I touched on this a bit a few years ago but didn’t really post what I wanted to due to time constraints. I blog from my phone so blog posts usually take a few weeks to write 🙂.
Anyway, this post is my demonstration of how laws of logic also apply to probability theory.
In Boolean logic, we can use a truth table to demonstrate that two logic statements are equivalent.
So if you want to prove the statement P v ~Q (“P or Not Q”) is equal to Q v ~P, you would draw up a truth table and see if the columns for each statement are equal:
As you can see, the P v ~Q column doesn’t match the Q v ~P column. This logically proves that P v ~Q is not equal to Q v ~P.
DeMorgan’s Law is the proof that shows that the statement ~ (P ^ Q) is equal to the statement ~P v ~Q, proving that you can logically substitute one statement for the other:
This is a visual representation of DeMorgan’s Law from Wikipedia:
This exact methodology can be used to prove that probability theory is just Boolean logic that incorporates uncertainty; the columns will match. Boolean “and” is equal to mathematical multiplication and Boolean “or” is equal to mathematical addition. Though both come with a caveat for mutual exclusion.
So for example. Flipping a coin for heads or tails is mutually exclusive; you either get heads or tails. But if you’re flipping more than one coin, then heads or tails is no longer mutually exclusive. This is easy, you just go from multiplying your probabilities to adding them (though if you’re flipping more than one coin and want heads and tails, you keep with multiplying probabilities) If you flip two coins, the probability of getting head or tails is 100%.
But what if you flip three (or more) coins? It can’t be 150% chance of getting heads or tails! But notice what is going on in this addition: you’re effectively counting getting both heads and tails twice. Necessarily, you’re either going to get two heads with three coins being flipped at the same time (thus one instance of heads and tails) or two tails (thus the second instance of heads and tails).
So you have to subtract the second heads and tails when doing probabilistic “or”. In other words, P v Q becomes P + Q – P * Q. Indeed, if you go back to flipping only one coin for heads or tails, because this one coin cannot be both heads and tails, P * Q is zero.
Here we see DeMorgan’s replicating when substituting 100% and 0% for “true” and “false”:
Sure, this works with 0%/100%, but what about other probabilities?
As you can see, DeMorgan’s Law holds up even when using probabilities that aren’t equal to 0%/100%.
Thus my dilettante proof that probability theory is Boolean logic expanded to account for uncertainty 🙂