# [FOM] the axiom of determinancy

Alexander, Samuel alexander at math.ohio-state.edu
Sat Jun 11 01:09:26 EDT 2011

```Dear Adam,

I assume you mean the Gale-Stewart game for the set X of computable sequences.  If so, a winning strategy for player 2 for this game is as follows:  on the nth move (assuming it is player 2's turn), ignore everything player 1 did, and play the nth busy beaver number.  Accordingly, player 1 has no winning strategy.

One might also ask about the case when strategies are required to be computable (though this is a departure from the Axiom of Determinacy).  It is not difficult to see that neither player has a computable strategy.  For example, player 2 does not have a computable strategy, because as long as player 1 uses any computable strategy and player 2 does likewise, then the resulting sequence will be computable.

You might be interested in a short note of mine, "A paradox related to the Turing Test", which you can read here (page 90):  http://www.kent.ac.uk/secl/philosophy/jw/TheReasoner/vol5/TheReasoner-5(6).pdf

-Sam Alexander
________________________________________
From: fom-bounces at cs.nyu.edu [fom-bounces at cs.nyu.edu] on behalf of addamo at wp.pl [addamo at wp.pl]
Sent: Friday, June 10, 2011 11:48 AM
To: FOM
Subject: [FOM] the axiom of determinancy

I have the following question:

Let us take the axiom of determinancy (AD). And let X be the set of
sequences of the values
play is
deteremined as the AD says. Is it true that there is a winning strategy for
a player II?
Is it true that it doesn't exist a winning strategy for the player I? And
what is the connection between