Forum Archive :
Puzzles
Lowest probability of winning

From: 
masque de Z 
Address: 
(none) 
Date: 
12 April 2012 
Subject: 
Smallest backgammon nonzero game equity ever possible 
Forum: 
2+2 Backgammon Forum 
What is the smallest win probability in backgammon over 0 of course. 1
point game say. Just for fun consider boards that say white has
astronomically small probability to win. Do you think you can solve this?
Does it have a trivial answer or not so fast ...?


Bill Robertie writes:
A pretty clear answer, I think. We can eliminate all contact positions,
since they're 'easy' to win, relatively speaking. For noncontacts, how
about this:
13 14 15 16 17 18 19 20 21 22 23 24 O's home
++++++++++++++
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   X O 
   
   
    ++
     2 
    ++
++++++++++++++
12 11 10 9 8 7 6 5 4 3 2 1 X's home
O has 15 men on his 3point. Black has 15 men on his 21point. Black on
roll wins with 15 consectutive big doubles, getting 60 crossovers, while
White responds with 15 consecutive 21s. It's not unique since Black
doesn't have to throw 66 every roll, just most of the time.


uberkuber writes:
Nice one Bill! After that, you start tossing white checkers on 1pt and 2
pt until it's a lock and you rollback 1 pip.


TomCowley writes:
Yeah, I thought about this while we were down and you can actually improve
by moving one checker to the 1 point. This position is basically a parlay
of
1) Black must get off in 15 rolls
2) White must not get off in 14 rolls (he always gets off in 15)
You can't make 1 harder without contact (or it being impossible for white
not to win), but you can make 2 harder, because as is, he can roll 61 the
first 2 times, or 62 once, and still not get off in 14. If you move a
checker up to the 1, then any 3161 or 3262 will let you to bear off 2 by
the 14th at the latest. I think moving 2 men to the 2 is also equally
difficult (both requiring exactly 14 21s in a row to avoid being able to
win in 14).




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