A bet on a statement P is an arrangement whereby the bettor wins a sum W if P is true and loses a sum L if P is false.  Some terminology is helpful:

Example: Let "the Saints will beat the Rams" be P.  Suppose you and I arrange a bet so that if the Saints beat the Rams you give me $3, and if they lose I give you $2.  Then, the stake is $5; the odds are 2/3 (2 to 3); the betting quotient is 2/5.
NOTE that in this example, I am the bettor on P; you are the bettor on -P.

The expected value of a bet for a bettor is the sum of the quantities obtained by multiplying the payoff (positive if one wins and negative if one loses) of each case by the probability of that case.
Example: in the Rams/Saints example, suppose that Pr(P) = 1/3.  Then, he expected values of the bet is $[3 (1/3) - 2 (2/3)], that is, $[1 - 4/3], that is -$1/3.
If one's expected value of a bet is

So, in the Rams/Saints example it would be unwise of me to bet because the expected value is negative.
One may wonder hat relation there must be between W and L for the above bet to be fair.  A little algebra gives us the solution:
The bet is fair just in case the expected value is 0, that is L(2/3) - W(1/3) = 0; that is, L(2/3) = W(1/3); so, L=W[(3/2) x (1/3)], that is, 2L =W.  So, for example,  if L=$2 and W=$4, then the bet is fair.
NOTES: Examples

Consider a bet on P such that I win $2 if P is true and lose $5 is it's false.  Determine, the odds, the stake, the betting quotient, and what Pr(P) must be for the bet to be fair.
i.    the odds are 5 to 2.
ii.    the stake is $7
iii.    the betting quotient is 5/7
iv.    Pr(P) = 5/7.

You flip a fair coin twice.  You win $8 if you get two heads; $6 if you get only one head, and lose $16 if you get no heads at all.  What's the expected value of the bet?
Pr(H&H) = 1/2 x 1/2 = 1/4
Pr(only one head) = Pr[(H1&-H2) v (-H1&H2)] = 1/2 x 1/2 + 1/2 x 1/2 = 1/4 + 1/4 = 1/2
Pr(no heads) = Pr (T&T) = 1/4.
So, the expected value is $(8 x 1/4 + 6 x 1/2 - 16 x 1/4) = $(2 + 3 - 4) = $1.  So,  the bet is adavantangeous.

The Dutch Book

If you make a series of bets such that you lose no matter what, then your bookie has made a Dutch Book against you.  It can be shown that  a Dutch Book can be made against you just in case the "probabilities" you assign do not obey the rules of probability calculus.  Here, instead of a proof, we'll look at an example.  Suppose you assign P a "probability" which is less than zero, say -1/2  Then, if W = -$3 and L =$1, the bet will be fair, for -1/2 x -3 = (1+1/2) x 1.  But if your wins are negative and your losses positive, you're sure to lose!  So, if you're ready to put your money where your mouth is, you'd better know some probability calculus!

Degrees of Belief

Intuitively, one's degree of belief in a statement P is one's level of certainty that P is true.  Obviously, our degree of belief that 2+3 = 5 is very high, but our degree of belief that the Rams will win the Superbowl is pretty low.  The notions developed above help us to measure exactly one's degree of belief in P is we assume that one is not risk averse, that is, is ready to bet  on the basis of the expected value of a bet on P.  In fact, if we can determine when one judges a bet on P fair, then we know what one thinks Pr(P) is, since Pr(P) is equal to the betting quotient, L/(W+L) (WHY?).