[EM] Probability of Condorcet Winners ...

[Craig Carey]

At 16:50 05.03.00 , Norman Petry wrote:
>In response to Craig Carey, Mike Ossipoff wrote (4 Mar 00):
...
>>>  plausibility to it?. Whereas for large problems, Condorcet can't find
>>>  an answer for
>>>  0.999999999999999999999999999999999999999999999 of all elections,
>>
...

>3) As truncation increases ("involvement" approaches 0), voting cycles
>become slightly more probable.
>
>4) There will be no Condorcet winner about 5% of the time, under most
>reasonable assumptions.  These few cases will require a suitable tiebreaker,
...
>
>I would be interested to know the assumptions that underlie Craig Carey's
>result for Condorcet's method.


I copy from a message I sent before.


At 02:35 19.November.99 , Craig Carey wrote:
...
:      A note on Condorcet's picking 0 winners when there should be 1
...
:Let's estimate F, where 10**(-F) equals the probability that the Condorcet
: method actually returns a single winner when there are 10**99 candidates.
:
:The approximate derivation here requires the drawing of a graph, and noting
: or saying that the directions of the arrows are quite random, and that
: F = log base 10 of approximately this: 10**99 times (1/2)**(10**99 - 1).
:
:So F = -3*(10**98) (approximately).
:
:So Condorcet rarely picks a winner for large enough elections.
:The next step in the argument is to wait for the defender of Condorcet to
: say that there are 'too many candidates'. That happens.

The correlations will be low.

...
:
:
:                     Bounding where Condorcet goes bad
:
:Maybe Condorcet is OK for 102 candidates but it is not OK for 103.


If people do not use Condorcet, then what Condorcet variant is used, Just
 something to show that a mathematical analysis with 10**99 candidates
 is a mismatch with the method's' typical use.