A team of three sinners plays a game against the devil. They confer on strategy beforehand; then they go into three separate rooms, and there is no more communication between them. The play in each room follows the same rules, as described below.

A move consists of choosing some numbers from the set $\{1,\dots,N\}$ where $N$ is a fixed natural number. A number can only be chosen once; thus the two players in the room, sinner and devil, are dividing a finite set between them, as in tic-tac-toe or hex. The devil moves first, and chooses two numbers at each turn. The sinner moves second, and chooses three numbers at each turn. Play continues until all $N$ numbers have been chosen.

The sinners win if there is at least one number which is chosen by all of them; otherwise the devil wins.

**Question:** Do the sinners have a winning strategy if $N$ is sufficiently large?

We are not interested in randomized strategies. A winning strategy is a

*deterministic*strategy which is sure to win against arbitrary play by the devil.I don't know if there's an affirmative answer for $N=\omega$, nor whether an affirmative answer for $N=\omega$ would imply an affirmative answer for some finite $N$.

You can ask the same question for larger teams, but a team of three is the smallest interesting case. A team of two wins automatically if $N\ge5$.

The devil has to move first, or else it's a trivial win for the sinners.

The $3$-to-$2$ biased game ($3$ moves for the sinners, $2$ for the devil) is the simplest interesting case. The $1$-to-$1$ game is an easy win for the devil, even against a team of two. The sinners can win the $2$-to-$1$ version (with teams of any finite size) if N is large enough, but that's not completely trivial.

I called the team players "sinners" because I didn't know what else to call them; "men" or "women" would be sexist, "persons" or "humans" sounds too stilted. I guess "mortals" would work but I didn't think of that until now.

I used the combinatorial-game-theory tag because it seemed right to me, but I see that the official explanation says it's for perfect-information games. Was that a sin?

**Edit for clarification:** Judging from the comments, my poetic imagery about the Prince of Darkness is confusing people. It doesn't matter what the devil knows or doesn't know, because I'm asking for strategies which win against **arbitrary play by the devil**, in other words, in the **worst case**.

If it helps, you may imagine that the devil, instead of playing maliciously, is making **random moves**. The players then have an excellent chance of winning. The question is whether, for some $N$, there is a strategy which wins with **probability one**.

To put it more prosaically, let $B=\{1,\dots,N\}$, and define a *strategy* as a function $\sigma:2^B\rightarrow2^B$ such that, for each $X\subseteq B$, we have $\sigma(X)\subseteq X$ and $|\sigma(X)|=\min\{3,|X|\}$. The question is whether, for sufficiently large $N$, there exist strategies $\sigma_1,\sigma_2,\sigma_3$ such that, whenever $S_i$ is the set of numbers obtained by the sinner in a $\sigma_i$-compatible play (whatever that means--definition left as an exercise for the reader), we have $S_1\cap S_2\cap S_3\ne\emptyset$.

**P.S. Shorter answer: Yes, the devil knows the sinners' strategies.**

**Edited again for further clarification**: I believe the following restatement is equivalent to the original question:

**Question (restated):** Do there exist (for some $N$) families $\mathcal W_1,\mathcal W_2,\mathcal W_3\subseteq2^{\{1,\dots,N\}}$ such that (1) player $i$ has a strategy which guarantees that he will get all elements of some set $W_i\in\mathcal W_i$, and (2) if $W_i\in\mathcal W_i$ for $i=1,2,3$, then $W_1\cap W_2\cap W_3\ne\emptyset$? (As before, the Adversary goes first, and picks two numbers at each turn; the Player goes second, and picks three numbers at each turn.)

I guess I should have stated it that way in the first place instead of confusing the issue by trying to make it a story problem. Sorry about that!

**Follow-up questions about Zackkenyon's answer:**

I haven't gotten into the details yet. I have a big-picture question about just what it is that you're proving.

Your proof starts in talking about a number $k$ and and a partition into sets $S_{k,i}$, without explaining where they come from or what they have to do with anything. First I have to introduce some notation. Then I'll try to state what I think you're doing, so you can tell me if I'm right or wrong.

A **strategy** for the sinners is an ordered triple $\sigma=(\sigma_1,\sigma_2.\sigma_3)$ where $\sigma_i$ tells sinner $i$ how to play in room $i$. Given a natural number $k$ and a partition $\mathcal S=\{S_1,S_2,\dots\}$ of $[N]$ into $k$-element sets, I'll say that a strategy $\sigma=(\sigma_1,\sigma_2,\sigma_3)$ has **property $Z(k,\mathcal S)$** if it guarantees that, regardless of how the devil plays, there will be some $i$ such that every number in $S_i$ is picked by sinner 3, while at least one number in $S_i$ is picked by both sinner 1 and sinner 2.

Clearly, if the strategy $\sigma$ has property $Z(k,\mathcal S)$ for some natural $k$ and some partition $\mathcal S$ into $k$-element sets, then $\sigma$ is a winning strategy. Maybe I'm missing something obvious, but I see no reason for the converse to hold; except, of course, that it holds vacuously if there **are** no winning strategies. Offhand, it seems to me that what you're doing is showing that no strategy can have the property $Z(k,\mathcal S)$.

Have I got that right? Does your argument show (I) that the sinners have no winning strategy of any kind, or does it just show (II) that a certain natural type of strategy (which most people think of first when trying to prove the affirmative) can't exist, while leaving open the possibility that some more exotic kind of strategy will do the trick? In case (I) I'd like to see some explanation of how that follows from what you wrote, or if there's some further part of the argument that you didn't write down. In case (II), that's a nice partial answer, but of course it doesn't dispose of the question.

To make the question clearer, let me try to define in general what it means for a triple $\sigma=\sigma_1,\sigma_2,\sigma_3$ to be a winning strategy for the sinners, so we can compare it to (what I understand to be) your approach. First lets consider what happens in room 3: using the strategy $\sigma_3$, sinner 3 acquires some set S of size $2N/5$. (No harm in assuming $N$ divisible by $5$.) Let $\mathcal S=\{S_1,S_2,\dots\}$ be the collection of all sets obtained in this way, with sinner 3 using $\sigma_3$ against all possible plays by the devil. Sinner 3 is guaranteed to get all elements of some $S_i$, but **which** $S_i$ he gets is up to the devil. Plainly, then, for the sinners to have a forced win, $\sigma_1$ and $\sigma_2$ have to guarantee that for **each** index $i$ there is at least one element of $S_i$ which is picked by both sinner 1 and sinner 2.

This is sort of like your $Z(k,\mathcal S)$ with $k=2N/5$, the big difference being that the $S_i$'s are not disjoint, which seems to really confuse the issue. Does your argument work if the $S_i$'s are not disjoint?