butch@inuxc.UUCP (R Meese) (01/03/84)
Here the answer to the backgammon problem submitted some
two weeks back. There will be other bg problem if there is
an interest. Mail me any bg problems or questions you may
have.
Butch Meese |:~>
inuxc!butch
24 23 22 21 20 19 18 17 16 15 14 13
'O' off |===========================================| 'O' PC= 18
O O O || O O O | | ||
O O O || O O O | | ||
|| O O O | | ||
|| | | ||
17 Point Match || | | ||
X-8 O-5 || |BAR| || cube@ 1
X dbl to 2? || | | ||
|| | | ||
X X || X | | ||
X X X || X X | | ||
X X X || X X X X | | ||
'X' off |===========================================| 'X' PC= 27
1 2 3 4 5 6 7 8 9 10 11 12
In this problem, the number of rolls to bear-off the remaining
pieces is more important than pip count. 'X' is behind in
pip count 27 to 18, but may be favored to win because of the
number of rolls needed to bear-off and being "on-roll"(X's
turn to roll).
There is no equation to calculate the average number of rolls
needed to bear-off a set position. Looking at both sides of the
of the board, you can estimate the number of rolls. 'X' has
7 pieces left and will bear-off in 4 rolls if he misses once and
5 rolls if misses twice without doubles and with doubles maybe
2 or 3 rolls. 'O' has 9 pieces left and can bear-off in 3 rolls
with 2 sets of doubles, 4 rolls with 1 set of doubles and
5 rolls without any doubles.
One method to determine X's chances to win is to play the position
a number of times. This position was played out 100 times
producing 73 to 27 games for 'X'. At 73%, 'X' has a double
and 'O' has a marginal take since the score in the match does
not come into play here.
These precentages are based on the fact 'X' was on-roll and
if 'O' was on-roll, 'O' would probably be favored.