Let's illustrate this with the following example:
(Impossible, probability = 0) ................................. [0.5 50/50 chance ] ...................... (Certain, probability = 1.0)
The above scenario generates three probability of a certain event, for example, the chance of rain tomorrow (say). Either, you can state that there is no chance or probability of rain tomorrow, in which case you will assign the probability of rain referred to as p(Rain) "0". Or, you are not sure at all, and you claim that the probability of rain is 50%, that is a 50% chance that it is going to rain and a 50% chance it is not going to rain, so we assign it the probability of rain or p(Rain) = 0.5; on the other hand, if it looks like that it is definitely going to rain tomorrow, then you state that the probability of rain tomorrow is 100%, and in terms of probability language we write, p(Rain) = 1.0; you can see from here that probability is "bounded", that is it can be at one end 0 and at the other end 1. What if we were able to free up so that the "chance" of something occurring versus not occurring would have a boundary of 0 at one end and infinity at the other end? This is useful for some estimation issues as we will see, and we call this likelihood, so let's explain that now. We use the same diagram as before but this time we use a ratio and we call this ratio "likelihood or odds" (we will visit Odds when we will discuss study designs such as case control study designs). So, here we go again with the rain example:
(Impossible, probability = 0) ................................. [0.5 50/50 chance ] ...................... (Certain, probability = 1.0)
Now we introduce the ratio of something happening versus something not happening. That measure is referred to as likelihood or Odds. So, in case where we stated that it was not going to rain at all, we express this in terms of probability of rain or p(rain) = 0, and probability of no rain or p(no rain) = 1.0. Then, the odds of rain as opposed to no rain tomorrow would be expressed in terms of p(rain) / p(no rain) and you can see we are dividing 0/1.0, and so the odds(rain) = 0. What happens when we have 50/50 chance of rain. Here p(rain) = 0.5 (50 out of 100), and equally p(no rain) = 50/100 or 0.5. Therefore here the odds(rain) = 0.5/0.5 = 1. In the extreme case, where we say that it is definitely going to rain, the p(rain)/(1 - p(rain) ) will be infinity as we will divide 1.0 by 0 and it is not estimable and given a value of inf. So, now we have this:
(Impossible, odds = 0) ............................ (fifty/fifty chance, odds = 1.0) .................... (Certain 100% chance, Odds ~ Inf)
So that is the concept of probability and odds that probability, and remember some rules (call them probability sutra):
- p(X) will lie between 0 and 1, it cannot be less than 0 (that is, there no such thing as negative probability) and it cannot be greater than 1.
- If there are mutually independent events that makes up our universe, let's say X and Y, then p(X) + p(Y) = 1; what this means is let's say you want to toss a coin. As you toss a coin, there can be only two outcomes, heads or tails. We will say p(head) + p(tail) = 1. If the probability of head and tail are equal then you know that p(head) = 0.5, and p(tail) = 0.5 and so together they are 1.0; can you now work out what it will be with the rain thing? p( it will rain) + p(it will not rain) = 1.0; you can vary the probabilities of raining versus not raining, they will always add up to one as these are mutually exclusive events.
- A third rule is this: if X and Y are "disjointed or mutually exclusive or independent events", then their joint probability of occurrence will be p(X)*p(Y). An example is to think in terms of tossing a fair coin. If you toss a fair coin, the probability of heads (shorthand:"H"), p(H) = 0.5 (you can see that the probability of tails or p(T) = 0.5 too). The first time you toss a coin (we call it number of trials as N, each act of tossing a coin is referred to as a trial), the probability of head is p(H) = 0.5. The second time you toss a coin, for that toss, the probability of head is still 0.5 (as these are two independent events). What is the chance that this time too, you will have a probability of head? This is where the third rule comes into play: as these are independent of each other, we will say that you will two heads in a row, say something like p(HH) = 0.5 * 0.5 = 0.25.
You can probably guess where I am getting with this. For any random event (such as a coin toss, or say something like the weather or rainfall), where there is no pattern, you will be able to multiply their individual probability and be able to predict what will the probability of an event happening over and over again (or finding a pattern) under conditions of random occurrence: just multiply their probabilities and you will get the value. So, if we were to guess say that the probability of each succeding year hotter than the previous one was 0.5, then if you were to base your prediction on the basis of just one year's experience (N = 1), you'd say p(next year hotter than this year) = 0.5; the odds will be 1.0, and we are none the wiser. Under the conditions of randomness, what is the probability that each succeeding year will be hotter than the previous year? (again assuming that these things are truly random or 50/50 chance), you can construct a chart like this: