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SNOWIE ROLLOUTS

Posted By: Timothy Chow
Date: Thursday, 7 January 2010, at 2:19 a.m.

In Response To: SNOWIE ROLLOUTS (Ray Kershaw)

Sorry, I realize now that I should have been clearer. Let's see if I can be more explicit.

First of all, let me confirm that you're asking about rollouts (as the subject line indicates). For evaluations things are subtly different. In particular, I believe that "live-cube" in the context of evaluations means that the cube action is assumed to be perfectly efficient (i.e., that we use the version of Janowski in which the cube is assumed to be always turned at the take/drop borderline), whereas in the context of rollouts it means that the position of the cube is being tracked as the game evolves.

And to clear up something that seems to have confused Matt Ryder: When I talked about "computing the equity...in a live-cube rollout," I didn't mean the cubeless equity. After all, if you're going to the trouble of doing a live-cube rollout, surely you're not primarily interested in the cubeless equity.

I confess I was surprised to read what Daniel Murphy said about Snowie live-cube rollouts. I had assumed that if one were doing a live-cube rollout, then (ignoring variance reduction) a double/pass would count as a simple win, rather than some estimated breakdown into win/gammon/backgammon. Daniel, are you saying that Snowie uses Janowski to compute the equity of a live-cube rollout? That seems very strange to me. Surely the point of doing a live-cube rollout is to sample the equity directly, and estimate it by averaging over all the games, rather than by using a quick-and-dirty surrogate like Janowski.

Anyway, the point I was making is that you shouldn't expect to be able to compute anything other than the cubeless equity from the 6 probabilities. If the bot were to report how many games ended with a single win (here I include double/drop as a single win) with the cube on 1, 2, 4, etc., and similarly for gammons and backgammons, then you could do a simple calculation to get the equity (again, ignoring variance reduction). You would just take the number of single wins with the cube on 1, plus twice the number of gammon wins with the cube on 1, plus three times the number of backgammon wins with the cube on 1, plus twice the number of single wins with the cube on 2, etc., minus all the corresponding figures for losses. But you can't get all this information from just 6 probabilities.

Messages In This Thread

• SNOWIE ROLLOUTS
  Ray Kershaw -- Tuesday, 5 January 2010, at 5:32 p.m.
  • SNOWIE ROLLOUTS
    Timothy Chow -- Tuesday, 5 January 2010, at 11:38 p.m.
    • SNOWIE ROLLOUTS
      Ray Kershaw -- Wednesday, 6 January 2010, at 12:03 p.m.
      • SNOWIE ROLLOUTS
        Daniel Murphy -- Wednesday, 6 January 2010, at 12:46 p.m.
      • SNOWIE ROLLOUTS
        Matt Ryder -- Wednesday, 6 January 2010, at 9:49 p.m.
      • SNOWIE ROLLOUTS
        Timothy Chow -- Thursday, 7 January 2010, at 2:19 a.m.
        • SNOWIE ROLLOUTS
          Daniel Murphy -- Thursday, 7 January 2010, at 4:08 a.m.
          • SNOWIE ROLLOUTS
            Timothy Chow -- Thursday, 7 January 2010, at 3:49 p.m.
            • SNOWIE ROLLOUTS
              Daniel Murphy -- Friday, 8 January 2010, at 2:38 a.m.
        • SNOWIE ROLLOUTS
          Matt Ryder -- Thursday, 7 January 2010, at 6:14 a.m.
          • SNOWIE ROLLOUTS
            Daniel Murphy -- Thursday, 7 January 2010, at 8:52 a.m.
            • SNOWIE ROLLOUTS
              Ray Kershaw -- Thursday, 7 January 2010, at 11:04 a.m.
  • SNOWIE ROLLOUTS
    Matt Ryder -- Wednesday, 6 January 2010, at 5:26 a.m.
  • SNOWIE ROLLOUTS
    Ray Kershaw -- Wednesday, 6 January 2010, at 10:38 a.m.