Election Opinion Poll Preferences: What's Possible for (a,b,c)?

  • Thread starter Thread starter Jameson
  • Start date Start date
Click For Summary
SUMMARY

In the discussion regarding election opinion poll preferences among candidates A, B, and C, it is established that certain combinations of voter preferences are impossible based on the fractions $a$, $b$, and $c$. Specifically, the combinations (0.61, 0.71, 0.71) and (0.68, 0.68, 0.68) are impossible due to the constraints of preference cycles. The analysis confirms that valid preference fractions must adhere to specific mathematical relationships, ruling out these combinations definitively.

PREREQUISITES
  • Understanding of preference cycles in voting theory
  • Familiarity with basic probability and fractions
  • Knowledge of election theory concepts
  • Ability to analyze mathematical inequalities
NEXT STEPS
  • Research the implications of preference cycles in voting systems
  • Study the mathematical foundations of voting theory
  • Explore the concept of Condorcet winners in elections
  • Learn about the Borda count method for ranking candidates
USEFUL FOR

Political scientists, statisticians, election analysts, and anyone interested in the mathematical modeling of voter preferences in elections.

Jameson
Insights Author
Gold Member
MHB
Messages
4,533
Reaction score
13
Three candidates A, B, C are contesting an election. In an opinion poll fraction $a$ of voters prefer A to B, fraction $b$ prefer B to C and fraction $c$ prefer C to A. then which of the following preferences are impossible for $(a,b,c)$?

1. (0.51,0.51,0.51)
2. (0.61,0.71,0.71)
3. (0.68,0.68,0.68)
4. (0.49,0.49,0.49)
--------------------
 
Physics news on Phys.org
No one answered this week's problem :(

Solution (from Stackexchange):
There are six possible preference orders for the candidates:

$d$: A>B>C
$e$: A>C>B
$f$: B>A>C
$g$: B>C>A
$h$: C>A>B
$i$: C>B>A

From this, $a = d + e + f$, $b = f + g + i$, and $c = e + h + i$.

$a + b +c = d + 2e + 2f + g + h + 2i \le 2(d + e + f + g + h + i) = 2(\text{# of voters})$.

In scenario (3), $a + b + c = 204\%$, which is impossible.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 0 ·
Replies
0
Views
3K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
606
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K