An infinite game possibly due to Ernst Specker I have a vague memory of an infinite game due to Ernst Specker with
the following properties:
(1) It is a two-person perfect information game, where the players
move alternately.
(2) The possible moves depend only on the current position.
(3) There is no winning strategy where each move is based only on
the current position
(4) There is a winning strategy (for one of the players) if the
strategy uses the current position and the history of previous moves.
Can someone provide a reference or explanation for this game?
 A: I don't know about the game attributed to Specker, but here is a
simple game with your desired features.
Let us call it the Chocolatier's game. There are two players,
the Chocolatier and the Glutton. To begin play, the Chocolatier
serves up finitely many unique and exquisite chocolate creations on
a platter, and then the Glutton chooses one of them to eat. Play
continues — at each stage the Chocolatier adds finitely many
additional chocolates to the platter, and the Glutton consumes one
of those available. The excess uneaten chocolates accumulate on the platter as play progresses.
After infinite play, the Glutton wins if every single chocolate
that was served was eventually consumed. Otherwise, the Chocolatier
wins.
The Glutton can easily win simply by paying attention to the order
in which the chocolates were added, and consuming them in that
order. If only finitely many chocolates are added at each stage, the Glutton should simply make sure to consume them before moving on to the
chocolates that were added at later stages.
Indeed, the Glutton can win even if the Chocolatier places countably many chocolates on each turn. At turn $n=2^k(2m+1)$, let the Glutton eat the $k$th
chocolate added at stage $m$, if any, and otherwise eat
arbitrarily.
But what about strategies that depend only on the current position,
that is, the assortment of chocolates on the platter?
In one sense, it is easy to see that there can be no such winning
strategy for the Glutton. If the Glutton will choose a particular
chocolate from a given assortment, then let the Chocolatier simply
replace it with an identical chocolate type on the next move, and again on all subsequent moves. If
the strategy does not know the history, then the Glutton will
choose it again every time, and the other chocolates will never be
eaten.
Perhaps it makes a more interesting game, however, to say that the Chocolatier loses if chocolate types are ever repeated. And for this version of the game, there are some interesting things to say.
If there are only countably many chocolate types available in all,
then again the Glutton has a winning strategy that depends only on
the current assortment on offer. Namely, the Glutton should fix an
enumeration of the possible chocolate types that might appear, and
at each stage select the chocolate that appears earliest in this
order — it is the tastiest-looking chocolate as defined by
that priority. With this strategy he will succeed in eating all the
chocolates, because at the limit, if any chocolate was left, there
would have to be a tastiest-looking one (earliest in the
enumeration), and this would have been eaten once the finitely many
tastier chocolates had been consumed.
If the Chocolatier were uncountably creative, however, and able to
serve up uncountably many different chocolate types, then this
argument breaks down. Indeed, in this case I claim there is no
winning strategy for the Glutton that depends only on the chocolate
assortment on offer. And indeed, there is no such strategy that
works even if we should insist that the Chocolatier present an
assortment only of size two at each stage.
To see this, suppose the Glutton will follow a fixed strategy that
selects a particular chocolate to eat from amongst any two
chocolates.
For any given chocolate $c$, if there are infinitely many others
$d_n$ that would be preferred to it by the strategy, if presented
as a pair $\{c,d_n\}$, then the Chocolatier can present these pairs
$\{c,d_0\}$, $\{c,d_1\}$, $\{c,d_2\}$, in turn. At each stage,
$d_n$ would be consumed and the Chocolatier can present
$\{c,d_{n+1}\}$. In the end, the inferior chocolate $c$ would never
have been selected and so the Chocolatier will win.
So if this is a winning strategy for the Glutton, then we may
assume that every chocolate is preferred to all but finitely many
of the others. But this is simply impossible with an uncountable
set. To see this, take any countably infinite set of chocolates and
close under the finite witnesses of strictly preferred chocolates.
One thereby constructs a countably infinite set $D$ of chocolates
so that any chocolate in $D$ is preferred to any chocolate not in
$D$. Since there were uncountably many chocolate types, there is
some $c\notin D$. This contradicts our assumption that there are
only finitely many chocolates preferred to a given chocolate.
Let me remark that in the version of the game where we do not allow
the Chocolatier to repeat chocolate types (as opposed to saying that this is allowed, but causes the Chocolatier to lose), then we should really
include the list of already-consumed chocolates as part of the
position, and this complicates the arguments above. (I asked a followup question about this at The Chocolatier's game.) I suspect but
do not yet know that the Glutton can have no winning strategy that
depends only on these more general kinds of positions in the uncountable case.
