Math Question - Probability

[MarcusHawkins]

Re: Math Question - Probability

Sorry if I mistook what you said earlier, I thought you were essentially recapping what I said in my first paragraph of the OP, but you were much more on the right track than I was.

No, you did not mistake what you said earlier, I was indeed taking what you said in the OP before I realized what was going on later on in the thread!

BUT, I KNEW there is a connection somehow, and thanks to ben's formula, I think I found it! I hope he double checks to make sure I'm right, and if not, I'm sure he'll think of something more solid! :P

Thanks nonetheless!

EDIT: I JUST started taking AP Calculus AB, (essentially first year college calculus in high school) and I am learning how to use limits correctly and master that skill.
After THAT, I will be learning how to use integrals correctly, so it should be fun!

[SKG_Scintill]

Re: Math Question - Probability

graphs are cute

for 1-5

for 1-k


courtesy of wolframalpha, wonderful site

edit: by the way, are we just neglecting my formula because it isn't elegant enough? :P
edit v2: rewrote it this way, is it elegant enough?

[benguino]

Re: Math Question - Probability

No, you did not mistake what you said earlier, I was indeed taking what you said in the OP before I realized what was going on later on in the thread!

BUT, I KNEW there is a connection somehow, and thanks to ben's formula, I think I found it! I hope he double checks to make sure I'm right, and if not, I'm sure he'll think of something more solid! :P

Thanks nonetheless!

EDIT: I JUST started taking AP Calculus AB, (essentially first year college calculus in high school) and I am learning how to use limits correctly and master that skill.
After THAT, I will be learning how to use integrals correctly, so it should be fun!

So I let m be the minimum value and M be the maximum value. Here's the updated calculations taking that into account: http://puu.sh/wU7Se/8517d7e64e.pdf

The formula isn't quite as nice...at least it doesn't look like it simplifies to what we thought it might be:


EDIT: Made it a tad bit nicer:

[benguino]

Re: Math Question - Probability

graphs are cute

for 1-5

for 1-k


courtesy of wolframalpha, wonderful site

edit: by the way, are we just neglecting my formula because it isn't elegant enough? :P
edit v2: rewrote it this way, is it elegant enough?

That rewrite looks a lot nicer
Like I said before, I didn't mean to neglect your formula...I'm just trying to solve the "continuous" case instead of the discrete case :P

[benguino]

Re: Math Question - Probability

EDIT: I JUST started taking AP Calculus AB, (essentially first year college calculus in high school) and I am learning how to use limits correctly and master that skill.
After THAT, I will be learning how to use integrals correctly, so it should be fun!

Nice The stuff in the link I put is still a bit beyond the scope of most single-variable Calculus courses. If you want to know more how to do the kind of work in the link, you can peruse an introductory calculus-based probability/statistics book after you feel comfortable with basis derivatives and integrals

[SKG_Scintill]

Re: Math Question - Probability

as a bit of visualization for my approach

in case I: you take a number from 1-5 twice, so I put the 2 numbers as seperate axes
the squares show how many times the number is the highest
you can see that the occurrences of number N are N squared minus (N-1) squared
aka: n^2 - (n-1)^2

same in case III where I put it as 3 axes and the occurrences of N are N cubed minus (N-1) cubed
aka: n^3 - (n-1)^3

case II is where I multiplied the occurrences of case I by the weight of the number
obviously I can't visualize the same for case III as it would be a 4d shape

at this point I've done n(n^2-(n-1)^2) for case II
so I did a summation for all the numbers N to get the sum of all amounts
with that sum I now have to divide it by the amount of combinations, which is just the size of case I
I tried "re-ordering the blocks" so they'd fit neatly inside eachother, but that doesn't seem to have a simple way (pretty obvious since the averages are decimal numbers)

for the latest rewritten formula I approached it like this:
take case I, make it one dimension greater, so it's [the highest number]^3, then subtract the sum of the smaller numbers squared (*1 to make it little cubes, but that's redundant)

so... summarized
average is ( [the highest number] to the power of ([amount of times you take a number] + 1) minus the sum of [all smaller numbers] to the power of [amount of times you take a number] ) divided by [the highest number] to the power of [amount of times you take a number]
average = ( [k] ^ ([m]+1) - sigma [from n=0 to n = k-1] of n ^ [m]) / k^m
also to be written as

...

no idea if that helps, just having fun
you can see my steps, maybe go a different route