What Are the Chances?

If making decisions is part of your job, you know that some decisions are far from straightforward. Sometimes we make decisions based on partial information or guesses. For example, when we must choose between A and B, we make estimates of the probabilities of A's success and B's success. Then we pick the one we think best. Sometimes we do this without realizing that this is the process we're using. That's where it gets interesting, because most of us, as humans, are really bad at making decisions that involve probabilities.

This post People make a common mistake
when forming estimates of
the probability of multiple
events occurringprovides an example of a simple decision problem involving probabilities. It's a problem that causes many of us to take a wrong turn. I begin by posing the problem in semi-realistic terms fairly close to something that actually could come up at work. After posing the problem, I take a short excursion into a little refresher of some basic probability methods. Then I'll return to the semi-realistic problem to apply those methods. Should be fun.

A semi-realistic problem

Suppose we're making a risk plan. The risk we're planning for is the risk that members of our team will be stricken with COVID during the next month. For this scenario, an important fact is that our team members are all classified as essential, and we all work in the same suite of offices, using the same break room. That's why we're concerned about COVID.

Our plan is to find a way to meet our deadlines as best we can, even if some of our team fall ill. So we do a bit of cross training and we prepare for several scenarios. Here's our plan:

• If A falls ill, B and C will take over A's responsibilities
• If B falls ill, C and A will take over B's responsibilities
• If C falls ill, A and B will take over C's responsibilities

We estimate that the probability of anybody falling ill during the next month is 10%. So the question is: How well will our plan work?

One thing is clear: if all of A, B, and C fall ill, our plan won't work at all. So our task is to work out the probability that all three of A, B, and C will fall ill.

A little refresher about probabilities

Our semi-realistic problem has three events: A falls ill, B falls ill, and C falls ill. That's a little complicated, so let's take a look at a two-event problem. One event is this:

Event X: When the Sun sets tonight, at the moment of sunset the sun will be obscured by clouds.

The probability: 1/5

And the second event is this:

Event Y: Right after sunset, I roll a die, and it lands so that either one or two spots are showing.

The probability: 1/3

So the question is this: what is the probability that both Events X and Y occur? Call that XY.

Solution: P(X) is 1/5. P(Y) is 1/3. So P(XY) = P(X) times P(Y) = 1/15.

Explanation: X occurs one-fifth of the time. For the other times, it doesn't matter what I get when I roll the die. Only when X occurs does the die matter. So right from the beginning, we're down to P(X)=1/5 of the time as an upper limit for P(XY). But the probability that Y occurs is only 1/3. So P(XY) = 1/3 of 1/5 of the time. That is P(XY) = P(X)*P(Y) = 1/15.

So here's what we learn from this: If Event X and Event Y are independent of each other, P(XY) = P(X) * P(Y). This isn't a proof. It's a plausibility argument. I'll leave the proof to the mathematicians.

It's also plausible to suggest that the number of independent events doesn't matter. If I add a third event Z:

Event Z: I flip a coin and it lands heads side up

The probability: 1/2

Then the probability of XYZ = P(XYZ) = P(X)*P(Y)*P(Z) = 1/15 * 1/2 = 1/30.

The generalization of what we've learned is this:

The probability that every one of a set of independent events occurs is the product of the probabilities of the members of the set of events.

Now we're ready to return to our semi-realistic problem.

Solving our semi-realistic problem

To compute the probability that our risk plan will fail totally, many would argue as follows. If P(A), P(B), and P(C) are respectively the probabilities for A, B, and C falling ill, then the probability of all three falling ill would be P(A,B,C) = P(A)*P(B)*P(C) = 0.1*0.1*0.1 = 0.001. So by this calculation, our plan will be workable 999 times out of 1000.

Unfortunately, in the real world, that would be wrong. By a lot.

What's wrong with our reasoning is this. To find the probabilities of all of A, B, and C falling ill, we multiplied their respective probabilities together. The theorem we're applying (actually misapplying) is, as we saw above:

The probability that every one of a set of independent events occurs is the product of the probabilities of the members of the set.

But in this case, the probabilities of the three events aren't independent.

That is, the probability that B falls ill, given that A is ill, is no longer just 10%, because all three work in the same office. One can infect the others. We might not know what the probability actually is, but it's more than 10%. The same is true for C. The probability that C falls ill, given that A is ill, is greater than 10%. Computing the actual result is difficult, but we can tell right away that the probability of all three falling ill is much larger than 1/1000.

Last words

If you surf the Web much, you'll find numerous examples of misapplication of our probability rule for independent events. The rule works well for independent events. But to apply it to multiple events needs justification. Verify that the events are independent before using the method.  Top    Next Issue

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