What is the probability that Penn will abolish football (as it has boxing) within 5 years?
What was the probability of a healthreform bill in spring 2009?
What is the probability that the Republicans will have a majority in the House after the 2012 election? the Senate?
Other examples:
Intrade
Iowa Electronic Markets
NWS
National Hurricane Center
What is probability?
A numerical measure of the strength of a
belief in a certain proposition: p(proposition).
Theories: frequency, logical, personal.
Rules of coherence: addition, multiplication, conditional probability, independence.
The proportion of times it might have happened in the past that it actually did, e.g., p("I get run over crossing 38th street after class") =
(Number of times people got run over crossing 38th st.)  (Number of times people crossed 38th st.)But why this denominator? why the numerator?
But how often can we apply this in the real world?
(not A is called the "complement" of A)
rain  not rain 
I.e., if one of the propositions is true, that "excludes" the possibility of the other being true: the two propositions "mutually exclude" each other.
When propositions A and B are mutually exclusive: p(A or B) = p(A) + p(B)
e.g. p(yes) = p(femaleyes or maleyes) = p(maleyes) + p(femaleyes)

For example, the expected value of "$10 if a coin comes up heads" is $5.
What is the EV of "$10 if a coin comes up heads twice (in 2 flips)"?
EV is (roughly) the average value if the bet were repeated.
The value of a bet on one event should not change when you break it into two events.
For example, the expected value of (and willingness to pay for)
$10 if "Red card"
should be the same as:
$10 if "Heart"
and
$10 if "Diamond"
Suppose you are willing to pay or accept $4 for $10 if "Red card"
$2.50 for $10 if "Heart"
$2.50 for $10 if "Diamond"
You have a ticket for the first bet. I buy it from you for $4. You are now ahead by $4.
Suppose you are willing to pay or accept $4 for $10 if "Red card"
$2.50 for $10 if "Heart"
$2.50 for $10 if "Diamond"
You have a ticket for the first bet. I buy it from you for $4. You are now ahead by $4.
Then I sell you tickets for the second and third bets for $2.50 each. You are now behind by $1.
Suppose you are willing to pay or accept $4 for $10 if "Red card"
$2.50 for $10 if "Heart"
$2.50 for $10 if "Diamond"
You have a ticket for the first bet. I buy it from you for $4. You are now ahead by $4.
Then I sell you tickets for the second and third bets for $2.50 each. You are now behind by $1.
Then I point out that the second and third bets together are the same as the first bet, so you are willing to trade those two tickets for a ticket for the first bet. You are still behind by $1, and we start over....
The conditional probability of proposition A given proposition B is the probability that we would assign to A if we knew that B were true, that is, the probability of A conditional on B being true. We write p(AB) or p(A/B). This does not mean "divided by".
For example, what is the probability that Obama wins if Perry is the nominee?
The Conditional Probability Rule is: p(A  B) = p(A and B) / p(B)
(This time the / does mean "divided by".)
In other words: p(A and B) / p(B) = p(A  B)
or:
p (A and B) = p(A  B) × p(B), the multiplication rule.
p (Obamawin and Perrynom) = p(Obamawin  Perrynom) × p(Perrynom)
Your choice should not change if you narrow down the space of possibilities to what matters.
Thus, if you prefer ($45, .20) to ($30, .25)
you should prefer
($45,.80) to ($30, 1.00). But:
U($45) × .20 = U($30) × .25 iff
U($45) × .80 = U($30) × 1.00
Estimate the probability of some event E.
Estimate the probability of E given some other event F, p(EF) Estimate p(EnotF) Estimate p(F) Compute E' as p(F)×p(EF) + p(notF)×p(EnotF) 
Are "It is raining" and "Melissa is mowing the lawn" independent?
Are "The first car to pass us will be a Dodge" and "The second car to pass us will be a BMW" independent?
When A and B are independent, then the multiplication rule can be simplified, because p(A  B) = p(A). Hence: p(A and B) = p(A) × p(B).
(1) p(H  D) = p(H and D) / p(D)
But we don't know p(H and D) or p(D). Let's try to use what we do know to find out what they are. The Multiplication Rule is:
(2) p(D and H) = p(D  H) × p(H)
We can change (D and H) to (H and D), since they mean the same thing:
(3) p(H and D) = p(D  H) × p(H)
So now we can work out p(H and D) from two things we know: p(D  H) and p(H). If we use (3) to change (1), we can get a form of Bayes's theorem:
(4) p(H  D) = p(D  H) × p(H) / p(D)
Now we just need to get p(D). We can use the fact that D must happen either with H or with notH:
(5) p(D) = p(D and H) + p(D and notH)
Then we can use the Multiplication Rule again to replace the "and" terms:
(6) p(D) = p(D  H) × p(H) + p(D  notH) × p(notH)
By replacing "p(D)" in (4) with our expression from (6) we get:
(7)
p(DH)×p(H) p(HD) =  [ p(DH)×p(H) + p(DnotH)×p(notH) ]
(we have 7)
p(DH)×p(H) p(HD) =  [ p(DH)×p(H) + p(DnotH)×p(notH) ]
So the doctor could use that formula to figure out whether a mammogram would be worthwhile. Putting in the numbers
.9×.1 .09 .09 p(H  D) =  =  =  = .33 [.9×.1 + .2×.9] .09+.18 .27
compute p(y) from p(ym), p(yf), and p(m):
p(y) = p(ym)×p(m) + p(yf)×(1p(m))
which is the same as p(y&m) + p(y&f).
Now suppose we want to comput p(my). By Bayes,
p(my) = p(m&y) / [p(m&y) + p(f&y)]
or p(my) = p(m&y) / p(y).
In other words, conditional assessment gives us the denominator of Bayes's theorem, for this calculation.
p(D  H) × p(H) p(H  D) =  [ p(D  H) × p(H) + p(D  notH) × p(notH) ] p(D  notH) × p(notH) p(notH  D) =  [ p(D  H) × p(H) + p(D  notH) × p(notH) ] p(H  D) p(D  H) × p(H) p(D  H) p(H)  =  =  ×  p(notH  D) p(D  notH) × p(notH) p(D  notH) p(notH) posterior odds = diagnostic ratio × prior odds log(posterior odds) = log(diagnostic ratio) + log(prior odds)