TITLE: A game on integers
QUESTION [37 upvotes]: $A$ and $B$ take turns to pick integers: $A$ picks one integer and then $B$ picks $k > 1$ integers ($k$ being fixed). A player cannot pick a number that his opponent has picked. If $A$ has $5$ integers that form an arithmetic sequence, then $A$ wins. $B$'s goal is to prevent that from happening. Who has a winning strategy (after finite number of turns)? Does it depend on $k$?

REPLY [4 votes]: The minimalist strategy for $A$ of "don't think, just pick the smallest unchosen non-negative integer!" has a simple counter strategy for $B$ which is interesting to analyze as to how far it gets before it eventually fails. I wonder how good it is.
The number of turns for $A$ before winning increases exponentially with $k$ (provided $A$ uses this minimalist strategy and $B$ uses the one I will describe.) This is true also for thee term arithmetic progressions.
Consider the game where $A$ needs to get an arithmetic progression of length $m$ to win and $B$ gets $k$ choices each turn. Then I think my strategy for $B$ matched with the minimalist strategy for $A$ will have $A$ win on a turn no smaller than  $k^c$ where $$c=\log_{\frac{m}{m-1}}(m-1).$$
A better strategy (for three term sequences) for $A$ would win in  $k+2$ turns (this could easily be improved to $\frac{k}3$ turns.)  Roughly, on the first $k$ turns $A$ selects available even numbers $a_1,\cdots,a_{k}$ then, whatever choices $B$ has made so far, it is possible to pick an even $a_{k+1}$ so that none of $\frac{a_{k+1}-a_i}2$ have been chosen yet. $B$ must leave one of those unchosen on her turn so $A$ picks that next turn and wins.
How to extend this to longer progressions is not clear however that argument shows that wlog $A$ gets three free moves on the first turn (we could arrange that the entire progression determined by $a_i,a_{k+1}$ is free.) 
Below is a strategy for  $B$ if she knows for sure that $A$ will blindly follow the minimalist strategy. The moves are all known in advance so it is just a question of when $A$ wins. For ease of analysis I will allow $k \gt 0$ to be any real number and interpret this as: on turn $t$ if $B$ has previously chosen $m$ integers she has  $\lfloor kt \rfloor-m$ moves available and can make none or some or all and hold the rest for future turns. 
The strategy is that she will pick in order the non-negative integers with a $4$ in their base $5$ expansion and let $A$ work up through the rest one at a time. This blocks any $5$ term progressions as long as $B$ stays ahead. However $B$ will save her moves and only pick enough numbers that $A$ will not win on the next turn (which may turn out to take more moves than she has).
When $B$ loses it will because $A$ was able to play a number in some interval $[5^j-5^{j-1},5^j-1].$ This will happen in the event that $$\frac{5^{j-1}-4^{j-1}}{4^{j-1}} \leq k \lt \frac{5^{j}-4^{j}}{4^{j}}.$$ The exact winning move depends on where $k$ falls in that range.
For example, If $k \geq \frac{61}{64} =0.953125$ then $A$ at some point picks $125$ without having won. No problems occur until after the turn when $A$ picks $468=3333_5.$ At this point anything in $[469,624]$ would be a winner for $A$ on the next move. But $B$ has moves in reserve, certainly enough to choose all of $[629,499].$ However to be able to go on and choose all of $[500,634]$ requires $k \geq \frac{369}{256}=1.44140625.$ If so, then everything is good until at least $2499.$