Math Is Fun Forum

figured it myself:

Yes:

The easiest starting point to work the intersection out are the three equations for x in post #1:

The above is true regardless which of the three equations you pair up.

However, finding the intersection is not the objective although it helps me in my real-life scenario to know where it is. Lets put it that way: I have a choice of lines and I choose those three that give me the best possible triangle. And that is one that gives me bottommost yellow dot (as per post #4) with a high y-coordinate.

Indeed, the fs are identical.

The question is:
We have three lines that form a triangle where the lines intersect. The y-coordinates of the three intersactions are all different. I want to draw a vertical line that intersects with these three lines.

I want to find a vertical line where the lowest of the three intersects is the highest (as per post # attachment).

Can this be done mathematically? I can do this programmatically by calculating the three triangle points, "drawing" a vertical line through each point, then comparing the y-coordinates of the each vertical line's intersections  with the lines forming the triangle.

Thank you so much for introducing me to GeoGebra! Fantastic for developing and documenting ideas!

I know it's obvious when looking at the chart. But why?

Or R02, see new screenshot.

All three lines cross for

.

Within given constraints, I choose the parameters with the objective of a high minimum

. So far I have been doing this from experience.

We know quite a few things about the lines' coefficients:

are positive (and therefore all lines' y-intersects >0)

is negative. I am pretty certain that, through choice, q1 and q2 are always positive because negative slopes give me low r.

Let's zoom in on the q's:

We know:

I choose a scenario where :

Sorry. I got so used to my naming conventions I forgot to clarify.

are functions.

is the only variable, all others are constants.

is a constant replacing a more complex expression of

.

I have already solved

. However, there are 3 different possible values depending on which two equations I try to equalize. (That is in the nature of the problem and quite correct.)

Lets call the 3 possible values

. In an ideal world all three lines 

would intersect in one point thus giving me just one

.

In the real world the don't, which is why I have to choose a make-do value for

, which fulfils certain criteria.

Having thought about this a wee bit more, I need to rephrase what I am looking for: I want to find the

for which the lowest 

is the greatest.

I'll post a screenshot to illustrate what I am looking for in the next post.

Started with IBM System/370 -> half a dozen other architectures I don't remember -> 8051

I am getting a bit out of my depth here. Expectation - is that the same as EV? My limited experience with this is just in relation to Roulette or BlackJack wager requirements. (You determine the EV from the house edge.)

Anyway, I am quite happy to retire this thread now. I have taken a few things from it apart from the hypergeometric distribution function: I even installed GeoGebra, and I am switching my wiki to LaTeX-based MathJAX.

Many thanks for the inspirations.

I shall be back with probability questions about my favourite dice game. No money involved but I'd like to get an edge.

Sorry, I' d been away for a while.

I am not quite sure I understand how you arrived at the formula in post #21.

You use the variable bets=3 in the formula, which is the number of bets in the WIN market. (Going back to my question and example in post #18, there are two sets of bets: 3 bets in the WIN market and 2 bets in the PLACE market.)

What I can't see is how we can possibly apply the same formula because it only takes into consideration the number of bets in one market.

I think you can only generalize the combined WIN and PLACE bet problem by putting the horses into different groups.

Variables
r # of runners
p # of places
h # of hits
bw # horses with a win bet and no place bet
bp # horses with a place bet and no win bet
bb # horses with both a place bet and a win bet

Idea
Treat 1st position  separately and combine with probabilities of the remaining positions. This should enable us to use the hypergeometric distribution function again:

Scenarios

WIN and 0 places

WIN and 1 place

no WIN and 0 places

no WIN and 1 place

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

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.

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.

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!

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

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)