I read on Robertie's Advanced Backgammon something similar. I think it was last
position of the book, bearoff position; player A, probably a little bit drunk if I
remember correctly, doubled to 2, player B beavered to 4 and A rejected. Does that
mean that A loses 2 points ?
If this is the case, an interesting situation arises: A owns a 2 cube and is in a
desperate position, losing 100% backgammons. Suppose A doubles to 4, B beavers and
A rejects. A loses only 4 points instead of 6. Is that correct ? Or am I missing
something ? If so, this could be a good trick sometimes :-)
Thanks,
Carlo Melzi.
--
Carlo Melzi, Department of Computer Engineering, Politecnico di Milano
cam...@tin.it http://pmair1.elet.polimi.it
Yep, that is right.
>If this is the case, an interesting situation arises: A owns a 2 cube and
is in a
>desperate position, losing 100% backgammons. Suppose A doubles to 4, B
beavers and
>A rejects. A loses only 4 points instead of 6. Is that correct ? Or am I
missing
>something ? If so, this could be a good trick sometimes :-)
Correct. But if B does not beaver, you'd lose 12 instead of 6, so you must
be pretty
sure ( > 75% ) that B will beaver, which may be a lot to hope for. However,
the joy
of telling him that he just threw away 2 points if he did, that will be
worth
something too.
--
Stein
Carlo Melzi wrote:
>
> Question is: can a beaver be rejected ?
>
> I read on Robertie's Advanced Backgammon something similar. I think it was last
> position of the book, bearoff position; player A, probably a little bit drunk if I
> remember correctly, doubled to 2, player B beavered to 4 and A rejected. Does that
> mean that A loses 2 points ?
>
> If this is the case, an interesting situation arises: A owns a 2 cube and is in a
> desperate position, losing 100% backgammons. Suppose A doubles to 4, B beavers and
> A rejects. A loses only 4 points instead of 6. Is that correct ? Or am I missing
> something ? If so, this could be a good trick sometimes :-)
I hope that I'm not missing anything, but I think the problem is not
special with beavers, it's like with any cube: The player owning the
cube after a previous double by his opponent should not redouble when
a (back)gammon is 100% sure, because his or her opponent will always
reject (unless he or she is drunk ;-)) and thus only lose single. So,
in the particular case, B should simply not redouble, be it a beaver
or normal.
Greetings,
Peter aka the juggler on FIBS
--
---------------------------------------
Peter Schneider
Software Development and System Support
Breitenfelder Str. 64, D-20251 Hamburg
Phone: +49 40 460 69 614 ISDN
---------------------------------------
Yes. If B is awake and doesn't beaver, A loses 12 instead of 6.
--
Julian Hayward 'Booles' on FIBS jul...@ratbag.demon.co.uk
+44-1344-640656 http://www.ratbag.demon.co.uk/
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