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In my probability lecture today, we were talking about the sample space for the experiment rolling two dice and having the sum be 2. My professor stressed that the order of the dice matters as if it did not, it would no longer be a problem of equal likelihood. For example, if 34 denotes rolling a 3 then 4, the sample space $\Omega$ = {11, 12, 21, 13, 31, ...} and the set of possible elementary outcomes that would satisfy the event would be E = {14, 23, 32, 41}. Could anyone explain to me why order matters in this problem? Why couldn't $\Omega$ = {11, 12, 13, ...} and E = {14, 23}?

A similar problem arose when I was solving the famous birthday problem in probability, "Probability that at least 2 people have the same birthday." Why does order matter in some problems but not in others?

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    $\begingroup$ "order matters" is a non-formal statement with dubious meaning. $\endgroup$ – Masacroso Sep 14 '17 at 4:14
  • $\begingroup$ Please use MathJax to format your posts. $\endgroup$ – Chase Ryan Taylor Sep 14 '17 at 4:41
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It is not really that order matters in this problem, it is just more straightforward to calculate the probabilities if we distinguish the two dice.

If the dice are identical, then the possible outcomes for two dice are $$\{1,1\}, \{1,2\}, \{1,3\}, \{1,4\}, \{1,5\}, \{1,6\}, \{2,2\}, \{2,3\}, \{2,4\}, \{2,5\}, \{2,6\}, \{3,3\}, \{3,4\}, \{3,5\}, \{3,6\}, \{4,4\}, \{4,5\}, \{4,6\}, \{5,5\}, \{5,6\}, \{6,6\}$$ but these events are not equally likely: the events in which the dice differ can each happen in two different ways, while the events with two of the same number can happen in one way only. For example, there are three ways to roll a 4, one way corresponding to the event $\{2,2\}$ and two ways corresponding to the event $\{1,3\}$.

This is simpler to deal with if the dice are distinct, because then the space of outcomes has a full 36 elements, and we can just count the events $(1,3), (2,2), (3,1)$; each of these can only happen in one way.

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Let's do a simpler example. Suppose we flip two coins. What are the possible outcomes, and their associated probabilities?

Well, we can both being heads, both being tails, and one being heads and one being tails.

Now, as far as the probabilities of each of these outcomes goes: if we are not careful, we would say that since there are 3 possible outcomes, each has a probability of $\frac{1}{3}$ ... which is wrong.

So, what you can do instead is to order the outcomes of the two coins, even though you toss them at the same time, for with such an imposed order, the possible outcomes are: HH, HT, TH, TT ... and now you see that the probability of both being heads is $\frac{1}{4}$, for both being tails also $\frac{1}{4}$, but for one heads and one tails it is $\frac{1}{2}$

Finally, the 'order' is not necessarily a temporal order, but really a way to distinguish the different outcomes of the different variables. That is' if we call the coins 'Alice' and 'Bob', then the possible outcomes are:

\begin{array}{cc} Alice & Bob\\ \hline H& H\\ H & T\\ T & H\\ T&T\\ \end{array}

And you get the same result.

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Compare rolling two dice and having the sum be $3$. The sum of $3$ is twice as likely as a sum of $2$ because it can be achieved by $12$ and $21$ compared to $11$. There is only one unordered pair that sums to $3$, but there are two ordered pairs that sum to $3$.

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  • $\begingroup$ My confusion is, if we say 12 and 21 are different, then shouldn't there be a 11 and 11? $\endgroup$ – Space20 Sep 14 '17 at 4:04
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    $\begingroup$ @Space20 Pretend you're actually rolling the dice. "12" means you roll one die, it comes up 1 and then you roll a second die and it comes up 2. This is different than "21" which means you roll one die, it comes up 2 and then your roll a second die and it comes up 1. 11 means you roll a die and it comes up 1 and then you roll a second die and it comes up 1. If you reverse the order then nothing changes. You're still rolling 1 each die roll. $\endgroup$ – FullofDill Sep 14 '17 at 4:29

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