simultaneously, thinking that each of them has a chance of survival since only one will be shot. But they have been notified that in such an event, the one with the lowest number (well ordering) will be shot. Thus the one with the lowest number among the conspirators drops out. Then the one with the lowest number among those remaining drops out. And so on. – Michael Hardy Nov 26 '23 at 23:19
  • OK. You confused the issue by changing "If a murderer is certain of death, he will not attempt to escape" to "no one will try to escape unless it is impossible that he or she or they will be shot." Even the original puzzle seems to require the unstated assumption that the guard is sure to hit his intended target. – bof Nov 27 '23 at 00:59
  • 1
    @AsafKaragila OK, if you assume (or the prisoners assume) that the guard is a sure shot, he never misses, his gun never jams, he can even pick out the lowest-numbered guy out of the trillions of escaping prisoners. The trouble with story problems is that mathematicians seldom take the trouble to make their stories make sense, or to state their assumptions explicitly as they would in a straignt mathematical problem. – bof Nov 27 '23 at 04:03
  • 1
    @AsafKaragila The OP further confuses the issue by changing the problem to "no one will try to escape unless it is impossible that he or she or they will be shot." In that case no well-ordering or AC is needed; all the guard has to say is "if any of you try to escape I'm going to kill one of you." Only one choice has to be made. – bof Nov 27 '23 at 04:07

2 Answers2

2

With such shared/common knowledge problems, even when AC isn't involved, I think a crucial first step is to get away from the "story" version. For example, since this is only interesting once the set of murderers is uncountable (since it needs to be non-wellorderable), talking about probabilities isn't going to be ultimately helpful.

Here's a first stab at a formalization:

Consider, for $X$ a set, a two-player game $\mathsf{Shoot}_X$ played as follows:

  • First, player $1$ plays a function $f:\mathcal{P}(X)\rightarrow\mathcal{P}(X)$ satisfying $f(S)\subseteq S$ for all $S\subseteq X$.

  • Next, player $2$ plays a function $g: X\rightarrow 2$.

  • At this point, the game is done: player $2$ wins iff $f(g^{-1}(1))$ has more than one element.

The intuition here is that player $1$ is announcing their shooting strategy ("If the set of runners is $S$, I'll 'randomly' shoot someone in $f(S)$") and player $2$ is saying which murderers (= elements of $X$) will run. The condition that $f(g^{-1}(1))$ has more than one element corresponds to every runner having a chance of surviving.

This formalization more-or-less trivializes things (since the game itself is basically $\mathsf{AC}$ in disguise): a winning strategy for player $1$ must be a choice function for $\mathcal{P}_{\not=\emptyset}(X)$, and so the existence of a winning strategy for player $1$ for all $X$ is automatically equivalent to choice. In fact, I think this highlights the importance of making the infinitary version of the game more precise.

This does, however, raise an interesting follow-up question:

Is $\mathsf{ZF}$ + $\neg\mathsf{AC}$ + "Every $\mathsf{Shoot}_X$-game is determined" consistent?

I suspect there's an easy argument that the answer is negative (= that from a failure of choice we can build an undetermined shooting game), but I don't immediately see it.

Noah Schweber
  • 18,932
  • 1
    Even better would be removing the shooting aspect. In times like these, is discussing shooting and killing people—and making it acceptable they are labelled murderers—the type of behaviour to condone? Make it a game like tag, and the violent emotive aspects can be lost without mathematical change. – David Roberts Nov 26 '23 at 09:54
  • Player I can win $\mathsf{Shoot}_X$ every time by playing $f(S)=\varnothing$ for all $S\subseteq X$. – bof Nov 26 '23 at 12:31
1

Yes.

The problem's statement can be reformulated as follows. Suppose that for any set $X$ there exists a function $f: (2^X\setminus \{\emptyset\})\to X$ mapping any nonempty subset (of escapees) to one of its elements (which guarantees that a participant will refuse to participate in the escape). Is it equivalent to the Axiom Of Choice?

The Axiom of Choice states that for any set $X$ of nonempty sets $Y_i\in X$ there exists a function mapping any set $Y_i$ to an element of the set.

But the equivalence is fairly trivial. Indeed, the AOC follows by constructing the function $f$ for $\bigcup Y_i$ and applying it only to the sets $Y_i,$ while the function function $f: (2^X\setminus \{\emptyset\})\to X$ mapping any subset to its element is created by applying the AOC to all nonempty subsets of $X.$

Michael Hardy
  • 11,922
  • 11
  • 81
  • 119
Станислав Крымский
  • 43
  • The assertion that for any set $X$ there is a function mapping each nonempty subset of $X$ to one of its elements is equivalent to the axiom of choice because it is the axiom of choice. But I don't see how this relates to your silly story about murderers. – bof Nov 26 '23 at 12:13