Re: Math Question - Probability
 

Makes sense heh

I think I found the problem with my formula -
My Nth root of .5 * 5 formula only works for picking values 0 - 5, I don't know how to factor in starting at 1. What it's multiplied by determines the asymptote and I can't Add a constant, or the result will go over 5.

Edit:// Ok I was close

I think it's (xth-root of .5)*4 +1

Re: Math Question - Probability
 

that doesn't work

using 1 to 5, doing it twice before taking highest

average: 95 / 25 = 3.8

using your formula:
0.5^(1/2)*4+1 = 3.8284~~~

Re: Math Question - Probability
 

I see, and since all answers will be rational numbers, using the square root function will fail. I can match one really close to the actual probabilities, but really close is like horseshoes and hand grenades in math lol

Thanks for going back and forth on this, I'm going to toy with the equation you posted a bit more.

Re: Math Question - Probability
 

I came up with this: http://puu.sh/wU5Jx/8ad6c883ab.pdf

Got the equation (5n+1)/(n+1)

If there's any errors or questions about my work, let me know. Or if someone finds a more elegant solution, I'd also be up for seeing it.

EDIT: Here's a graph

Re: Math Question - Probability
 

Oh, so the formula is essentially (Max value*n + Min value)/(n + Min value), if I'm understanding this right?

Re: Math Question - Probability
 

Plugging 2 into this equation gives 11/3 or 3.666....

Is there an error in what Scintill showed earlier to show an answer of exactly 3.8?

Re: Math Question - Probability
 

It looks like that is the case if you're looking at the discrete probability (the only options are 1, 2, 3, 4, 5) as opposed to the continuous probability (you can pick any real number between 1 and 5). So I'd expect it to be close but not exactly the same.

Re: Math Question - Probability
 

It seems that might be the case...wolframalpha won't do the integral in the general case for me...so I'll have to compute it by hand. I can double check.

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.

Re: Math Question - Probability
 

Gotcha, thanks for the explanation.
I really need to take a calculus course sometime to better understand integrals

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!

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?

Re: Math Question - Probability
 

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:

Re: Math Question - Probability
 

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

Re: Math Question - Probability
 

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 :)

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