This is actually a really interesting problem because depending on how you set it up, you'll get very different answers. To illustrate, notice the difference be

This is actually a really interesting problem because depending on how you set it up, you'll get very different answers. To illustrate, notice the difference between these questions: 1) Suppose I flip a coin 3 times and repeat that experiment on 10 different occasions. What's the probability that I get all heads at least once? 2) Suppose I flip a coin 30 times (same total number of flips). What's the probability that I get at least 1 string of at least 3 heads? (1) is more like the binomial calculation you've set up, but I think (2) is closer to the question we want to be asking. This is important because, as it turns out, (2) is much more likely than (1). You can tell intuitively that there are more sequences of coin flips that would satisfy (2). For example "THH HTT ..." would satisfy (2) but not (1). So how should we pose the hurricane question? I propose: What's the probability that in a 100 year period, across 100 cities, there will be at least 1 city that experiences a 500-year flood for at least 3 contiguous years? Solving this analytically is non-trivial, but it's easy to simulate: // Scala val rand = scala.util.Random def coinFlip(p: Double): Boolean = rand.nextDouble <= p // Simulates 100 years and returns true if there's a contiguous string // of three events with p=1/500. val targetEvent = Seq(true,true,true) def threeIn100Years = Seq .fill(100)(coinFlip(1.0/500)) .containsSlice(targetEvent) // Simulate this 100,000,000 times val simulations = 100000000 val count = Stream .fill(simulations)(threeIn100Years) .foldLeft(0){case (acc, cur) => if(cur){acc + 1} else {acc}} // Estimated probability for a single city. val p = count/simulations.toDouble // The probability it will happen *at least* once in a population of // 100 cities. // This calculates the probability if *won't* happen 100 times, and // then takes the complement, giving us the final probability 1 - Math.pow(1 - p, 100) The final answer I get is 9.5E-5, so still very unlikely, but 100x more likely than the binomial calculation.