dare i say that we CAN put this one to rest??
cheers
jerry
Fair enough - I've preached that there's always more than one way to solve a problem....
dare i say that we CAN put this one to rest??
cheers
jerry
Actually, I have a similar question on probability that deals with a population, sample size, and matching.
While I don't claim to completely undertand the combin formulae, I DO understand that when you plug in the appropriate figures, the odds do match my state lottery.
Where my curiosity lays is in the odds of the $1,000,000 Pyramid game bonus round.
The pyramid contains a population of 59 selectable boxes which contain either a credit value or a pyramid symbol.
You are instructed to pick 7 of the boxes, hence the sample size is 7.
Now if there were only 7 pyramid symbols in the game, the combin formulae would work to give you the odds of hitting 7 out of 7, however; there are 11 hidden pyramids among the population of 59 boxes.
You get a bonus for each pyramid revealed, so what I'd like to know is what are the odds of finding 1 through 7 pyramids in a population of 59 squares with 11 hidden pyramids and 7 selected boxes.
Could we also invert the combin formulae so we show a 1 in ***XX odds summary rather than a decimal value?
Also, I'd like to get this into an Excel spreadsheet, so if you could help with the combin formulae, I'd be most thankful!
C.
I believe I figured it out with a bit of searching ... Amazing what one can do when they put their minds to it!
Guessing we're dealing with HyperGeometric Distributions here, and as luck would have it, excel does have a HYPGEOMDIST function (YEAH!)
Which of course now leads me to my next question, and that is, are HGDs additive? For example, if a population is made up of 3 or more elements instead of just successful and unsuccessful.
Can you calculate the HGD for each element and add them together for a combination?
For example, a population has 100 elements, 95 of which we're not concerned with, however, there are 2 successful elements X, 2 successful elements Y and 1 successful element Z in the population.
What I'd like to know is if we must have at least one of each (X,Y, and Z), what are the odds it we are able to make S # of selections?
This gets really weird for me, and I can't fathom how to deal with more than just successful vs. unsuccessful.
Any advice, formulae or bottles of strong spirits to help comprehend this?
Many Thanks!
C.
cet,
seems like you have your first problem under control. so to answer your second: in a hypergeometric probability you are dealing with success or failure only.
so to solve your example problem with 2 x's, 2 y's and 1 z in the set of 100 and a requirement that you draw one of each to win.
so try something like this:
use s=3 as a base to get a sense of how it would work, thus you would have to draw one x, one y and one z in three draws. try to simply count the number of way you could draw x,y,z out of the set and divide by the total number of sets of three draws possible (100C3). this might get you going.
i have to leave right now for an appointment, but i have an idea of a strategy and will post tomorrow if you are still at it.
cheers
jerry
i have solved this, if you are still watching start a new post on your question and i will post a solution.
cheers
jerry
I appreciate it Jerry. It's confounding when there are multiples of success available.
Currently I'm thinking about a population of 120 with 2 successful of x and y and 1 successful of z in the population.
We require 1 each of x, y, & z to be selected out of 12 picks from the population.
Frankly, I'm stumped! :P ... It took every brain cell just to get the concept of HyperGeometric Distribution. LOL!
I appreciate any suggestions you may have.
Regards,
C.
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