Brief Intro to Probability

Super Brief Introduction to Probability

To help understand Plantinga’s chapter on the Fine-Tuning Argument for God’s existence, as well as some of the subsequent chapters, please look over the following. I have tried to provide some explanation of the basics of probability. If you do not find my explanation all that helpful, find it confusing, or even just plain wrong, please let me know. 

There are three basic axioms of probability and then a bunch of very useful derivations or theorems. I am just going to present them altogether here and I will do so in the language of logic.

1. If a statement A cannot be false (A must be true), then the probability of A = 1. 

-Note that we write ‘the probability of A’ as P(A). So, for the last part of 1 above we write: P(A) = 1

-We can say it like this: For all necessarily true statements A, P(A) = 1

-Question: What are some examples of necessarily true statements?

2. If a statement A cannot be true (A must be false), then the probability of A = 0

            -For all necessarily false statements A, P(A) = 0

-Question: What are some examples of necessarily false statements?

3. From 1 and 2 above, it follows that for any statement, its probability is either 0, 1, or somewhere in between 0 and 1. 

            -For all statements A, P(A) is either 0 or P(A)  1

-So, it is not possible for a statement to have a probability less than 0 or to have a probability greater than 1. Those are simply impossible, they are definitionally nonsense. 

4. If A and B are mutually exclusive, then the probability of A or B is the probability of A plus the probability of B.

            -If A and B are mutually exclusive, then if A is true, B is not, and if B is true, then A is not. 

            -In other words, when 2 statements are mutually exclusive, they cannot both be true. 

-So, 4 tells us that the probability of mutually exclusive statements just is the sum of the probability of each statement

            -If A and B are mutually exclusive, then P(A or B) = P(A) + P(B)

Example: If A and B are mutually exclusive, and P(A) =.5 and the P(B) = .2, then the P(A or B) = .7 

5. From 1 and 4 we can derive the following: P(~A) = 1 – P(A)

            -1 above tells us that all necessarily true statements have a probability of 1. 

-So, consider this disjunctive statement (a disjunctive statement is a statement with 2 or sub-statements connected by an ‘or’): A or ~A

-(A or ~A) must be true. They are mutually exclusive (both cannot be true) and they are jointly exhaustive (one must be true).

-So, from 1 above, we know that P(A or ~A) = 1

            -We also know from 4 above that P(A or B) = P(A) + P(B)

                        -So, we know that P(A or ~A) = P(A) + P(~A)

-Since we already showed that P(A or ~A) = 1 and we just showed that 

P(A or ~A) = P(A) + P(~A), we now know that P(A) + P(~A) = 1

-Simple math operations allow us to get the following: 

            P(~A) = 1 – P(A)

As well as:

            P(A) = 1 – P(~A)

Let’s sum up a bit. What we just showed was that when we have set of statements that are mutually exclusive and jointly exhaustive we can know a few interesting things about that set. We can know that not all of the statements can be true and one must be true. We can also know that the probability of one of the statements from the set is equal to 1 – the probability of the sum of the probability of the remaining statements. 

Example: Suppose our only statements are

F = God fine-tuned the universe

C = Chance fine-tuned the universe

And suppose that F and C are mutually exclusive (both cannot be true) and are jointly exhaustive (one must be true). We now know that P(F) = 1 – P(C). 

So, even if we do not know the precise probability of F or of C we know something interesting. We know that if P(C) is low, say in the range of 0 to .4, then the P(F) is between .6 and 1. 

Let’s continue: 

Now what about cases where we do not have mutually exclusive statements? That is, cases where our set of statements can all be true. In such cases we have a new rule

6. P(A or B) = P(A) + P(B) – P(A & B)

-In 4 above, we considered cases where it is not possible for both A and B to be true. If A is true, then B is not, and if B is true, then A is not. But now we are considering cases where it is possible for both A and B to be true. So, if A is true, B might very well be true also, and if B is true, A might very well be true also. 

-Example: Suppose we want to know what the probability is that student X will register for either Logic or History of Ancient and Medieval Philosophy (with no other info other than that X is permitted to register for any class). Suppose that 20 students typically register for Logic and 30 students typically register for History of A and M. Since there are 100 students, there is a .2 chance that X registers for Logic and a .3 chance that X registers for History. So, there is a .5 chance that X registers for either Logic or History. Not so fast, you say. It is also possible for X to register for both Logic and History. So, we need to know the probability of X registering for both classes. Why? Suppose we fail to consider that probability—the probability of registering for both classes. Now suppose that of the 20 students who typically register for Logic and the 30 students who typically register for History, 5 of them typically are in both classes. That means that our earlier claim that there is a .5 chance that X registers for either class is wrong. It turns out, given this new information that instead of 50 students usually taking Logic and History, there is really only 45 students. We counted 5 students twice in our original derivation of the probabilities. So, we have to account for that. To do so, we have to include that new information in our calculations. 

-So, we take the P(Logic) + P(History) and subtract the P(Logic & History). We know that P(Logic) = .2 and the P(History) = .3 and now we know that P(Logic & History) = .05. So, the P(Logic or History) = .2 + .3 -.05, which = .45 or 45%. So there is a 45% chance that X will take Logic or History. 

We can use the above info to think a bit harder about fine-tuning stuff. Suppose that 

F = God fine-tuned our universe

M = the multiverse fine-tuned our universe

Now given fine-tuning, it may seem that F and M are equivalent. That is, fine-tuning evidence does not support F over M or M over F. So, if F and M are mutually exclusive (both cannot be true) and jointly exhaustive (one must be true), then we’d have to assign F and M the same probability, namely, .5. 

But what if F and M are not mutually exclusive? That is, what if it is possible for both to be true? Perhaps God fine-tuned the multiverse so that it would produce our fine-tuned universe. We now have the following: 

P(F or M) = P(F) + P(M) – P(F & M)

Plantinga actually seems to favor something like this as do a lot of other writers on this topic. What this tells us is that we need the following sort of emendation to M:

M*=the multiverse fine-tuned the universe by chance alone

So, by thinking about the basic rules and principles of probability we can get a bit clearer about what the fine-tuning argument is saying. The multiverse hypothesis is not necessarily in competition with the design hypothesis. To make it in competition with the design hypothesis we have to state the multiverse hypothesis precisely. That’s what M* does. 

As we saw in the text and in class, the kinds of probability statements the fine-tuning argument works with are not quite like those we have been considering. In my next post, I will say something about them.