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I would like to calculate the probability of a decisive vote in USA presidential election. This is a follow up question to this one.

Is it ok to assume that in USA, there are two conditions of casting a decisive vote:

  • (A) your vote is needed to break a tie in your state's election
  • (B) the selection of your electors would alter the outcome of Electoral College voting

If so, is it correct to assume that the probability of decisive vote in USA is of conditional probability nature:

$P(decisive) = P(A) \!\cdot\!\ P(B)$

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  • $\begingroup$ I do not see anything conditional about $P(decisive) = P(A)\cdot P(B)$. $\endgroup$ – Dave Mar 5 at 14:46
  • $\begingroup$ @Dave Probability of a tie in Electoral College voting, given a tie +1 vote in your state. Smells like conditional probability. $\endgroup$ – Przemyslaw Remin Mar 5 at 14:52
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    $\begingroup$ So write out conditional probabilities...what exactly do you mean? What are $P(A)$ and $P(B)$ that wind up making $P(\text{decisive}) = P(A)\cdot P(B)$ a conditional probability? $\endgroup$ – Dave Mar 5 at 15:00
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    $\begingroup$ This may be too much of an oversimplification of what happens, because in the US electors are determined in different ways in different states: in most states, all electors go to the winner of the popular vote, but in a few there are districts that vote for their own electors. But regardless, you seem to be stating the most basic property of conditional probabilities; the application appears irrelevant. Note that (B) needs to be restated as "the selection of your electors would alter the outcome of Electoral College voting" (which is far likelier than resolving a tie). $\endgroup$ – whuber Mar 5 at 15:39
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    $\begingroup$ Exactly this analysis has been carried out by Gelman and coauthors in this published paper for the 2008 election here (stat.columbia.edu/~gelman/research/published/probdecisive2.pdf), and for 2016 in this blog post here (statmodeling.stat.columbia.edu/2016/11/07/…) $\endgroup$ – CloseToC Mar 5 at 17:08

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