Probability calculations involve a lot of really tricky problems involving the number of ways you can do things. Well we’ll only scratch the surface of this, (phew!), but I’m going to explain some of the basics.

Let’s start off by considering tossing two coins. Say they’re equally likely to come out heads (H) or tails (T). We’ll enumerate all possible outcomes: (HH, HT, TH, TT). There are 4. Because they’re all equally likely, the probability of each one is $\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$4$}$. Note you can also get this as we did above: in eqn 1.1.

Now we can ask, what’s the probability of getting two heads or two tails? We can express this as $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$+$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$)$}$. Because these outcomes are mutually exclusive, that is, you can’t have both at the same time, we just add the probabilities, as with our dart example above. So $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$+$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$+$}\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$4$}\colorbox[rgb]{1,1,1}{$+$}\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$4$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$2$}$.

Another way of seeing this is notice that there are two possible ways of being in the set $\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$+$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$T$}$, either getting $\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$H$}$ or getting $\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$T$}$. So the probability of getting $\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$H$}\colorbox[rgb]{1,1,1}{$+$}\colorbox[rgb]{1,1,1}{$T$}\colorbox[rgb]{1,1,1}{$T$}$ is 2/(total number of possible outcomes). This is 2/4 = 1/2. That is, the probability of hitting a pink square in the figure above, is 1/2.

Now let’s move on to throwing a pair of dice. Again let’s make the reasonable assumption that each die has equal likelihood of landing on any of its six sides. So how many possibilities are there? Let’s just start enumerating them and we’ll see a pattern. You can throw a 1 and a 1. Let’s notate this as (1,1) Or you can throw a 1 and a 2 that is (1,2), (1,3), (1,4), (1,5), (1,6). Now the next one is (2,1), (2,2) hopefully you get the idea. You go all the way up to (6,6). This is just $\colorbox[rgb]{1,1,1}{$6$}\colorbox[rgb]{1,1,1}{$\times $}\colorbox[rgb]{1,1,1}{$6$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$36$}$ possibilities all of which are equally probable and has a probability of $\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$36$}$.

1. Draw out the possible outcomes as a 6 by 6 square grid. Circle the outcomes that add up to 5, e.g. (4,1) (3,2) etc. They should lie on a 45 degree line.

So what’s the probability of throwing 2? There’s only one way of getting that, (1,1). So the probability of getting a 2 is $\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$36$}$. How about 3? There are two ways of doing that: (2,1) or (1,2). So that gives $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$3$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$2$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$36$}$. How about 4? There’s (1,3), (2,2), and (3,1). So the probability of this $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$4$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$3$}\colorbox[rgb]{1,1,1}{$/$}\colorbox[rgb]{1,1,1}{$36$}$.

1. Figure out the remaining probabilities. For example, what’s the probability of throwing a 5? Do this all the way up to 12.

2. What number are you most likely to throw?