ternary operator vote

[Bengt Richter]

On Wed, 12 Feb 2003 00:15:50 GMT, Andrew Koenig <ark at research.att.com> wrote:

>>> If "no change" is an alternative, you can express disapproval by
>>> voting for that alternative.
>
>Bengt> A: +1
>Bengt> B: +0
>Bengt> C: -0  # actual opinion, no way to express
>Bengt> D: -1  # ditto
>Bengt> no change: +1 means what?
>
>In approval voting, you are presented with a collection of
>alternatives, and you choose which of those alternatives you are
>willing to accept.
>
>You can choose more than one.
>
>So, if the alternatives are A, B, C, D, or no change, you could
>vote for no change (alone) if you want to express disapproval of
>any change.
>
But you are bypassing my point. I want to say I'm ok with A, but
"hell no" to C. Just a no_change +1 makes it seem like I'm equally ok
with A or no change, and indifferent to C.

IOW, even allowing only {-1,0,+1} for A-D gives 3**4 information states,
whereas {0,1} for A-D+No_change gives 2**5 : 81 vs 32, so obviously (ISTM)
a single no_change is no substitute for the expressiveness of -1,0,+1.

It's always (for n>=2 choices) going to be true that 3**n > 2**(n+1).

Regards,
Bengt Richter