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

**tails, you keep with multiplying probabilities) If you flip two coins, the probability of getting head**

__and__**tails is 100%.**

__or__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

**tails).**

__and__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

**tails, because this one coin cannot be both heads**

__or__**tails, P * Q is zero.**

__and__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 🙂