Math Question - Probability

[TheSaxRunner05]

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.

May I ask where the 3.2 constant comes from? I'm a little confused where that number comes from.

[ilikexd]

Re: Math Question - Probability

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

that's a shame, i had bruteforced 3, 3.8, and 4.2 as the answers for n=1,2,3 with very long arithmetic and assumed a relation there

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?

[TheSaxRunner05]

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.

[SKG_Scintill]

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

<?php

for ($m = 1; $m <= 10; $m++) {
	$res = 0;
	for ($n = 1; $n <= 5; $n++) {
		$part = $n * (pow($n,$m) - pow($n-1,$m));
		$res += $part;
	}
	$avg = $res / pow(5, $m);
	echo $avg . '<br/>';
}

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 =

[SKG_Scintill]

Re: Math Question - Probability

heh in my formula's case 0^0 = 1

[TheSaxRunner05]

Re: Math Question - Probability

heh in my formula's case 0^0 = 1

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

[SKG_Scintill]

Re: Math Question - Probability

that doesn't work

using 1 to 5, doing it twice before taking highest

[num1, num2]: highestnum
1,1: 1	2,1: 2	3,1: 3	4,1: 4	5,1: 5
1,2: 2	2,2: 2	3,2: 3	4,2: 4	5,2: 5
1,3: 3	2,3: 3	3,3: 3	4,3: 4	5,3: 5
1,4: 4	2,4: 4	3,4: 4	4,4: 4	5,4: 5
1,5: 5	2,5: 5	3,5: 5	4,5: 5	5,5: 5

[amount of combinations]
5^2 = 25

[number's amount of occurrences]
1: 1
2: 3
3: 5
4: 7
5: 9

[weight of numbers]
1: 1 * 1 = 1
2: 3 * 2 = 6
3: 5 * 3 = 15
4: 7 * 4 = 28
5: 9 * 5 = 45

[sum of weights]
1 + 6 + 15 + 28 + 45 = 95

[average] = [sum of weights] / [amount of combinations]

average: 95 / 25 = 3.8

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

[TheSaxRunner05]

Re: Math Question - Probability

that doesn't work

using 1 to 5, doing it twice before taking highest

[num1, num2]: highestnum
1,1: 1	2,1: 2	3,1: 3	4,1: 4	5,1: 5
1,2: 2	2,2: 2	3,2: 3	4,2: 4	5,2: 5
1,3: 3	2,3: 3	3,3: 3	4,3: 4	5,3: 5
1,4: 4	2,4: 4	3,4: 4	4,4: 4	5,4: 5
1,5: 5	2,5: 5	3,5: 5	4,5: 5	5,5: 5

[amount of combinations]
5^2 = 25

[number's amount of occurrences]
1: 1
2: 3
3: 5
4: 7
5: 9

[weight of numbers]
1: 1 * 1 = 1
2: 3 * 2 = 6
3: 5 * 3 = 15
4: 7 * 4 = 28
5: 9 * 5 = 45

[sum of weights]
1 + 6 + 15 + 28 + 45 = 95

[average] = [sum of weights] / [amount of combinations]

average: 95 / 25 = 3.8

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

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.

[benguino]

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

[MarcusHawkins]

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

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

[TheSaxRunner05]

Re: Math Question - Probability

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

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?

that doesn't work

using 1 to 5, doing it twice before taking highest

[num1, num2]: highestnum
1,1: 1	2,1: 2	3,1: 3	4,1: 4	5,1: 5
1,2: 2	2,2: 2	3,2: 3	4,2: 4	5,2: 5
1,3: 3	2,3: 3	3,3: 3	4,3: 4	5,3: 5
1,4: 4	2,4: 4	3,4: 4	4,4: 4	5,4: 5
1,5: 5	2,5: 5	3,5: 5	4,5: 5	5,5: 5

[amount of combinations]
5^2 = 25

[number's amount of occurrences]
1: 1
2: 3
3: 5
4: 7
5: 9

[weight of numbers]
1: 1 * 1 = 1
2: 3 * 2 = 6
3: 5 * 3 = 15
4: 7 * 4 = 28
5: 9 * 5 = 45

[sum of weights]
1 + 6 + 15 + 28 + 45 = 95

[average] = [sum of weights] / [amount of combinations]

average: 95 / 25 = 3.8

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

[benguino]

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?

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.

[benguino]

Re: Math Question - Probability

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

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.

[TheSaxRunner05]

Re: Math Question - Probability

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

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.

[TheSaxRunner05]

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.

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