Intelligence Squared Scoring and Winner Determination There is an NPR podcast called Intelligence Squared. Each episode is a broadcasting of a live debate on some contentious statement such as "The 2nd amendment is no longer relevant" or "Affirmative action on college campuses does more harm than good". Four representatives debate-- two for the motion and two against.
To determine which side wins, the audience is polled both before and after the debate. The side that gained more in terms of absolute percentage is deemed the winner. For example:
          For    Against  Undecided
 Before   18%      42%       40%
 After    23%      49%       28%

 Winner: Against team -- The motion is rejected.

Intuitively, I think this measure of success is biased and I am wondering how one would poll the audience to determine the winner in a fair way.
Three issues I immediately see with the current method:


*

*At the extremes, if one side starts with 100% agreement, they can only tie or lose.

*If there are no undecided, then the side with less initial agreement can be viewed as having a larger sample size from which to draw.

*The undecided side is not likely to be truly undecided. If we assume that the two sides are equally polarized, it seems our prior belief about the undecided population should be $\text{Beta}(\text{# For}, \text{# Against})$ if each was forced to take a side.
Given that we have to rely on audience polling, is there a more fair way to judge who wins?
 A: The issue of bias here seems to be that one side may be favored to win even if they don't have better debating skill, rather than the statistical concept of bias of an estimator. A natural approach would be to tackle this concern directly: use data from previous contests to fit a regression model 
\begin{equation}
p(\textrm{for}_\textrm{after},\textrm{against}_{\textrm{after}},\textrm{undecided}_{\textrm{after}} \mid \textrm{for}_\textrm{before},\textrm{against}_{\textrm{before}},\textrm{undecided}_{\textrm{before}})
\end{equation}
and set the winning rule as a function of the before-debate-poll so that the predictive probability of winning is $0.5$ for both teams. Note that there are still multiple choices for the decision rule as the outcome space is 2-dimensional but, if we trust the predictive model, this doesn't matter in terms of fairness of the contest. One could, e.g., just decide that the for-team wins if the For-Against ratio after the debate exceeds its predictive median (conditional on the before-poll). 
Ideas for building a predictive model
Initially I had in mind just some "black-box" model of the after-poll-numbers as a function of the  before-poll-numbers and noise. However, a better approach might be to borrow whuber's idea of considering the transition probabilities. Simplest (though maybe not realistic) approach would be to consider the transition probabilities as independent of the before-debate-poll numbers. For example, assume the transition probabilities are drawn from Dirichlet distributions:
\begin{align}
(P(\textrm{for} \mid \textrm{for before}),P(\textrm{ud} \mid \textrm{for before}),P(\textrm{ag} \mid \textrm{for before})) & \sim   Dir(a_{ff},a_{uf},a_{af}) \\
(P(\textrm{for} \mid \textrm{ud before}),P(\textrm{ud} \mid \textrm{ud before}),P(\textrm{ag} \mid \textrm{ud before})) & \sim   Dir(a_{fu},a_{uu},a_{au}) \\
(P(\textrm{for} \mid \textrm{ag before}),P(\textrm{ud} \mid \textrm{ag before}),P(\textrm{ag} \mid \textrm{ag before})) & \sim   Dir(a_{fa},a_{ua},a_{aa}),
\end{align}
where the $P$s are transition probabilities for individuals and the $a$s are hyperparameters that control how the transition probabilities vary from debate to another. The $a$s are learned from data of previous shows, either by optimizing point estimates (e.g. maximum a posteriori or maximum likelihood) these, or a full Bayesian solution that outputs a posterior distribtuion of the $a$s. One could also add some symmetry constraints if one wants to assume for and against behave similarly (before knowing the particular debate question) e.g., $a_{ff}=a_{aa}$, $a_{fu}=a_{au}$. 
Given posterior distributions or point estimates of $a$s, and the distribution of individuals in current before poll (that I now assumed to be independent of the transition probabilities), it is straightforward to simulate the distribution of after-debate-poll numbers, and then pick the median of, e.g., for/against-ratio as the winning threshold. 
