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Giant has six 20-sided dice, and by special rule, automatically loses initiative regardless of the die rolls.
Tess, on the other hand, has the recipe n(4) (8) (12) n(20) (X). Tess is thought to be a weak die -- and I guess it's true -- but she's not without some potential and can sometimes lead to interesting decisions. I've had at least four puzzle problems involving Tess on this blog in the past.
[I do think you can make much more effective recipes with null dice, but that's a subject for another post.]
Nulling out opponent's dice change the 2/3-calculation in interesting ways. Tess almost certainly wants her swing die to be high here, she's going to get the initiative and needs to maximize the chance she can make some non-null attacks at once before her larger non-null dice are gone, as they are almost certain to go in the first few turns.
It seems pretty likely that Giant will capture all three of Tess's non-null dice [If Tess finishes the game with uncaptured non-null dice, she has almost surely won, though we can consider the pathological cases where that doesn't happen at the end]. So, assuming Tess has no uncaptured non-null dice left at the end, the question is: how many of giant's dice will be nulled out, how many captured by Tess, and how many finish the game uncaptured?
So what are the victory conditions: Letting X stand for the value of the X swing, Giant starts out with 120 - (20+X) more sides, so his victory threshold begins at (100-x)2/3 = 66.667 - (2/3)*x If none of giant's dice are nulled out, he will have to finish with MORE than this many sides uncaptured -- that will take four dice if x< 10 and three dice if x>10 (when x is exactly 10, three dice is a tie for Giant, four dice is a win)
But every die nulled out shifts this threshold down by 2*20/3. the picture looks like this
HOW MANY DICE NULLED OUT BY TESS
Giant wins if at the end he still has
(3 dice sufficient if X>10)
1 die sufficent if X>10
66.67 - (2/3)*x
53.33 - (2/3)*X
40 - (2/3)*X
26.67 - (2/3)*X
13.33 - (2/3)*X
0 - (2/3)*X
-13.33 - (2/3)*X
YES X=10 & 3 dice left
YES, X=20 & 2 dice left
YES X=10 and 1 die remains
YES X=20 and no dice remain
Tess wins if she captures (NOT nulls out)
4 or more
4 or more
3 or more
if X < 10, 3 suffices
If X< 20, 3 suffices
if X < 10, 2 suffice
if X < 20 20, must capture both
Note it IS possible for Tess to win even with nulling out 4, 5 or 6 dice.
- With 4 dice nulled out and X less than 20, Tess wins if she captures the remaining 2. When X=20, Tess must capture the remaining two AND have at least one die left uncaptured at the end.
- With 5 dice nulled out, tess must capture the remaining die AND have non-null dice totalling more than (2/3)*X sides in order to win. It is just barely possible that this could happen in a game even without Giant's cooperation.
- With all six of giant's dice nulled out, tess must have non-null dice totalling more than (2/3)*(20+x) sides remaining. One could come up with ridiculous scenarios in which this occurs, but in all of them, unless Giant cooperates, the final turn could have resulted in at least one non-nulled capture.
Maybe there's a simpler way to look at this:
[Again the case when Giant has no dice left here assumes all of Tess's non-null dice were captured; the rare cases when this is not true are listed after the first table]
HOW MANY DICE GIANT HAS LEFT AT THE END OF THE ROND
NO MATTER WHAT
IF X greater than 10 OR
IF TWO DICE NULLED OUT
IF X greater than 10 and three dice nulled out OR
IF FIVE DICE NULLED OUT
IF ANY DIE NULLED OUT
IF FOUR DICE NULLED OUT
X=10, no dice nulled out
X=20 and 1 die nulled out
X=10, three dice nulled out
X=20, four dice nulled out
IF X less than 10 and no dice nulled out
IF X less than 20 and at most 1 die nulled out OR
IF X less than 10 and at most 3 dice nulled out OR
IF X less than 20 and at most four dice nulled out OR
IF no die nulled out
IF at most 2 dice nulled out
IF at most 3 dice nulled out