probabilities - which formula to apply?

Fibonacci · 2011-06-18 00:47:36

Tell us what level you are at.

I took a programming/maths degree 25 years ago.
However, I have always struggled with probabilities. My main problem is that I often don't know which formula to apply.

Terminology

In horse racing, there are two types of bets: WIN and PLACE
A WIN bet wins if the horse comes 1st. The horse wins.
A PLACE bet wins if the horse comes either 1st, 2nd or 3rd (which of the three positions makes no difference). The horse places.

Example

10-runner race; 2 place bets on horses A and B

Question

1) What are the chances of both place bets winning?
2) What are the chances of exactly one horse placing?

Attempt (at question 1)

Most importantly, what's the formula for this problem?

Last edited by Fibonacci (2011-06-18 00:55:48)

bobbym · 2011-06-18 01:06:42

Hi Fibonacci;

Welcome to the forum! That is what I am getting for both of them placing, 1 / 15. You did that right.

For the second part:

That is the chance that neither of them places.

That is the chance just one places.

If you are a programmer you should be able to answer these type questions rather easily. I learned programming because I never wanted to do any more probability or combinatorics.

You can run simulations or enumerate all the possibilities.

Bob · 2011-06-18 01:10:18

hi Fibonacci

I don't have a formula for this. I can never remember them. So I'd draw a 'tree diagram', which is at least a method. I'll have a go at one.

Assumption. All 10 horses have an equal chance of winning, or getting a place. (Wouldn't that make life simpler!)

So For part (a) I agree with you and bobbym .... = 1/15

For part (b) add together the probabilities on the second and third branches

= (21 + 21)/90 = 42/90 = 7/15

Bob

Last edited by Bob (2011-06-18 19:45:22)

bobbym · 2011-06-18 01:30:51

Hi Bob;

I think your neither placed is not correct. It should be (7/10)(6/9) = (42/90) = 7 / 15.

Bob · 2011-06-18 01:37:39

hi bobbym

Thanks. I'd realised that error as I read through what I'd said.

It's what, for cards, I call 'without replacement'. I'd copied my probs from the top half of the diagram (= 'with replacement') without thinking it through.

That'll teach me to rush! It was all the excitement of the race that did it!

Bob

bobbym · 2011-06-18 01:39:01

Hi Bob;

No problem, I make lots of mistakes in all types of problems.

Fibonacci · 2011-06-18 05:25:35

Thank you for your replies.

The tree diagram is quite usefol for visualizingthe different senarios. How did you generate it?

I need to program this, so need the formula.

Attempt at Formulae

r = 10, p = 3, b = 2 (# of runners, places, bets)

case 1 (both place):

I still can't quite get my head round it but that's the number of combinations in the frame divided by the number of combinations in the field.*

case 2 (exactly one places)

no horse places:

*Terminology
frame = placed horses; field = all runners

Last edited by Fibonacci (2011-06-19 00:22:11)

bobbym · 2011-06-18 05:38:10

Hi Fibonacci;

Why do want to complicate your life by throwing in the binomials? Why not just do it with fractions?

What do you mean by bets? And what do you want to do with them?

Fibonacci · 2011-06-18 06:04:14

bobbym wrote:

What do you mean by bets? And what do you want to do with them?

Going back to the example in my original post, I am looking at 2 specific horses (because I put a place bet on them).

bobbym wrote:

Why do want to complicate your life by throwing in the binomials? Why not just do it with fractions?

Because I am looking at generalizing the problem with a formula that I can enter in a spreadsheet. Unless I am missing something I can only do this by using the factorial function.

We have done case 2 (exact number of places) by subtraction. That's all fine well for small p and b, but how do I work out the exact number of hits for r = 16, p = 5 and b = 4 ? (i.e. 0, 1, 2, 3, or 4 horses placing)

bobbym · 2011-06-18 06:12:59

For the 2 horses to place:

That takes care of b = 2 and the rest of your parameters.

That's all fine well for small p and b, but how do I work out the exact number of hits for r = 16, p = 5 and b = 4 ? (i.e. 0, 1, 2, 3, or 4 horses placing)

That was not your original problem. This is more complicated.

You will have to use the hypergeometric distribution.

Where did he go?

Bob · 2011-06-18 19:23:18

hi fibonacci

The tree diagram is quite usefol for visualizingthe different senarios. How did you generate it?

(i) I start by analysing how many (probability) events there are. In this case, event 1 = horse A ends the race, event 2 = horse B ends the race.

(ii) Consider how many things could happen for each event. eg. For horse A it could get a place or not get a place.

(iii) Now calculate the probability for each branch based on

So for horse A gets a place, as you know, there are 3 places and 10 horses, so P = 3/10

If that happens, then for horse B there are 2 places left, and 9 horses, so P = 2/9

Note. You get a different probability when on the third branch, p = 3/9, because A was not placed this time.

(iv) To calculate the probability of emerging from the 'tree' at particular point, multiply all the probabilities along the branches.

so eg. P(A is not placed but B is placed) = 7/10 x 3/9 = 21/90

note: I left all fractions un-simplified at this stage because it is easier to do this check:

Do the end probabiliities add up to one?

6/90 + 21/90 + 21/90 + 42/90 = (6 + 21 + 21 + 42)/90 = 90/90 = 1 .................. 'tick'

And also, to get answers that involve using more than one branch, it's easier to do step (v)

(v) If you want an answer where more than one branch is a 'wanted' case, add up the end branch probabilities.

eg. P(exactly one horse gets a place) = P(A placed but not B) + P(A not placed but B is)

= 21/90 + 21/90 = 42/90 = 7/15

Hope that helps.

