Will the ratio of "No" to "Yes" be greater than the percentage of "Yes" when this question is settled?

And I thought the other question hurt my brain!

Phi on you for asking this! Well the inverse of Phi anyway.

:-)

I put equal predictions on both options because I don't want to be accused of trying to game the site 8^)

I can't bet on it! lol

Currently the rating is Yes 62, No 38, that would mean No/Yes = 38/62 = 0.613
which means 61.3 < 62 (the ratio is smaller than the value) so it would settle as "No"

Now I put my 100$H on "no", the rating is Yes 61, No 39, that would mean No/Yes = 39/61 = 0.639
which means 63.9 > 61 (the ratio is larger than the value) so it would settle as "Yes".
Hm. Damn.

chuckle

These questions are cool. I had to stop and figure out the dividing line.

A YES at or above 61.804% will settle as "NO" and a YES at or below 61.803% will settle as "YES."

My head hurts.

http://www.google.com/search?q=1%2Fphi

I wasn't ready to trust anyone, so I tried out this Java code to detect:

class calc
{public void main()
{double yes,no,finalyes=1;
double i;

for(i=1;i<100;i=i+0.01)
{yes =i;
no=(100-i);
double ratio=no/yes;
yes=yes/100;
if(ratio>yes)
finalyes=yes*100;
}

System.out.println("For Yes: Yes % must be:"+finalyes+"% or more");

}
}



It returned an output:

For Yes: Yes % must be:61.79999999999627% or more

However, I realised that would be gaming, so I modified it to this:

class calc
{public void main()
{double yes,no,finalyes=1;
double i;

for(i=1;i<100;i=i+0.01)
{yes =i;
no=(100-i);
double ratio=no/yes;
yes=yes/100;
if(ratio>yes)
finalyes=yes*100;
}

System.out.println("For No: Yes % must be:"+finalyes+"% or more");

}
}

For No: Yes % must be:61.79999999999627% or more

Which is a small error, since increments have been taken to be 0.01%, not 0.001 or less.

Hubdub will probably take 1%, which means:

For No: Yes % must be:62.0% or more

Am I someone with a lot of time to waste or not... (that's rhetorical, don't answer)

Rohan - You're assuming integers which is creating some roundoff error. Try algebra: let Yes=x, No=(1-x) then:
(1-x)/x=x
x^2+x-1=0
use quadratic formula and you find that x=-1.61803 or .618034
Since we're a positive kind of group on hubdub, you pick x=.618034 thus you're looking for yes of 61.8034% or more :)

import java.io.*;
class Quadratic /*This program solves a quadratic equation(ie. with two possible
solutions) in the form a*x^2 + b*x + c=0, where 'x' is unknown
and 'a', 'b' and 'c' are accepted from the user. It uses global
variables */

{double x1,x2,a,b,c,discriminant; //Global variables

public void main () throws IOException /*Main function, which accepts the values of
'a', 'b' and 'c', and also calls the
calculate function */

{BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("The equation should be in the form a*x^2 + b*x + c=0, where 'x' is unknown");
System.out.println("Enter 'a'");
a=Integer.parseInt(br.readLine()); //'a' is accepted
System.out.println("Enter 'b'");
b=Integer.parseInt(br.readLine()); //'b' is accepted
System.out.println("Enter 'c'");
c=Integer.parseInt(br.readLine()); //'c' is accepted
calculate(); //Calling calculate function
}

public void calculate () /*Calculate function, which calculates the value(s)
of 'x' by and prints them */
{discriminant=(b*b)-(4*a*c); //Calculates discriminant
if(discriminant<0) /*To prevent negative discriminant, as square root of a
negative number is impossible to find */
{System.out.println("Roots are negative, and thus there is no solution");}
else
{if(a!=0) //If 'a'=0, then the equation is a linear equation eg. x+4=0
{discriminant=Math.sqrt(discriminant);
x1=(-b+discriminant)/(2*a); //Calculating first value of 'x'
x2=(-b-discriminant)/(2*a); //Calculating second value of 'x'
}
else
{x1=-c/b; //Values of 'x' for a linear equation
x2=-c/b; //x2=x1, as a linear equation has only one solution
}
System.out.println("Solution: x= "+x1+" or x= "+x2);} //solution is printed
System.out.println();
}

}



Output:
The equation should be in the form a*x^2 + b*x + c=0, where 'x' is unknown

Enter 'a'

1
Enter 'b'

1
Enter 'c'

-1
Solution: x= 0.6180339887498949 or x= -1.618033988749895

Look like you're right

rohan, you must have to have too much time for something like this experimental approach. Why didn't you just visit triathematician's link? Bang, there's your answer ... and it must be right; why else should he call himself this way?

let's game!!

Rohan, if you have time to do that much programming to figure out a Hubdub question, I might be able to scare up a project or two for you ... pro bono of course 8^). How's your SQL?

Okay, my head hurts, my eyes hurt and because of that my stomach now hurts, hahahaha, so can anyone explain to me RIGHT NOW, at this very second, with it being 61% YES and 39% No ... WHAT am I supposed to bet? I am sooooo utterly lost man :-D
Hehehehehe
These questions are fantastic, I just don't know what to do, hahaha :-D

And please hurry, I have only 7hrs till the question suspends and I need to make $50,000 quick..... Big Jimbo Da Clam wants his money back!!!

lucidstates

Ananaverageamerican: I'm 16. How good do you think my SQL is?

Rohan,

I have no clue. My older son is 16 and he's learning SQL in High School, but yeah, he's not that good at it yet. Not to mention that I was just teasing ... and that's pretty impressive Java coding for a 16 year old (unless you downloaded it all from the 'net 8^). I encourage you to study SQL (and XML, which is a text file that acts like a database), these days code is rarely significant absent databases; even if you are writing things like operating systems.

rohan you let the proverbial cat out of the bag i was enjoying you're, and satyaki's, input without anyone knowing how old you were

ananaverageamerican: I don't know any SQL. And I wrote all that Java myself. You think that's good, look at the program I put on satyaki's profile.

meanderingsearcher:LOL