Generate Random Boolean Probability
Solution 1
Well, one way is Random.Next(100) <= 20 ? true : false
, using the integer value of NextInt to force your own probability. I can't speak to the true 'randomness' of this method though.
More detailed example:
Random gen = new Random();
int prob = gen.Next(100);
return prob <= 20;
Solution 2
You generate a random number up to 100 exclusive and see if it's less than a given percent. Example:
if(random.Next(100) < 40) {
// will be true 40% of the time
}
More generally, for a probability of X/Y, use an idiom like:
if(random.Next(Y) < X)
Solution 3
Here is an extension method that will provide a random bool with specified probability (in percentage) of being true;
public static bool NextBool(this Random r, int truePercentage = 50)
{
return r.NextDouble() < truePercentage / 100.0;
}
you can use this like
Random r = new Random();
r.NextBool(); // returns true or false with equal probability
r.NextBool(20); // 20% chance to be true;
r.NextBool(100); // always return true
r.NextBool(0); // always return false
Solution 4
Assuming your probability is represented as double
between 0.0 and 1.0, I would implement it more simply like this:
Random rand = new Random();
...
double trueProbability = 0.2;
bool result = rand.NextDouble() < trueProbability;
result
will be true
with the probability given by trueProbability
http://msdn.microsoft.com/en-us/library/system.random.nextdouble(v=vs.110).aspx
If this isn't "random enough", you can take a look at RNGCryptoServiceProvider
:
Solution 5
I think it can help you
Random gen = new Random();
bool result = gen.Next(100) < 50 ? true : false;
Dark Side
Updated on July 09, 2022Comments
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Dark Side almost 2 years
I only know how I can generate a random boolean value (true/false). The default probability is 50:50
But how can I generate a true false value with my own probability? Let's say it returns true with a probability of 40:60 or 20:80 etc...