Here's the reality test: would you ever expect to win $2.50 on a single toss of a $5 coin? Nope. The probability of winning $5 is 0.5, but that's not the same as expecting to win half the value of the coin. No bank is going to credit you for $2.50 in collateral based on an un-tossed coin.
Ross's dirty trick is in trying to use his false statistical metaphor to fool us into thinking that he has a rigorous mathematical rationale for what follows. But he doesn't.
I can only guess that he's trying to spin O'Donnell's vote tally in the Republican primary, into a winning margin for the general election. Well .... the real math isn't that encouraging.
For example, the actual party affiliation numbers in Delaware are not encouraging for any Republican -- they're outnumbered by more than 3:2 by Democrats, with unafiliated voters comprising an additional 23.4%.
An O'Donnell victory would have to combine a significant difference in turnout between R and D voters (unlikely, since O'Donnell is admittedly "scary" to Democrats, and they're more likely to show up to vote against her, than against a Mike Castle); and she will also need to attract a significant majority of the "other" voters, who are not registered R or D -- and in any case we must assume that independents will show up in sufficient numbers to make up the difference between registered R and D voters.
Look at the one statistic he does provide is a 25% "intensity" gap. Note that this is not a measure of likely voter turnout, but rather a measure of how much they like their candidate. As noted, though -- O'Donnell is polarizing, and the intensity of Democrat dislike of her, is at least a counterbalance to their lack of enthusiasm for Coons.
Staying out of the statistical weeds, what do you think of this observation by Ross...
No one can say at this point if Christine O'Donnell will win in the general election. According to the polls, the odds are still against her. The Democrat party, however, is reportedly spending a significant amount of money on the Coons campaign. If O'Donnell has no chance of winning, what are they worried about? Why waste their money?
I’m going to track your expectation.
I think many non-Delawareans don't realize that only 3 out of 10 registered Delaware voters are Republican (and they have lower turnout rate than the Democrats). And Christine O'Donnell got only slightly more than half of the Republicans. So this "mandate" of hers is from 16% of the voters. That's not much of a base to carry one through a general election.
Expected Value is a measure of an ucertain event and its associated payout. In this case, the uncertain event is the tossing of a coin. The payout is $5 or $0. The expected value is the probability-weighted average of all the outcomes.
Another term that might make more sense to you is the Certain Equivalent. In other words, how much money would it be worth to you to NOT play the game? If you were given the choice of getting a certain $2.50 or to flip the coin and take the result, what would you choose? What if you were offered $2.00 or the coin toss? $3.00? For the coin toss, the certain equivalent would be $2.50, because over repeated tossings of the coin, the average would be $2.50.
Look at post #4. This is the application of expected value. If Casle would vote with the GOP on 5 bills and stab them in the back on 45, while O'Donnell is good for voting 100% for the GOP but only has a small chance of winning, what is the better bet?
Of course, the counter argument is that the first vote to caucus is the only vote that counts, because the rest is moot if the GOP doesn't control the committees, but then O'Donnell wasn't the only candidate to flip a strong win into a strong lose that week. Dino Rossi also fell dramatically against Patty Murray, putting us 2 seats down, so Delaware voters did not cost us a sure majority in the Senate.
-PJ