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BGonline.org Forums
Simborg's Chouette Rule
Posted By: Timothy Chow In Response To: Simborg's Chouette Rule (Stick)
Date: Wednesday, 28 October 2009, at 5:39 p.m.
All right, since nobody has yet answered my question as to what the rule is, here's my attempt to reconstruct the rule from what people have been saying.
The rule applies when the player not on roll—call this Player B—has 4 or fewer checkers remaining, and the player on roll—call this Player A—has no checkers off yet. (And perhaps the rule applies only when there is no contact remaining?) Player A rolls the dice. If there is no way to play the roll in a way that bears off a checker, then Player A does not play the roll immediately. Instead, Player B rolls. If Player B does not roll doubles (and here's a point that I'm not clear about; what if Player B does roll doubles but does not bear off all checkers? Does Player B have to play the doubles immediately?) then Player A rolls again, and then can play the combination of both rolls in any way he likes (or does he have to play the first roll first and the second roll second? Maybe it doesn't matter). Then the game continues. Presumably, if we're still in a situation where the rule applies, then we reset to the initial state.
So for example, suppose Player B has just two checkers left. Player A traps them both behind a six-prime, but rolls 44, and Player B jokers both checkers out. The rule now applies. Player A rolls, can't immediately bear a checker, so Player B rolls. Say Player B rolls 21. Then Player A rolls again, and gets to play both rolls at once. Right? And then we lather, rinse, and repeat. Conceivably, this could continue for several turns. Or am I misunderstanding the rule?
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