Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Probability distribution of horses

Status
Not open for further replies.

mhamini

Member level 1
Joined
Aug 23, 2005
Messages
40
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
Iran-Tehran
Activity points
1,690
A horse race is going to take place with six runners.
The race is over 5 furlongs (1000 meters) and for each of the six contestants it is known that their probable times at this distance are:

horse 1: 57.00 sec
horse 2: 57.20 sec
horse 3: 57.35 sec
horse 4: 57.80 sec
horse 5: 58.10 sec
horse 6: 59.50 sec

But, as is always the case in horse races, these times are uncertain, so the outcome is unknown.
In fact each of the above times is accurate by plus or minus 0.50 seconds, i.e. for horse "1" there is a Gaussian distribution with mean 57.00 and standard deviation 0.5, for horse "2" there is a Gaussian with mean 57.20 and st. dev. 0.5 and so on.

What is the probability for each horse to win the race ?

There is an easy (but a little slow) answer that can be derived by Monte Carlo simulation using random numbers, but it's not what I 'm asking for.
Does anyone know a functional approximation for the winner's pdf ?
 

The final results can be expressed by a bunch of integrals.
Assume that the random variables and the corresponding density functions are:

X₁---f₁(x₁),
X₂---f₂(x₂),
...
X_{n}---f_{n}(x_{n}).

Obviously, if n=6, you can write down the density functions as you know the means and the standard deviations.
Suppose that you want to calculate the probability that X₁ win this game. Then you want to calculate the probability of following events:

X₁-X₂ < 0,
X₁-X₃ < 0,
...
X₁-X_{n} < 0,

which are

P(X₁-X₂<0),
P(X₁-X₃<0),
...
P(X₁-X_{n}<0),

and then you multiply them together to get the probability you want:

P(X₁-X₂<0)P(X₁-X₃<0)...P(X₁-X_{n}<0).

Let's only calculate one of them. The others can be similarly obtained.

P(X₁-X₂<0)
= ∫∫_{x₁-x₂<0}f₁(x₁)f₂(x₂)dx₁dx₂
= ∫_{-∞}^{∞}f₂(x₂)dx₂∫_{-∞}^{x₂}f₁(x₁)dx₁.

Oh, well, this is all I can get to. I don't have a good idea to carry out this integral.
 

    mhamini

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top