Advanced independence (cloned)

Advanced Independence

You may recall that certain rules of probability only hold for independent events, and others only hold for dependent events. You may also recall that I said: if the situation doesn’t let you use a rule, you should change the way you are looking at the situation so that the rules apply again.

That all works, as we saw in the example in the previous section. But there is also another option, if you’re willing to “break the rules” a bit.

Breaking the rules

Here’s the deal: breaking the rules is bad because when you break the rules, you wind up overcounting or undercounting. But in some cases, the easiest thing to do is break the rules, and just keep track of the overcounting or undercounting, then correct for it at the end. (This is a pretty good example of enlightened brute force, by the way.)

An example

For example, let’s revisit (yet again) the sample problem in which there’s a chance of rain on Monday and a chance or rain on Tuesday, and you want to know the chance that it will rain on either or both days. This time, let’s change the chance of rain to 1/3 on either day (because it makes some of the calculations easier).

We’ve already solved this problem by carefully reframing it in terms of independent and dependent events, of course. But our new way of solving it works well too. It goes like this:

There is a 1/3 chance of rain on Monday, and a 1/3 chance of rain on Tuesday. If we want the total chance that either or both happens, we can start by adding 1/3 + 1/3. That’s because we want the chance of rain on Monday or Tuesday, and or means add.

Now, of course, we are breaking a rule here, but here’s how we deal with it. We notice that our method runs into trouble when it rains on both days. The issue is that we counted rain on Monday as part of the first 1/3, and we counted rain on Tuesday as part of the second 1/3 chance. If it rains on both days, then we’ve double-counted.

If that seems confusing, try thinking of it this way. We want it to rain on either day or both. When it rained on Monday, we got what we wanted. When it rained on Tuesday, we got what we wanted again. We got what we wanted twice. But we were only trying to count one win, not two.

So to get the right answer, we start with the 2/3 we calculated earlier, then we notice that the chance of rain on both days (1/9) got counted twice. So we have to subtract one of those 1/9th’s from our total. We get 2/3 - 1/9 = 6/9 - 1/9 = 5/9, which is the same answer we got earlier by being much more careful with the rules.

What to do

Like I mentioned earlier, there aren’t any quizzes for this unit. Instead, the takeaway here is that in complicated probability problems (which is, let’s face it, most of them), it’s usually pretty easy to come of with two different ways of solving the problem: one in which you work out the cases carefully then grind out the arithmetic, and another in which you attack the problem a little sloppily at first, then go back and correct for overcounting and undercounting.

Crucially, neither of these methods is always better than the other. It depends on the problem (and perhaps also on your comfort with the various parts of each process).

You might also find it helpful to look at the first few paragraphs of the Wikipedia entry on the Inclusion-Exclusion Principle.