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### Horse Racing Odds

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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.

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

* Odds are listed here as (chances for):(chances against), which
is consistent with the notation used in the discussion.  Historically,
in horse racing odds are expressed as (chances against):(chances for).
So what we have shown as odds of 1:3 would be shown on a racing sheet
as 3:1.

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/
```
Associated Topics:
High School Probability

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