One of the examples of conditional probability that’s used in the textbook for the stats class I’m teaching seems wrong to me. It involves sports (which I have 0 interest in), but I don’t think that’s where my confusion is coming from. Here’s the situation, paraphrased from the textbook:
Suppose Roger Federer makes 56% of his first serves, faulting on the first serve 44% of the time. When he faults on his first serve, he faults on the second serve 2% of the time. Converting these percentages as probabilities, what is the probability of a double fault? Let F1 denote the event “faulting on the first serve,” and F2 denote the event “faulting on the second serve.”
The textbook solution is as follows:
P(F2|F1) = P(F2 and F1)/P(F1) = 0.02, so
P(F2 and F1) = 0.02 * 0.44 = 0.0088.
Why isn’t the probability 0.02, since a double fault, by definition, can only happen if you fault the first time? That is, F1 must occur for there to even be a double-fault situation. So the probability of double faulting should be taken as the probability of faulting the second time, given that you faulted the first time (if you don’t fault the first time, you don’t double fault).
0.02 is indeed the probability of a double fault given a first fault. But the question isn’t asking for that conditional probability; the question is asking for the overall probability of a double fault. And thus, the question has you calculate P(F2 and F1) = P(F2) = 0.0088, rather than P(F2 | F1) = 0.02.
In other words, the 0.02 they give you isn’t P(F2); it’s just P(F2 | F1). The problem is asking you to recover P(F2) from this and P(F1).
Think of it this way:
Suppose he gets up for a first serve 100 times.
He makes 56 of those, and misses 44.
For those 44 misses, he gets a second serve. He misses that second serve 2% of the time, or 0.88 times. Thus he double faults 0.88 times out of 100, which give the probability of double faulting equal to 0.0088.
Just think, if he serves 100 points, we know that 56 times, the ball goes into play. The other 44 times, he has to serve again. 2% of those 44 serves will be faults, resulting in a double fault. 2% off 44 is 0.88.
Therefore, the likelihood of him double faulting is 0.88 out of 100, or 0.0088, just like your equation shows.
On further inspection, it seems that maybe you’re asking why we would want to calculate the probability of double faulting before he gets his first fault, to which the only answer I can think of is “Why the heck not?”. It is calculable, and of interest (to some).
Maybe that’s it. I can’t otherwise explain my thickheadedness about this problem (other than I’m just flat stupid, but the Monty Hall problem makes sense to me, so who knows). I think Indistinguishable’s explanation made it more clear to me. P(F2|F1) is the probability of double faulting, given he’s faulted the first time. What the textbook is after is the a priori (I’m using this term loosely) probability of faulting, the probability of double faulting before he’s faulted the first time.
When a player is serving in tennis he has two chances to get the ball in the opposite court. If he blows both chances, he loses the point. So usually a player will hit his first serve really hard. That makes it difficult to return, but also makes it more likely he’ll hit the net or hit it out of bounds. If he faults on the first serve, the second serve will be much more conservative. It will be easier for his opponent to return, but much less likely to cause a fault.
This is where you get the two different probabilities.
Federer faults on 44% of his first serves, but he only faults on 2% of his second serves.
You can also rephrase the question to be: “When Federer is serving, how likely is it that he will lose the point on the serve? In other words, how likely is it that he will lose the point without his opponent even taking a swing at the ball?”
That does make it seem a little easier to understand. I don’t watch sports, tennis included, and assumed that the server got as many tries as he wanted, but couldn’t score unless the serve was successful (hence the motive to try to serve well).
Another way to look at it is to take sports completely out of it and rework the problem.
A fair coin is flipped. If it lands on heads, you flip it again. What are the odds of flipping two heads in a row?
You probably know the answer to this automatically, right? 25%.
But it’s because of the conditional probability of 50% x 50%, that you arrive at that answer. You don’t automatically skip the first head throw in starting the calculation. The probability of that first toss is half of what you’re trying to calculate in the first place.
Same with the double fault in tennis. The second fault cannot happen until the first fault does. So it must be factored into any calculation you make.
The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A), read as the probability of B given A. The formula for conditional probability is:
The Venn Diagram below illustrates P(A), P(B), and P(A and B). What two sections would have to be divided to find P(B|A)? Answer