Math Question - Probability

[TheSaxRunner05]

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?

[Dinglesberry]

Re: Math Question - Probability

N/2

EDIT: i think it might be π

[TheSaxRunner05]

Re: Math Question - Probability

N/2

EDIT: i think it might be π

If it was N, I'd have an expected max value of 6 after 6 rolls, but no single result
can go over 5

[TheSaxRunner05]

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

[inDheart]

Re: Math Question - Probability

what is the "best" result?

[TheSaxRunner05]

Re: Math Question - Probability

what is the "best" result?

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.

[SKG_Scintill]

Re: Math Question - Probability

just tried it empirically with some quick code

$array = array();
for ($i = 0; $i < 1000; $i++) {
	$highest = 0;
	for ($j = 0; $j < 100; $j++) {
		$rand = mt_rand(1,5);
		if ($rand > $highest) {
			$highest = $rand;
		}
	}
	array_push($array, $highest);
}
$sum = 0;
foreach ($array as $num) {
	$sum += $num;
}
echo $sum / sizeof($array);

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

[TheSaxRunner05]

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?

[TheSaxRunner05]

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.

[MarcusHawkins]

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.

[TheSaxRunner05]

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.

I don't think you quite get what I'm asking, see my later posts in the thread to see what I mean.

I think I figured it out though.

[SKG_Scintill]

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

[ilikexd]

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 N-1 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.

[ilikexd]

Re: Math Question - Probability

It should be the EV of the highest number among N terms is the highest number among N-1 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.

[SKG_Scintill]

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