# Heads, Tails And Beans |S|=5

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.