This is the line of thought I followed:
As a consequence,
If #4 were true, then #2 must be true (by exclusion), but this would imply that #4 itself is false. Then,
If #5 were true,
This feels a bit underdetermined to me, probably meaning that I'm missing some hidden hints that nail things down further. But here's one thing we can do, which meets the explicitly given requirements:
where we take
Ask the following question of all three guards:
Now the number of Yeses (Y) will be between 0 and 3 inclusive.
If Y=1, go through that door. The position may either be
in which case you go to heaven, or it may be one of
in which case you go to hell.
If Y=2, namely
then pick one of the Yeses at random and ask the utterer the same question again. ...
7S can be forcibly won with five HCP:
If West leads a spade, South takes the ace. If West leads a heart, North ruffs and leads a spade to the ace. Otherwise South ruffs low and leads the ace of spades.
Imagine the family tree of such a population. It might look something like this
a <------Generation A - one person
b b b
ccc ccc ccc <------Generation C - nine people
The solution to the paradox is that
By shifting the burden by one generation back,
Also, it is important to consider the following
If they are ...
(Assisted 7NT): Given optimal card distribution, just have the opponents discard the high cards:
(Assisted 7♠): The ♠A can eat ♠K and ♠Q at most, so we must also have the ♠J, for a total or
(Guaranteed 7NT): First answer, which assumed that I cannot arrange the opponents' cards.
(Guaranteed 7NT): Edited in after OP clarified that I get to arrange the ...
Let's call everyone by their initial, except that since there are two As we'll use N for Alyin and A for Alayna. Then the answer is
If you want to check my work, put the following into a Python interpreter and verify that you get a bunch of Trues out (I do):