Convert Percent to Odds Ratio and Reverse - formula

Question

Easy softball for someone to knock out of the park: what is the correct coding formula for computing an odds ratio to a percent, and a percent to an odds ratio? Example:
Odds to Percent
1 in 8192 = 0.0001220
Percent to Odds
0.0001220 = 1 in 8192
Programming language is irrelevant.

Answer 1

odds_to_percent(a, b):
return str(float(a)/b)
percent_to_odds(x):
return '1 in ' + str(1.0/x)
Not sure what exactly you need. There are many decisions about rounding and co.

Answer 2

Here's what I came up with:
public struct Odds
{
private int _a;
private int _b;
public int A
{
get
{
return _a;
}
}
public int B
{
get
{
return _b;
}
}
public Odds(int a, int b)
{
this._a = a;
this._b = b;
}
public Odds(double percent)
{
Odds odds = FromPercent(percent);
this._a = odds.A;
this._b = odds.B;
}
public Odds Invert()
{
return new Odds(_b, _a);
}
public double ToPercent()
{
return ToPercent(_a, _b);
}
public static double ToPercent(int a, int b)
{
return ((double)a / (b + a));
}
public static Odds FromPercent(double percent)
{
int multiplier = GetDecimalMultiplier(percent);
return new Odds(1, (multiplier - (int)(percent * multiplier)) / (int)(percent * multiplier));
}
private static int GetDecimalMultiplier(double n)
{
if (n > 0.01)
{
return 100;
}
if (n > 0.001)
{
return 100000;
}
if (n > 0.0001)
{
return 100000000;
}
throw new Exception("Number too small!");
}
public override string ToString()
{
return A + "-" + B;
}
}
Odds to Percent:
public static double ToPercent(int a, int b)
{
return ((double)a / (b + a));
}
Percent to Odds(May not be perfect)
public static Odds FromPercent(double percent)
{
int multiplier = GetDecimalMultiplier(percent);
return new Odds(1, (multiplier - (int)(percent * multiplier)) / (int)(percent * multiplier));
}
private static int GetDecimalMultiplier(double n)
{
if (n > 0.01)
{
return 100;
}
if (n > 0.001)
{
return 100000;
}
if (n > 0.0001)
{
return 100000000;
}
throw new Exception("Number too small!");
}