# Bayes' formula application to the probability of a horse winning, depending on the jockey

I have a problem that I have been racking my brain to figure this out and I just don't have the background to know if I am correct or not so I am hoping that you can help me out, OK here it goes so bear with me. I am going to try and make it as simple and clear as possible. I am going to start with a horse racing example and then go into my dilemma

Horse A and Horse B are going to race against each other in a to horse race.

They have raced each other 12 times before and Horse A has won five times and horse B has won seven times. So horse A has won 41.7% of the time but....

There are two Jockeys RED and BLUE

3 out of 5 of those wins came when Blue Jockey was riding horse A. However Blue Jockey rode him only once on the days horse A lost.

Today Blue Jockey is riding horse A so we get a 60% chance if we just use that Blue jockey 3 out of five fact. But I know we need to consider the total amount of races as well

So far I have this then

P(horse A wins given blue jockey rides) = .60 X .417 /.333

because 60% of 3 of the 5 wins were when Blue was riding. 41.7% win over all and .333 cause Blue Jockey has ridden the horse 4 times out of twelve races.

Right?

Now here is where my question with changes

Two horses A & B but five jockeys. so lets just use some quick made up stats for the sake of time.

Horse A has a 66% chance to win but when Jockey Black rides the horse A, horse A wins 70.5% of the time.

now the colors of the jockeys are put in a jar to be pulled out to see who is going to ride horse A. the following are the chances in % of color being pulled out

Red 19% Black 21% Green 13% Blue 37% Pink 10%

Would my formula to see the actual percentage that horse A wins be:

 Horse A wins given Black rides = .705 X .66/.21


which then gives me 2.21 which is WHAT?? what does that mean?

AND

if it were a different rider lets say pink and for the sake of simplicity same win% would the denominator then be .10

This is causing me to lose sleep , just kidding(kinda) :)

• Unless I have to run the formula again with the 75% * 21% at the numerator? hmmmmm still having doubts over here. grrrrr – Beto Aug 4 '14 at 14:31

i suspect you are asking how to combine data in a coherent way to predict outcomes. i will suggest an approach to do so: we need a statistical model for observed data. For example, and following the horse racing application, suppose we observe the results of many horse races. In each race we assume two horses, horse A vs horse B, to keep it simple. Any of 5 jockeys can ride the horses. Maybe jockey #1 tends to ride horse A more often. Maybe jockey #2 is an unskilled jockey, etc. That will all come out in the data analysis. A possible model is the logistic regression:

$${\hat p}=\frac {1}{1+exp[-(\beta_0+\beta_1A_1+...+\beta_5A_5+\beta_6B_1+...+\beta_{10}B_{5})] }$$

where $\beta_0,...,\beta_{10}$ are values that will be estimated when fitting the model to a data set.

For each race we record the winning horse and the jockey of each horse. We do that by specifying:

$A_k=1$ if jockey $k$ rode horse A and $=0$ if not.

$B_k=1$ if jockey $k$ rode horse B and $=0$ if not.

If we code the data so that horse A wins & B loses is $=1$, and B wins & A loses is $=0$ then the result of the model is to predict the probability that A wins given jockey 3 on A and 1 on B. That would be the expression

$${\hat p}=\frac {1}{1+exp[-(\beta_0+\beta_3+\beta_6) ]}$$

Note that the $\beta's$ may be negative. In real horse race data, the impact of a jockey on the outcome is limited so all the A's and B's will come out near 0. But if they had a big effect, the coefficients would indicate that. The value of $\beta_0$ measures the ability of the horses apart from the jockeys. If, on the other hand, you are looking to fit your example into a theorem of Baye's type problem, you might let us know and start here:

http://en.wikipedia.org/wiki/Bayes%27_theorem

• Could I also use Bayes here but by adding the different data as B ? – Beto Aug 7 '14 at 20:41

Keep track of your probabilities in a table:

$$\begin{array}{c|cccccc} & \text{R} & \text{Black} & \text{G} & \text{Bl}& \text{P} &\text{Total} \\ \hline \text{A wins} & & 0.1481 & & & & 0.667 \\ \text{A loses} & & 0.0619 & & & & 0.333 \\ \text{Total} & 0.19 & 0.21 & 0.13 & 0.37 & 0.10 \\ \end{array}$$

You cannot create conditional probabilities at random and expect the results to make sense. And you are not using correct formulas.

I placed the given information in the Total row and column. The 2x5 table is the probability of the intersection of the row and column events. These 10 cells must sum to 1 and row and column sums must add up. Then we can fill in 2 cells from $P(A \text{ wins } | \text{ Black is A's jockey})=0.705.$ Then $$P(A \text{ wins and Black is A's jockey})=0.705*0.21=0.1481$$

And 0.21-0.1481. Those are the only cells we can fill in. The only other probabilities we can compute is $$P( \text{Black is A's jockey } | \text{ A wins})=0.1481/0.667=0.2220$$ and the correspondind one for given that A loses.