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POAlgorithm for probability over 100%
 

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    <p>I am in need of an algorithm (in Java, but the theory should be pretty general) for some sort of probability... thing. I don't even know what to call it, which is why I haven't had any luck Googling.</p> <p>To describe a bit better, I have a task that needs to be done X number of times when a function is called. Sometimes, there's only a 1 in 10 chance of that happening, meaning that (approximately) one out of every ten times that function is called that task will actually be performed - easy enough to do with random(). Sometimes it will be 2 in 10, or maybe even 10 in 10. Still easy enough, straightforward conditional, even if it's not always "X in 10"</p> <p>The problem is, it can have <em>greater</em> than a 1 in 1 chance. It may be 15 in 10 - half the time it runs once, the rest it runs <em>twice</em>. Or 10 in 5, where it runs twice (approximately) every time called. As you can see, this has now ventured out of simple inequality testing.</p> <p>So, what I'm looking for is some algorithm that, based on random numbers, will return a run <em>count</em> based on how frequently (1 in 10, 200%, 0.75, whatever format works) something ought to happen. If anyone can provide any leads on this - like, maybe an actual term to search for - it'd be much appreciated!</p> <p>EDIT: No wonder I can't find many results, everyone's as confused as I am!</p> <p>First and foremost, no real "maximum" value. Not formally defined. If the algorithm can extract that from the probability value it receives, that's great.</p> <p>It also definitely needs to be random, making it inherently imperfect. If you flip a coin 10 times, you may very well end up with 8 heads, even though it theoretically ought to be perfectly even! And that's alright. In fact, it's the point.</p> <p>I could tell you <em>why</em> I need it... but that would be in violation of proper object-oriented modularity practices :) External parties need only know the interface; it accepts a value (probably a float, "0.75" seems like it would work best) and returns an int. If you call it 100 times, with a probability of 1 in 5, the sum of its returned values should average out to 20.</p>
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    1. COOne of us doesn't get probabilities.
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      1. USDenys Séguret
    2. COFirst, answer this: Why?
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    3. CO`while (likelihood > 10) { likelihood -= 10; routine.run(); }` then do your regular equality test.
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