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 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum