Horse Racing Odds

Date: 02/21/2002 at 14:06:37
From: Kevin Flynn
Subject: Probabilities

In any case of probabilites of an event, the individual probabilities 
of the situation must add up to one. However, in horse racing the odds 
often add up to more than one.  How come?

Date: 02/22/2002 at 01:33:55
From: Doctor Twe
Subject: Re: probabilities

Hi Kevin - thanks for writing to Dr. Math.

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.

- Doctor TWE, The Math Forum
  http://mathforum.com/dr.math/