Here we’ll look at some more probability calculations before moving on to permutations, combinations and the binomial theorem.

### More Probability…

If you recall from the last part, we used set notation to describe a general way of calculating simple probability: \(P(A)={{|A|} \over {|S|}}.\) Where \(A\) is a set of *events *and \(S\) is a set of all possible *outcomes*.

Here’s a sample space set describing two dice:

\[S=\{1,2,3,4,5,6\}^2\]

…for all \(|S|=6^2=36\) possible outcomes. This set contains repeated values and ordering, so let’s directly enumerate the whole set:

\begin{array}

((1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1) \\

(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2),(2,6),(6,2) \\

(3,3),(3,4),(4,3),(3,5),(5,3),(3,6),(6,3) \\

(4,4),(4,5),(5,4),(4,6),(6,4) \\

(5,5),(5,6),(6,5) \\

(6,6)

\end{array}

Each tuple represents a probable roll of a pair of dice. Normally we roll a pair of dice at the same time so that we don’t care what the order is – a five and a three are read the same as a three and a five, for example. However, if we roll each dice individually you can see that the order counts – I may roll a one first followed by a six, or the other way around.

Notice how the doubles occur only once in the above table but the different rolls occur twice – (3,6) and (6,3), for example. This leads to a counter-intuitive probability problem. You’d naturally think that any pair of numbers (including the doubles) would have an equal chance, but this is not so as can be seen here:

\[A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}\]

…which is an event set for all the possible double throws. If we calculate the probability:

\[P(A) = {{|A|} \over {|S|}} = {6 \over 36} \approx 0.17\]

…a 0.17 or 17% chance of rolling any double, and as there are 30 possible outcomes for rolling two different numbers:

\[P(A) = {{|A|} \over {|S|}} = {30 \over 36} \approx 0.83\]

…or an 83% chance of rolling anything other than a double. Finally, you can see that you are twice as likely to roll a differing pair of dice than a double:

\[A=\{(6,6)\} {\;\;\;\;} B=\{(5,3),(3,5)\}\]

\[P(A) = {1 \over 36} \approx 0.03 {\;\;\;\;} P(B) = {2 \over 36} = {1 \over 18} \approx 0.06\]

…a one in thirty-six chance of rolling a particular double verses a one in eighteen chance for rolling two different numbers. This explains why doubles are so highly prized in dice games.

### The Sum Of Probabilities

If set \(S\) contains all of the possible outcomes from some random event, like tossing a coin, then if we sum up all of the probabilities of those outcomes we get the number 1 or 100%.

We can see this easily with a coin sample set: \(S=\{\text{Heads,Tails}\}\). I think that we can all work out that there is a 50% chance that the outcome is either heads or tails. The sum of both these probabilities is, of course, 100%. Mathematically, we can write it like this:

\[\sum _{x \in S} p(x) = 1\]

…where the function \(p(x)\) returns a probability value between 0 and 1, inclusive \((p(x)=[0,1])\), for any event \(x\) inside of set \(S\). The *sigma* \(\sum\) symbol simply means to add all of the probabilities for all possible events together:

\[p(\text{Heads}) + p(\text{Tails}) = 0.5 + 0.5 = 1\]

The event set is a subset of the sample space set. If \(A\) is the event set then \(A \subseteq S\). This, logically, means that the events inside of set \(A\) can be any of the subsets of the power set of \(S\), \(\wp(S)\). Including the empty-set, \(\emptyset\). In that case, we haven’t flipped a coin so we have no chance of winning or losing! 🙂

All of the possible event sets (subsets) \(A\) from a heads/tails set \(S\) give us these probabilities for \(p(x)\):

\[p(\emptyset)=0 {\;\;} p(\text{Heads})=0.5 {\;\;} p(\text{Tails})=0.5 {\;\;} p(\{\text{Heads, Tails}\})=1\]

…and therefore:

\[P(A) = \sum _{x \in A} p(x)\]

In the next lesson we’ll look at the factorial function, its interaction with the prime numbers and ways to solve it for positive real numbers.

© Doc Mike Finnley 2019