Digression: Some Probability Ideas

This page is not a comprehensive introduction to probability theory, even remotely! Rather it is a recapitulation of some ideas (which the author hopes are correct) associated with the Dice Game Problem and by extension other matters of chance.

Terminology: Since the main sequence source page is concerned with dice, I'll define the act of selecting a die face at random as a throw, and the result of a throw is a roll (a number from 1 to 6). So I throw some dice and I roll some sixes. This terminology is arbitrary but serves to differentiate the act of observation (throw) from the result (roll).

Idea 0: The Allowed Range of Probability Values
The probability of some event under consideration is either 0 (no chance) or 1 (sure thing) or somewhere in between. This is a useful restriction because a calculated probability that is less than 0 or greater than 1 is quite probably in error!
 
Idea 1: Complementary Probabilities
Probability of some event or outcome A, P(A), is related to the probability of non-outcome of A, P(!A). (I'll use ! before an event to indicate negation of that event. ! after an integer or variable is the usual factorial.) Since probabilities are usually on [0, 1], we have P(A) + P(!A) = 1 or P(!A) = 1 - P(A). This is a surprisingly useful idea.

Example: The probability of rolling a 6 on one throw of a die is 1/6. P(r6) = 1/6; P(!r6) = 5/6.

Example: For t throws, the probability of never once rolling a 6 is (5/6)t. P(throwing at least one 6) is therefore 1 - (5/6)t and this is much simpler than the equivalent expression derived by "counting ways" per Idea 2.

P(A) = 1 - P(!A)

Idea 2: Product of Independent Probabilities for AND Questions
The ideas of 'independent' versus 'dependent' events are confusing in probability theory (to me), in comparison with how they are used commonly. Consider these two possible events which assume I'm at my house thinking about what to do:

C: I walk outside and hang the clothes on the clothes-line.
S: I walk into the garage and sweep up the sawdust.

The events are independent and we have several possible outcomes of "seeing what I do", as follows:
C happens (never mind whether S happens) = P(C). Suppose this is 1/4.
S happens (never mind whether C happens) = P(S). Suppose this is 1/3.
Both C and S happen = P(C&S) = P(C)*P(S) = 1/12.
Either C happens or S happens or both happen = P(C) + P(S) - P(C&S) = 1/2.

The last case can be thought of as correcting an over-estimate of the probability. Since P(C) is independent of P(S) and vice versa, adding their probabilities will count the probability that both occur twice. Therefore we subtract one of these countings to get the proper result. This counting of ways simply means: We count the number of possible outcomes, we count the number of desired outcomes, and set the probability of the desired outcome as the ratio of desired to possible outcomes. But we have to take care not to count certain outcomes more than once.

Now let's consider the following two events:

D: I drive from my home to the post office parking lot.
W: I walk from the parking lot into the post office.

In this case W is dependent on D because I only have the option to walk inside if I first drive to the parking lot. Suppose P(D) is 1/4 and P(W) is 1/3, but we require that W can only happen if D happens. In fact a better way to write this is P(D) = 1/4, P(W | D) = 1/3, and P(W | !D) = 0. This last means there is no way W can happen if "Not-D" happens. If I don't drive to the parking lot I can't possibly then walk inside.

Now we have the same questions to compare with the independent case above:
Outcome D (never mind whether W happens) has probability 1/4 just as before.
Outcome W (only possible if D happens) has probability P(D)*P(W|D) = 1/12.
Probability both D and W happen = P(D)*P(W). This will be 1/12 as in the previous line.
Either D happens or W happens or both happen = P(D) = 1/4.

Again the last line requires some thought. W will happen only if D happens. But if D happens then the result will be true regardless of whether W happens, so W is irrelevant to the total question. In this second example I've looked at how a dependency can completely alter the manner in which probabilities are calculated in relation to the first example.

Notice that the third case, P(D and W), comes out 1/12 just as above where C and S were independent of one another. This is interesting because there is (in both cases) a dependence built into the use of the word 'and'. That is, we have a subtle distinction between the nature of the events and whether they depend upon one another. In the first case with C and S independent, the outcome question "Do C and S both happen?" ties them together. In the second case "Do D and W both happen" also ties together two probabilities, but W is already tied to D by it's definition. The nature of the question we ask about the final outcome has built into it the nature of the constituent events.

The main point here is that independent events (including the numbers rolled on dice) are not too bad to work with in questions of probability. The particular result of note is that the probabilities of independent events are multiplied to arrive at the probability that they both occur.

The probability of several independent events all happening is the product of the probabilities of the individual events.
This is the "AND" rule which corresponds to multiplication of probabilities.

Idea 3 Sum of Independent Probabilities for OR Questions
The probability of a final outcome that can be arrived at by a multitude of possible pathways is the sum of the probabilities of each of those pathways. This is a reiteration of the notion of 'counting ways' that something may occur.

Example: What is the probability of three thrown dice summing their rolls to 16? The following are the only ways of rolling a 16 from three s dice:
5 + 5 + 6 with probability (1/6)3 (That is there is only one way of rolling this sequence in 216 possibilities.)
5 + 6 + 5 with probability (1/6)3
6 + 5 + 6 with probability (1/6)3
6 + 6 + 4 with probability (1/6)3
6 + 4 + 6 with probability (1/6)3
4 + 6 + 6 with probability (1/6)3

Sum of probabilities = P(3 dice sum to 16) = 1/36.

The probability of multiple-pathway events is the sum of the probabilities of the paths.
This is the "OR" rule which corresponds to addition of probabilities.


Idea 4 Differentiating between Discrete and Continuous Probability Distributions
Discrete probability can be defined as: Certain results have non-zero probabilities. There are an integer number of dice and the probability of rolling all 6's is ((1 way of success) divided by (6 raised to the number of dice = the number of possible throws)). This number may be small but it is always non-zero.

In contrast, continuous probability has the property that the probability of a precise result is always zero. Probabilities only become non-zero over ranges of possible outcomes. In this case we have to integrate a probability density function over some range to arrive at a probability of an outcome in that range.

Example: What is the temperature on the surface of a stove?

If we have a thermometer with a digital readout and one digit after the decimal point, then our temperature reading is discretized. The probability might be .000009 of measuring a temperature of 124.2 degrees.

But suppose we have a very precise thermometer. (Since this page is about the idealizations of mathematics it's fair to ignore further complications associated with measurement. If this were a page on quantum mechanics then I could not get away with this simplification.) With a perfect thermometer the stove will have a zero probability of being at 94.78221133400 degrees but it might have say a .00037 probability of having a temperature between 94.4 and 94.8 degrees Celsius.

Idea 5. Probability Distribution Function, Cumulative Distribution Function
Expression for weighted expectation. Includes PDF, CDF -> 1.

Examples:

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