07262017, 02:38 PM  #1 
The Doctor
Join Date: Apr 2006
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Math Question  Probability
Let's say you had a number line with all real numbers between 1 and 5. If you were to randomly select a bunch of numbers within this range, the average result would become closer to three the larger the sample size became. So the expected value of this scenario is 3.
My question though, is what if you take N random numbers generated and keep the best result? What is the relationship between N and the expected maximum result within the list of outputs? 
07262017, 02:43 PM  #2 
YOOOOoooo
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Re: Math Question  Probability
N/2
EDIT: i think it might be π Last edited by Dinglesberry; 07262017 at 02:43 PM.. 
07262017, 02:48 PM  #3 
The Doctor
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Re: Math Question  Probability
If it was N, I'd have an expected max value of 6 after 6 rolls, but no single result
can go over 5 
07262017, 02:53 PM  #4 
The Doctor
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Re: Math Question  Probability
I'm not quite sure how to get there, but my intuition is that if you graphed the answer, it would have a horizontal asymptote at y=5

07262017, 03:06 PM  #5 
gradually receded
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Re: Math Question  Probability
what is the "best" result?

07262017, 03:11 PM  #6 
The Doctor
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Re: Math Question  Probability
Sorry that isn't very mathy lol
Later I spell it out a bit better, saying the expected maximum value, I mean the expected maximum number from the pool of N random real numbers between 1 and 5. So if you picked 100 random real numbers between 1 and 5, how close to 5 is the expected maximum value? How do I find an equation for the maximum expected value for N random numbers. 
07262017, 03:23 PM  #7  
Spun a twirly fruitcake,
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Re: Math Question  Probability
just tried it empirically with some quick code
the $j < [num] is the amount of numbers I take before I take the best result here's the result for different amounts 1: 3.089 2: 3.811 3: 4.162 4: 4.436 5: 4.577 10: 4.892 100: 5 so yes, it goes to 5 doesn't give a mathematical relationship but whatever I was bored
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Last edited by SKG_Scintill; 07262017 at 03:25 PM.. 

07262017, 03:34 PM  #8 
The Doctor
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Re: Math Question  Probability
I found something close 
Y = nth root(.5) * (Maximum value) Does this seem correct? It has a max value of 5, but N=2 gives 3.53 Edit:// If you pick two numbers, you're asking, at what value are the product of these two probabilities 50%? So having two probabilities of sqrt .5 (.7071...) multiplied together gives probability .5. The same would go for 3, 4, 5, or N numbers. Is this reasoning sound? Last edited by TheSaxRunner05; 07262017 at 03:53 PM.. Reason: Fixed formula 
07262017, 03:55 PM  #9 
The Doctor
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Re: Math Question  Probability
I edited my old post to hell, so I'm making a new post. My previous error was including the minimum and maximum value in the formula.

07262017, 04:40 PM  #10 
Constantly Improving!

Re: Math Question  Probability
Simple.
(Max Value + Min Value) / 2 Of course, you can also create (Min Value + Max Value) / 2 and receive the same result. EDIT: I think I misread your question, and I apologize. HOWEVER, this is good for finding the average of all possible numbers from your lowest to highest amount, whatever that may be.
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Last edited by MarcusHawkins; 07262017 at 04:51 PM.. 
07262017, 04:48 PM  #11  
The Doctor
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Re: Math Question  Probability
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I think I figured it out though. 

07262017, 05:04 PM  #12  
Spun a twirly fruitcake,
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Re: Math Question  Probability
I'm breaking my head lmao
so I may be going completely in the wrong direction this *should* be the average of your situation you can see the way it expands for every extra random number you take before you take the highest one I can't get that expansion written as a product summation but fk man I only did high school math fun tho would like to see someone else just go, you know, the easy route
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Last edited by SKG_Scintill; 07262017 at 05:06 PM.. 

07262017, 05:16 PM  #13 

Re: Math Question  Probability
The EV of the highest number of N terms between 1 and 5 is the EV of the highest number of N1 terms + 3/(2^N)
So when you start with 1 term between 1 and 5, you start with 3/(2^0)=3 3+3/(2^1)=4.5 4.5+3/(2^2)=5.25 etc. Oh never mind, that's just wrong. I need to adjust the numbers. e: This ends up being the answer for numbers between 0 and 6 though. Last edited by ilikexd; 07262017 at 05:30 PM.. 
07262017, 05:24 PM  #14 

Re: Math Question  Probability
It should be the EV of the highest number among N terms is the highest number among N1 terms + 3.2/(2^N). It doesn't apply to the first term though, which is just 3.
3+3.2/(2^2) 3+3.2/4=3.8 3.8+3.2/(2^4) 3.8+3.2/8=4.2 etc. It should asymptotically approach 5 too. 
07262017, 05:54 PM  #15  
Spun a twirly fruitcake,
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Re: Math Question  Probability
got it down to this
k is the highest number you go to (from 1 to "5") m is the amount of times you take a random number before taking the highest number ...now I gotta think... edit: darn nathan is that it? :P edit v2: mine works for the first term edit v3: looking at yours nathan, I think that's one of the first things that got me troubled, because that approach stops working from 4 and onward I believe
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Last edited by SKG_Scintill; 07262017 at 05:58 PM.. 

07262017, 05:59 PM  #16  
The Doctor
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Re: Math Question  Probability
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07262017, 06:05 PM  #17  

Re: Math Question  Probability
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why don't you just run your program a bunch of times and take the average results so an equation can be fitted to them? 

07262017, 06:13 PM  #18 
The Doctor
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Re: Math Question  Probability
Here's my line of thinking: The probability of a single random number being above or below 3 is 50%. There is 50% of the range on the left and 50% of the range to the right.
It gets more complicated when you add a second probability. I'm looking for the product of two numbers that equals .5, so I take the sqr root of .5  or .707 This number means the highest of two individual random numbers has a 50% chance of being in the top 29.3% percentile of all possible random numbers. If I take the .707 and multiply by 5, the max value, I get ~3.535 for an expected value. For three numbers, I'm asking what three probabilities multiplied together reach 50% probability. The cubic root of .5 is .793, multiplied by 5 is ~3.968. If this line if reasoning is incorrect, it would be helpful to me to know where the logical flaw lies, so I better understand it. 
07262017, 06:47 PM  #19  
Spun a twirly fruitcake,
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Re: Math Question  Probability
@Nathan: I made a code that does my formula to give you the average number instead
using 1 to 5, up to 10 times before taking the highest number here are the results: 3 3.8 4.2 4.4336 4.584 4.68704 4.76064 4.81477376 4.85544192 4.88647424 edit: tried to use wolframalpha to see what it could do with my formula, but it's giving me some harmonic number series, which I know nothing of here's the formula I know nothing of k, n and m being variables, H being... something mathematical edit v2: for now I'll keep it at this formula, reposting it so I can at least sleep k is highest number (1 to "5") n is current number in summation m is amount of times before taking highest number average =
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07262017, 09:01 PM  #20  
Spun a twirly fruitcake,
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Re: Math Question  Probability
heh in my formula's case 0^0 = 1
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