I apologize if this question is a bit of a read. (You might want to get a frosty beverage.) Professor Alan Natapoff of MIT demonstrated, if 9 Voters are districted into 3 electoral districts of 3 Voters each and the election is run like the U.S. electoral college where each district gets one electoral vote, the odds of a Voter casting the tie breaking vote in Their district is$${u/2}$$ where $${u}$$ is the probability of other Voters voting for Candidate A times the probability of other Voters voting for Candidate B times 4. Meanwhile, the chances of that district casting the tie breaking electoral vote (i.e., the probability the other districts will tie) is, according to the Professor,$$(\frac {1}{2})\frac {(u^2)(u+3)}{4}$$
While I can see how the Voter-casting-the-tie-breaker-for-the-district probability is determined (binomial probability), I do not see how the probability for the district-casting-the-tie-breaker-amongst-the-other-districts is determined.
It seems convoluted though it seems to be correct. Presumably at some state the argument says $0\le u \le 1$ so the probability of having the tie-breaking vote overall can be no more than $\frac14$
Assuming i.i.d. voting and suppose $p$ is the probability of another voter supporting A and $q=(1-p)$ of supporting B, then the probability the other two voters in this district being split is $2pq = \frac u2$ where $u=4pq$.
The probability of another district voting for A is $p^3+3p^2q$ and for B is $q^3+3q^2p$, so the probability the other two districts are split is $2(p^3+3p^2q)(q^3+3q^2p)$ $=-8p^6+24p^5-18p^4-4p^3+6p^2$ $=2p^2q^2(4pq+3)$ $=\frac12 \frac{{{u}^{2}} \left( u+3\right) }{4}$