Thanks e-reuter@uiuc.edu (Erik Reuter) for pointing out the error in
my post -- I plead sleep deprivation after an all-nighter in AC :-).
I did realize the asymmetry of the situation when I awoke this
morning, but Erik had already caught and corrected my error before I
could --- that's what makes this group so great.  [I really did try to
post this at 8AM, but my ISP was constipated or something & wouldn't
take the article.]

It's obvious that if one player makes a mistake according to FToP,
then *some* player must benefit (since the game is zero sum).  What
Andy Morton pointed out was that it's possible for a player to
actually *lose* expectation when another player makes an FToP mistake,
which contradicts FToP (which claims you gain whenever some player
makes a play they should not make if they had full knowledge of the
situation). So FToP does not hold, in general, for multiway pots.

To recap: Player A bets, B calls and C is deciding whether to call or
fold with p units in the pot (including A and B's bets).  In the
card(s) to come, A has a outs, B has b, and C has c.  In the original
situation given by Andy, A had a made hand, so if C folds A gets the
additional benefit of C's outs, and B gets none.  That's why B prefers
that C call whenever the FToP says he shouldn't.  But then, B actually
prefers that C always calls since none of C's outs impact B.

Eric rewrote the theorem with this in mind in terms of the "Best Hand".
That captures the idea that the outs of the folding hand go to the
best hand, but's that's not true in general.  For example, Player A
has two small pair, Player B has a flush draw, and Player C has top pair.
Now if any of C's outs complete B's Flush (but not A's Full House),
the Theorem could apply to either A and B (it can only make a difference
for one of them in any given instance).

Morton's Theorem [Precise version]: Suppose Player A has bet one unit
    and has a>0 outs to win the pot, Player B has called with b>0
    outs, and Player C is to act with c>0 outs.  Then, ignoring future
    action, either there is a pot size p such that exactly one of A or
    B prefers that C fold even though C does not have pot odds for the
    call (violating FToP), or C's outs are divided among A and B
    exactly according to their respective chances of winning the pot.

Morton's Theorem [Informal version]: In almost any multiway situation,
    if the final player to call has any outs, then there is some
    player that prefers that C fold even though C does not have pot
    odds for the call.

The special condition (weasel words) in the precise version of the
theorem come about only in special cases.  Suppose with 42 cards to
come a=24, b=12 and c=6.  If c'=4 of c's outs go to A and c"=2 go to
B, then a/c' = (a+b)/c and b/c" = (a+b)/c, so there is no pot size
where the FToP fails.  However, I haven't (with a short search) been
able to come up with a situation like this.

I've thought a little about how to extend this result to include
implied odds, but I think I'll wait until I've fully caught up on
sleep before posting about it :-).

--jazbo