Is the probability that the GOP will steal control of the House in the November elections 18.4%?

Is the probability that the GOP will steal control of the House in the November elections 18.4%?


538's hybridization and weighting of polling and other data say there is a 84.6% chance that the Democrats take control of the House in 2018.


Yet the political markets are selling those buy contracts at 66 cents -- political investors think the chance the Dems actually secure this victory is only 66% or 18.4% less.


Can these very different estimates be reconciled?


One hypothesis: Nate Silver's shop does NOT estimate the likelihood of the GOP STEALING votes. When these polls are conducted, people who believe themselves registered state their voting intention, but many of them do not yet know their registrations and thus votes will be suppressed, or that their mail-in votes will be thrown out on a variety of illicit technicalities.


Is 84.6-66=18.6% the (extraordinary) measure of the likelihood that the GOP will "steal" the House in 2018? This is not the probability that they will TRY to steal the election -- that probability may be 100%. It is the probability that they will SUCCEED.


Alternative hypothesis: 18.4% is the measure of the likelihood that the final three weeks of key campaigns will see massive infusions of GOP financing and attack and other ads that will convert massive numbers of voters across many swing districts.


Of course, the 18.4% gap could be explained by a combination of these two, failures of 538's methods, and failure of the political markets. We just don't know.