In probability, I often have trouble determining which situations to take as distinct outcomes for calculation. For instance, if we have a die with its faces numbered $1, 2, 2, 3, 3, 6$ and we roll it twice.

We get $2$ on the first roll and $3$ on the second. Again rolling it twice we get $2$ and $3$. But the $2$ we got the second time is not the same $2$ as the first one. Its the other $2$ inscribed on the face of the die (the die has two $2$'s).

So, for the purpose of calculating the probability that the sum of the two rolls in a die will be a certain number $4$, say, will these two situations constitute distinct outcomes?

This is just a simplified, distilled example of a persistent problem I face in probability. Is there any way to think about outcomes that can make this clearer?


What you are asking is whether $(2,2)$ is a distinct outcome from $(2,2'), (2',2), \text{or } (2',2')$ where the result of $2$ and $2'$ are each from different faces of the die.

The answer is: it depends.   You can consider them four distinct outcomes, each with a probability mass of $1/36$, or you can consider them a single outcome with probability mass of $1/9$.   The choice of which model to use depends on which makes it easier to calculate the probability of your event.   As long as your model weighs the probability of its outcomes accurately it will not matter.

$$\begin{align} \mathsf P\big(\{(1,3), (2,2), (3,1)\}\big) & =\; \dfrac{2+4+2}{36} \\[2ex] \mathsf P\big(\{(1,3), (1,3'), (2,2), (2,2'), (2',2), (2',2'), (3,1), (3',1)\}\big) & =\; \dfrac{8}{36} \end{align}$$

The advantage of drilling down to a model of hidden distinct outcomes of equal probability is then you just need to count numbers of outcomes within an event; as you don't need to worry about assigning weights.

  • $\begingroup$ I'm afraid my problem is still not answered. What you have effectively told me is that $(2, 2)$ and $(2, 2')$ would be counted as distinct outcomes. Assigning weights is effectively 'merging' the outcomes into one compound outcome. What I want to know is why we call $(2, 2)$ and $(2, 2')$ different outcomes at all? Ideally, I was expecting a precise definition of an outcome, and then an explanation of how it resolves the specific problem in my question. $\endgroup$ – Gerard Mar 17 '15 at 12:18
  • $\begingroup$ An outcome is simply a distinct result of the experiment; in this case the ordered pair of results of two independent dice tosses. The sample space of the experiment is the set of all possible outcomes. Whether you consider two faces with the same number to be distinct or not is entirely your choice. $\endgroup$ – Graham Kemp Mar 17 '15 at 12:52

Outcomes are members of a sample space, also called a probability space; events are subsets of the sample space, also called the probability space. Thus if you throw a die twice, then among the outcomes are these three: $(3,1), (2,2),(1,3)$. Those are three outcomes. But the set $\{(3,1), (2,2),(1,3)\}$ is just one event, which can be described as the event that the sum is $4$.

  • $\begingroup$ I'm sorry, but I don't see how this answers my question. I still don't know how to figure out if two situations constitute two outcomes or a single one. The sample problem I presented in my question is still not resolved. $\endgroup$ – Gerard Mar 17 '15 at 5:09

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