I have two options here:
- take the poison with my small dice and keep the X=20 safe for another turn
- avoid the poison by rerolling the X=20. If I do this, it certainly makes more sense to take the X=12 than the (6).
I have two options here:
I guess the natural thing to do here is take the shadow die worth 10 sides. But if you don't, and it rolls out of range, you could also win by capturing both 6-sided dice while losing at most 12 sides, and if it doesn't roll out of range, you can take it on the next turn.
Which plan makes more sense? Which is mathematically better?
I think you could tweak the values on the dice to make this a still more interesting problem, but this is how it just presented itself:
I'm likely to win this round, but does randomlife have any moves with a chance of winning? and if there are, what's the best one?
Does devious have any chance of winning in this position? Or even coming out with a tie? What should he do?
I'm probably going to lose this round no matter what, but what is irilyth's best move to minimize my chances of survival?
You know, of course, that he has to take my poison to win, and so he can't afford to lose 18 sides. But should he capture my poison with his I(10), with a 40% chance of being captured -- and with the I(8) almost certain to be lost after that -- or is it better to capture with the X=20, with only a 25% chance of being captured (though if it is captured, he's lost for sure.)
It's the almost above that makes me pause. How likely would it be to escape if the I(10) rolled low enough to be captured? The die that rerolls when capturing the I(10) would have to roll a 1, which happens with probability 20% or 25%, AND the I(8) would have to roll higher than my only other remaining die, which would have value 2 or 4. This doesn't sound like it's enough to trim that 40% chance of the I(10) being captured into a less than 25% chance of losing.