Can any one explain the mathematics involved in horse racing betting odds?

Question

Horse racing betting odds, are full of mathematical terms. It becomes a painful task to calculate horse racing betting odds. Can you explain me the mathematics involved in horse racing betting odds?

Answer 1

Let me first clarify the difference between probability and odds, just
to be sure we're using the same terminology. The probability of an
event is:

             (Chances for)
     P(x) = ---------------
            (Total chances)

So, for example, the probability of drawing an ace in a single deck of
52 cards is 4/52 = 1/13 (or about 0.077 = 7.7%).

Odds, on the other hand, are given as:

     (Chances for) : (Chances against)

Of course, (Total chances) = (Chances for) + (Chances against), so
we can determine (Chances against) as (Total chances) - (Chances for).
The odds of drawing an ace in a deck of cards are
4:(52-4) = 4:48 = 1:12.

On to your question.

When the odds are converted to probabilities, they usually add up to
more than 1 to give the house the "edge" - that's how the track makes
its money.

Let's take a simple example. Suppose we have a series of 12 races with
4 horses, given the following betting odds (for simplicity, we'll only
consider the odds of winning each race):

     Horse     Odds   Probability (from odds)
     -----     ----   -----------------------
     Horse A   1:1    1/(1+1) = 1/2 =  6/12
     Horse B   2:1    1/(1+2) = 1/3 =  4/12
     Horse C   3:1    1/(1+3) = 1/4 =  3/12
     Horse D   5:1    1/(1+5) = 1/6 =  2/12
                                      -----
                  Total Probability = 15/12 > 1

Now suppose I were to bet $1 on each horse for each race. In order
for me to break even on each horse, horse A would have to win 6 of the
12 races - then I'd win +$6 on the races A won and lose -$6 on the
races A lost. Similarly, horse B would have to win 4 of the 12 races
for me to break even -- I'd win 2 * $4 = +$8 (because of the 2:1 odds)
on the wins but lose -$8 on the losses. Horse C would have to win 3 of
the 12 races (3*$3 = +$9 on the wins, -$9 on the losses), and horse D
would have to win 2 of the 12 races (2*$5 = +$10 on the wins, -$10 on
the losses). Of course, this means that all together, they have to
have 15 wins in 12 races, so somewhere they're going to fall 3 short
of my "break even" requirement.

If, for example, horse A only wins 5 races and horse C only wins 2
races, then I've lost -$2 on horse A (+$5, -$7) and -$6 on horse C
(2*$2 = +$4, -$10). The house has just collected $8 from my pocket.

As long as no horse wins more often than its "probability" (based on
odds), the house wins. Of course, it is possible that horses D and B
will win 4 races each, horse B will win 3 races, and horse A will only  
win 1 race. In this case, I will lose -$10 (+$1, -$11) on horse A,
break even on horses B and C, and win +$12 (4*$5 = +$20, -$8) on horse
D for a net winning of $2 - But you can "bet" that that won't happen
too often.  ;-)

I hope this helps. If you have any more questions, write back.