Any time there is even a ghost of a chance of snow, my three kids get very excited (ok, let’s face it, being a southern transplant to NY, so do I). We go to https://www.snowdaycalculator.com in the hopes it will gives us some idea of how likely it is that school will be canceled. Here in NYC, my wife tells me it used to be rare to never, but, of late, it seems the threshold is about 8 inches of snow. Less than that and everyone wakes up early and trudges through. More than that and we sleep late and go sledding.
This last winter we had one snow day early on and then nothing but beautiful weather. My kids informed me at the time that they really didn’t want to have more than 3 snow days, because after that, they need to make it up. I have no idea if this is true, but it is the set up to a nice probability problem.
It goes like this: if the chance any random winter day being a snow day is p, what is the chance that you have exactly the sweet spot of 3 snow days in a season where there are n school days? For the example below, we’ll assume the chance of a snow day is 1 in 20, and that there are 60 school days in winter. If we make a couple of assumptions, the calculation of whether there will be exactly 3 days becomes a standard probability question that is solvable using a statistical distribution called the Binomial Distribution.
But first things first. Let’s assume (almost certainly wrongly) that whether or not a particular day is a snow day is completely independent of whether there were snow days on any prior days. Also, lets assume that the chances of 1 in 20 (5%) apply equally for the first day of winter as they do for the middle of winter or the end of winter. So, in other words, the chance of 1 in 20 stays the same throughout the winter and doesnt vary based on the kind of winter we have had and doesnt even vary based on weather we had 2 feet of snow and canceled school the day before.
The first important concept is that, given these assumptions of independence and equal probability, the chance of having, say, two snow days in a row is just the chance of 1 snow day squared. In other words, the chance that we have a snow day the first 2 days of winter is (.05)*(.05)=.0025, or 1 in 400. So, the chances of the first two days of winter off is 1 in 400. Extending that to the required 3 days, we have .05*.05*.05*, which is 1 in 8,000, an exceedingly small number, implying that only once in 8,000 years would we be lucky enough to have the first three days as snow days.
But that would not be good unless there were no snow days the rest of the season. Therefore, we need to account for the chances that the other 57 days are *not* snow days. That number is .05^3 (as before) multiplied by the chance of no snow day (.95) to the power of the number of days that occurs, which is 57. Therefore, the chance that we have 3 snow days followed by 57 non-snow days is (.05^3)*(.95^57)= about 1 in 150,000. It’s looking really grim right now. On average, it will take 150,000 winters before we have one where the first three days are snow days and we have no more the rest of the season.
But luckily, we are trying to figure out something more likely, and that is the chance it will be a snow day on *any* of 3 of the 60 days of winter. We don’t care at all if it is the first three, the last three, or any three in the middle, and they dont need to be consecutive. It turns out that the chance we calculated, of 1 in 150,000 is the correct chance of a snow day on a particular set of 3 days no matter which particular set we choose. For example, the chance of a snow day on January 1, February 1, and March 1, but not at all the rest of the winter is also 150,000, as is the chance of a snow day on Jan 31, Feb 28, and March 31 (this is true because we assume that the chance of a snow day is the same and snow days are independent of one another).
This realization allows us to solve the original question of the chance of exactly three snow days simply by adding up all the different ways we can choose 3 days out of 60 and multiplying it by the chance of 1 in 150,000. Luckily for us, it turns out there are quite a few ways to choose 3 days out of 60 — about 34,000 ways. This is computed by figuring out that there are 60 ways to choose the first snow day, 59 ways to choose the second and 58 ways to choose the third (60*59*58) and dividing this by 6 to account for the fact that the selection doesn’t need to distinguish between the first second and third days-they’re all just snow days and can be ordered 123 132 213 231 321 or 312 (i.e., in 6 ways). Net-net, the chance of exactly three days being a snow day is about 34000/150000, or a little more than 1 in 5.
If you want to learn more about the general formulation of this type of problem, lookup up the Binomial Distribution (Wikipedia can be a good start: https://en.wikipedia.org/wiki/Binomial_distribution ).