by Jonathan ‘AngleWyrm’ Wooldridge
Here we are in the present, then something happens and we are in a new present. We can remember a history of the old versions, but we don’t have a record of what’s yet to come. Randomness is a word I use to describe that unknown frontier of the future, and probability is for me the study of the future.
I have here one fair green die. What do I know about its future? There are six possible outcomes.
Which is all well and good, but I’m interested in how it relates to me and the experiments I’m conducting in my secret laboratory. If I’m going to roll that die, then I want something to hope for. Like for example winning.
So let’s define a new state of affairs, a categorical judgement of the outcome as either that hottie Success, or her goth sibling Failure, who turns out to be the far more interesting of the two.
With the notion of grouping outcomes into categories of success or failure comes the idea of chances. How many of the possible outcomes belong to the set Success, and how many to Failure?
Let’s score any result of 3 or more as a success.
$$ P(success) = \frac{4}{6} =\frac{2}{3} \text{ of the 6 outcomes; } P(failure)=\frac{2}{6}=\frac{1}{3} \text{ of the 6 outcomes} $$
Thus I can measure how the judgement is likely to play out, and use that vision of the future to inform my choices in the present. Time travel at it’s best.
I’ve noticed whenever there’s a die beckoning to be tossed, I usually do that more than once. Sooner or later Success pops in and gives me a smooch, and Failure has something to say about it.
<aside> 💡 As you add more tries to an experiment, it becomes less likely they all turn out failures.
</aside>
Whereas Success replies with only a vague and non-committal “it’s complicated.”
So at the coffee shop on a date with Failure, I have a question: How much less likely? Failure sips on her Not from a Jedi coffee mug. “Let me introduce you to a friend of mine, one fair red die.”
$$ P(\text{all fail}) = P(\text{red failure}) \times P(\text{green failure}) = \frac{1}{3}\times\frac{1}{3} = \frac{1}{9} \text{ of the 36 outcomes} $$
Which leaves the remaining 8/9 of the 36 outcomes for Success, who is gracious, and lets her sister divvy up the outcomes, because Failure has a keen eye for such things.