BGonline.org Forums

NACK'S POSITION

Posted By: Nack Ballard
Date: Sunday, 23 May 2010, at 7:04 p.m.

In Response To: NACK'S POSITION (David Rockwell)

The original position is shown here. (Thanks to Stick for kindly changing the graphically flawed Gnu diagram to an XG diagram).

David's estimate of White's chances of winning without hitting:

Step #1 - White rolls an immediate 11 (1/36)
Step #2 - Black rolls 1-1 every time for 14 rolls while White ... (1/36)^ 14
Step #3 - rolls 66 every time for 13 rolls bring 13 checkers into the home board (1/36)^13
Step #4 - White rolls 6-6 six straight rolls while Black ... (1/36)^6
Step #5 - rolls 1x five straight rolls (10/36)^5

Equals approximately 2.0 E-56

Now, in Step #3, you actually need two crossovers less than you would have with all 66s. This calculation understates the odds because you may still win if White doesn't roll this well in Step #3. Remember that every time you roll low, Black has to add another 11 roll. These extra, less likely, sequences will raise the odds a little bit. Also, White may be able to win with a few large doublets other than 66 every time in the bearoff. I will make a wild guess that the number above should be adjusted to 1.0 E-52

You have 51 zeros before you get to the 1.

This may not be much more accurate, but...

Black: I do the same for Step #5. However, for Step #2 I think Black rolls 1-1 only 13 times.
White: Same for Step #1. For Steps #3 and #4, let's try giving White 44 55 or 66 for eleven rolls, then only 66 for four rolls, then 44 55 or 66 again for three final rolls. She can do it nowhere near with all such combinations, but I'm "compensating" for the possibility of mostly 66 combos with smaller numbers plus the small additional chance of accomplishing the win in more than nineteen rolls (to Black's eighteen).

When I say "compensating," it's really overcompensating, so I'd consider this to be an optimistic appraisal of White's chances:

(1/36)^(1+13+4) * (3/36)^(11+3) * (10/36)^5 = (3^14 * 10^5) / 36^37 =
1 / (8.9 * 10^44)

If we pessimistically give White only 55 66 (wherever she had 44 55 66), then her chances drop to

(2^14 * 10^5) / 36^37 =
1 / (2.6 * 10^47)

If my conceptual guesstimates and arithmetic are right, White's chances lie between the above two numbers (i.e., the denominator is between 45 and 48 digits), probably closer to the latter.

In other words, incorporating my estimate and Havard's (which I haven't checked), if a bot could be programmed to play the first position diagrammed here over and over and could play so incredibly fast as to have White win it once per second, the same bot would be able to produce a White win without hitting in this position only once every hundred million years.

Nack

Messages In This Thread

• NACK'S POSITION
  Ray Kershaw -- Thursday, 20 May 2010, at 9:30 p.m.
  • NACK'S POSITION
    David Rockwell -- Friday, 21 May 2010, at 12:41 a.m.
    • NACK'S POSITION
      Nack Ballard -- Sunday, 23 May 2010, at 7:04 p.m.
      • NACK'S POSITION
        Ray Kershaw -- Monday, 24 May 2010, at 9:45 a.m.
        • NACK'S POSITION
          Nack Ballard -- Monday, 24 May 2010, at 9:12 p.m.
          • NACK'S POSITION
            Nack Ballard -- Monday, 24 May 2010, at 9:33 p.m.