# The Old-School Liberal

## Ron Paul: The Most Electible Republican

Posted by Poorsummary on December 2, 2007

The 2008 presidential election is rapidly approaching, and Republicans are going to have a difficult time holding onto the presidency after the Bush administration. Prediction markets like Intrade provide a fairly accurate assessment of the probability of each candidate winning the primary nomination, as well as the probability of each candidate becoming president. In determining which candidate to send to a general election, however, primary voters should choose the candidate most likely to win the general election, given that they receive the primary nomination.

Although Intrade does not provide conditional probabilities, they are easy enough to compute*:

Given a probability space $\scriptstyle (\Omega, F, P)$ and two events $\scriptstyle A, B\,\in\, F$ with P(B) > 0, the conditional probability of A given B is defined by

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.\,$

Using the above formula, the probability of Ron Paul winning the presidency, given that he wins the Republican nomination is as follows:

P(Presidency | Primary) = P(Presidency ∩ Primary) / P(Primary)

Plugging in the current probabilities according to intrade, you get:

P(Presidency | Primary) = .035 / .052 = 0.673

Therefore, the probability of Ron Paul winning the presidency, given that he wins the primary is roughly 67%– hardly a sure thing, but the highest of any other Republican candidate. This is in line with my speculation that Ron Paul may be the most “centrist,” or have the broadest appeal across the political spectrum. Dan at Ron Paul Graphs has been tracking the Intrade data, and you can see that Ron Paul is the most electable (i.e. most likely to win the general election, given victory at the primaries) of all the Republican candidates, and is becoming more so:

Republicans clearly need to start thinking more about winning the general election, and decide whether they are willing to hand the presidency to Hillary, in order to continue supporting the neoconservative war-mongering of the Bush era.

*To illustrate, suppose I draw a card from a complete deck of cards (without Jokers). I ask you to tell me the probability that I am holding the Jack of Spades. Knowing that there are 52 cards in the deck, you tell me 1/52 or 1.92%. Now suppose that before asking you the question, I told you I was holding a Spade. Since you know a quarter of the cards are spades, you know that there is a 1 in 13 chance I am holding the Jack of Spades, or 7.69%.

Though this may seem obvious intuitively, you could also calculate the odds of my holding a Jack given that I am holding a Spade using the definition of conditional probability:

Let A = I am holding the Jack of Spades.

Let B = I am holding a Spade.

Then P( A | B) = the probability that I am holding the Jack of Spades, given I am holding a Spade,

and P( A ∩ B) = the probability that I am holding the Jack of Spades and I am holding a spade.

Setting up the equation as outlined above, you get:

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.\,$
P( A | B) = (1/52) / (1/4) = 1/13

…which is what we calculated before.