Bob

Last edited by Bob (2011-06-18 19:40:11)

Fibonacci · 2011-06-18 23:55:12

bob bundy wrote:

The tree diagram is quite usefol for visualizingthe different senarios. How did you generate it?
(i) I start by analysing how many (probability) events there are. [...]

Thank you, Bob. I meant, "How did you create the graphic?"

Bob · 2011-06-19 00:37:41

hi

Oh, right. Mis-understood you there.

I've got some software called 'Sketchpad' which I use a lot for all sorts of diagrams.

But it costs money, so you might like geogebra which is similar and available as a free download. Google it and you'll find the site.

Then I do a screen shot, edit and crop, and save as a gif for up-loading.

Bob

Fibonacci · 2011-06-19 01:23:49

bobbym wrote:

[...] how do I work out the exact number of hits for r = 16, p = 5 and b = 4 ? (i.e. 0, 1, 2, 3, or 4 horses placing)
You will have to use the hypergeometric distribution.

Ingenious!

So, going back to my variables
r # of runners, p # of places, b # of bets, h # of hits (winning bets)

Curiously, before I saw your post, I had tinkered around with an example and came to the conclusion that the formula should be:

It took me a while to convince myself that hypergeometric distribution is the right formula because the example on Wikepedia is about black and white marbles picked from an urn. And I thought this was different because I had to take into consideration the different possbile position of horses in the race outcome.

I realize now that my attempt is actually equivalent to the hypergeometric distribution, just needlessly complicated!

Many thanks!

Last edited by Fibonacci (2011-06-19 01:25:51)

bobbym · 2011-06-19 01:29:05

Hi;

Yes, this is just a disguised ball and urn problem.

For instance to solve the simplest problem:

bets ( horses you have in the race ) = 2
number that finish = 0
runners = 10
places = 3

Plugging in:

The explanation is 2 bets with 10 horses and you are paid for the first 3 positions and none of your horses finished in the money.

For a tougher example:

bets ( horses you have in the race ) = 3
number that finish = 1
runners = 16
places = 4

This says you have 3 horses in a race with 16 horses and 4 places are paid with only one of your horses coming in the money.

Fibonacci · 2011-06-19 04:09:26

Hello bobbym,
Have I got hallucinations? I thought a couple of hours ago I had seen you write that I had reworked your formula incorrectly?

My version:

Your version:

I hadn't noticed but our formulae are actually equivalent.

I concur with your examples. I get the same results.

Last week I was trying to work out something more complicated: the probabilities of a combination of win bets and place bets (on the same race). It can be fairly easily done with a tree diagram but we probably won't find a neat formula for it!

bobbym · 2011-06-19 04:21:44

Hi Fibonacci;

Yes, I make a lot of changes to my posts. There are sometimes incorrect statements because I am using the post as a worksheet. I eventually work out most of the kinks. I am sorry for that confusion.

Last week I was trying to work out something more complicated: the probabilities of a combination of win bets and place bets (on the same race). It can be fairly easily done with a tree diagram but we probably won't find a neat formula for it!

State exactly what you want and provide an example and I will try to come up with a formula.

Fibonacci · 2011-06-19 06:46:19

Combination of WIN and PLACE bets on the same race

r = 10, p = 3

WIN : b = 3 (horses A or B or C to win)
PLACE : b = 2 (horses A or B to place)

possible outcomes:

a WIN bet wins + 2 places (= C wins and A+B place)
a WIN bet wins + 1 places
a WIN bet wins + 0 places
no WIN bet wins + 2 places
no WIN bet wins + 1 places
no WIN bet wins + 0 places

Many thanks. I am enjoying the discussion.

Fibonacci · 2011-06-19 06:49:57

bob bundy wrote:

you might like geogebra which is similar and available as a free download

Interesting. It's part of my Linux Debian distro.

Bob · 2011-06-19 06:53:21

Excellent! Have fun with it. bobbym has done several clever thing recently (search bobbym and geogebra) so you'll get a good start.

Bob

bobbym · 2011-06-19 18:45:05

Hi Fibonacci;

For the first one:

a WIN bet wins + 2 places (= C wins and A+B place)

I presume it could be B wins and A and C place etc.

You use the same formula for this one.

Fibonacci · 2011-06-20 08:46:59

Fibonacci wrote:

r = 10, p = 3
WIN : b = 3 (horses A or B or C to win)
PLACE : b = 2 (horses A or B to place)
a WIN bet wins + 2 places

Looking at the first 3 positions, valid outcomes (where x is any horse but A,B, C):
ABx
AxB
BAx
BxA
CAB
CBA

Fibonacci wrote:

a WIN bet wins + 1 place

CAx
CxA
CBx
CxB
Axx (where x can be C)
Bxx

Note: there is no place bet on C.
Note 2: if A or B win, this means a place in the place market, too.

bobbym · 2011-06-20 12:15:19

Hi;

I can compute all the positions where ABC are in the first 3 places or just AB, AC,BC, A, B, C. Will that help?

Fibonacci · 2011-06-21 07:07:52

Thank, I'd be interested.

bobbym · 2011-06-21 17:05:21

Hi Fibonacci;

For the first part.

So far I have only tested for bets = 3, so I would not change that parameter. You can vary the other two.

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probabilities - which formula to apply?

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