So here’s my homework problem: On the original WiiFit there were five activities for testing mental and physical agility, one of which I really disliked. Two of the five were chosen at random each day. On WiiFitPlus, there are two sets of five activities each, with one exercise drawn at random from the two disparate sets, each of which has a test I really dislike. Am I more likely under the WiiFit or under the WiiFitPlus routine to get a day with one of the tests I can’t stand? Here, my reasoning.
To start with, I’m making the assumption that the selection of tests is uniformly random, that no test is more or less likely to be chosen from those available, and that a particular test being picked yesterday does not make it more or less likely to be picked today. I’m also assuming that on the WiiFit, whatever is picked as the first test can not be picked as the second one. Certainly I never saw a duplicate test when I used the WiiFit and I used it long enough that it’d be most unlikely to have missed that.
In the terms used for studying probability, we’d call the “experiment” the process of getting one of the tests from those available. Each experiment has one of several possible outcomes: getting (for WiiFit) the Balance, Steadiness, Single Leg Balance, Agility, or hated Walking tests, for example. Getting any one of them, such as the Walking Test, implies also not getting any of the others; and this may seem like a trivial thing to throw out, but not getting any of the others implies getting the one left behind.
Also in the terms of probability, an “event” is some collection of possible outcomes of an experiment. For example, “getting the Walking Test on the first try” is an event. So is “getting the Walking Test or the Agility Test” on the first try. So is, for that matter, “getting any kind of test on the first try”. Events are wonderful things, extending as widely or as narrowly as we need them to. If every outcome is equally likely, and I am aware that I am begging the question by throwing in that phrase but if you don’t let me do that we’ll take forever to get back to the same spot, then the probability of an event is the number of outcomes which satisfy the event divided by the number of possible outcomes. (If you see things to nitpick about this, please take my word that you see them because I’m avoiding technical details that would answer the nitpicks. If you don’t see things to nitpick about this, you’re missing part of what’s fun about mathematical reasoning: try to find some!)
If you know the probability of two events, then you can figure out the probability of both the first and the second event being simultaneously satisfied — for example, the probability of the first test being the Walking Test and simultaneously the second test being (say) Either The Walking Or The Stillness Test. That’s the probability of the first event being satisfied multiplied by the probability that the second event is satisfied under the assumption that the first is also satisfied. (This is not a circular definition, but it’s a good exercise to think of why it’s not.)
The probability that the first event is satisfied, or the second event is satisfied, or both together are — and we call this satisfying the first or second event; mathematicians typically default to or meaning “one or the other or both at once” — is the probability of the first event being satisfied plus the probability of the second event being satisfied minus the probability that both the first and the second event are simultaneously satisfied. Why the subtraction? It’s a matter of not over-counting outcomes that satisfy the first or second or both events.
For example, going back to the default example of rolling a die, suppose the first event is “rolling an odd number”, and the second event is “rolling a prime number”. The first event is satisfied by three outcomes: rolling a one, three, or five. The second event is also satisfied by three outcomes: rolling a two, three, or five. So if we try to figure the probability of “rolling an odd number or rolling a prime number” and just add the three outcomes that satisfy “rolling an odd number” to the three that satisfy “rolling a prime number” we get six outcomes. But “rolling an odd number or rolling a prime number” is satisfied by only four outcomes: one, two, three, or five. And, sure enough, there are two outcomes that satisfy both “rolling an odd number” and “rolling a prime number” simultaneously. This doesn’t prove the rule, but probably satisfies you that the rule is right, and you can wonder later whether that actually makes it a kind of proof.
What’s the chance of getting the Walking Test on WiiFit, where it might come up as the first test or the second? As the first test, there are five equally likely tests to come up, and one of them is the hated Walking Test, so the chance it’ll be the first is one in five. Easy enough. What about as the second test? There are four possible tests to pick from — whatever was first can’t be drawn again so there are only four candidates — and therefore the chance is either one in four, if the first test wasn’t the Walking Test, or else the chance is zero in four, if the first test was the Walking Test. The chance that the Walking Test comes up as the second is therefore the one in four chance that it comes up on the second test under the assumption that it did not come up as the first test, which has a probability of four in five. This is starting to sound like a mess. It’s a resolvable one, though: if we do our arithmetic correctly we’ll get to the right answer.
We have one chance in five that the Walking Test came up as the first exercise. We had four chances in five that it did not come up as the first exercise. However, if it did not come up as the first exercise, then we had one chance in four that it came up as the second exercise. It can’t come up as both the first test and the second test on the same day, so the probability of that happening is zero. So the chance that it comes up as the first or as the second is the chance it came up as the first, plus the or chance it came up as the second. So the total chance of a lousy day is .
(With a flash of insight we could also do this another way around and avoid this considering of different cases and messy addition, trading the hard work of being clever for the hard work of arithmetic. I don’t want to go that route but those who have such an insight I invite to write about it.)
For WiiFitPlus the problem starts out looking awfully similar. There’s one chance in five that the Walking Test will come up as the first exercise. There’s one chance in five — regardless of whether the Walking Test came up — that the hated Prediction Test will come up as the second exercise. At first glance we just have to add the probability that the first test is the unliked one to the probability that the second test is the other unliked one. But it is possible to get both the Walking Test and the Prediction Test the same day, which is demoralizing, but is really just the one spoiled day. The chance of getting the Walking Test the first test and the Prediction Test the second is going to be or .
To have the right probability of getting the unwanted test on the first round, or on the second, or both, we have to add the probability of getting the unwanted the first time, , to the probability of getting the unwanted test the second time, , and subtract the probability of getting unwanted tests both the first and second time, , This works out to or or , which is less than the WiiFit chance of an unpleasant day of or .
So my conclusion is that WiiFitPlus makes things slightly better: there are going to be fewer days in which one of the hated tests comes up under its division. However, it manages this by sometimes giving days in which both hated tests come up. But whether you would regard this as an improvement depends on whether you think the day is made twice as bad by getting both unwanted tests.