Why is sum of 2 rolls is not calculated another way?

Question

Question: what is the probability of getting a sum of 5 using a pair of dice?

I am aware of the solution being, choosing one face for each dice - that gives total $36$ possibilities in the sample space

For the event space, only the sum of $(1,4), (4,1), (2,3), (3,2)$ equals to 5, so probability of getting a sum of $5$ is $4/36$.

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Why is the solution not approached in another way?

That is, since the sum of 2 dice can only lead to $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$ i.e. $11$ possibilites, the probability of getting 5 as sum is $1/11$. Why is this method wrong, and how is the approach wrong?

I want to know the fundamental behind this. I am confused and not getting the intution behind. Thanks!

Answer 1

Note that the sum as you said can be {$1 \cdots 11$}, but you haven't taken into account that both rolls are distinguishable according to the exercise. Therefore, for the number $5$, pairs of rolls could be $(1,4),(2,3),(3,2),(4,1)$. This is called compositions of a number, and is one of the basic tools of combinatorics.
What you have done is just choosing just one of the possible sums, but for each sum there are many possiblities depending on the pair of rolls. That is why not only it has to be taken into account the possible sums (10 results), but all possible pairs of results.

Answer 2

If you replace the phrase you have used, "....total $36$ possibilities in the sample space..." by "...total $36$ possibilities in the equiprobable sample space..." the flaw in the second approach should become self-evident, as a sample space of $11$ is certainly not equiprobable. different sums can occur in a different number of ways.

The two die rolls are independent, and thus produce $6\times6 = 36$ equiprobable outcomes. Writing the outcomes from die $1$ and die $2$ as ordered pairs these are

$(1,1)\;(1,2)\;(1,3)\;(1,4)\;(1,5)\;(1,6)$
$(2,1)\;(2,2)\;(2,3)\;(2,4)\;(2,5)\;(2,6)$
$(3,1)\;(3,2)\;(3,3)\;(3,4)\;(3,5)\;(3,6)$
$(4,1)\;(4,2)\;(4,3)\;(4,4)\;(4,5)\;(4,6)$
$(5,1)\;(5,2)\;(5,3)\;(5,4)\;(5,5)\;(5,6)$
$(6,1)\;(6,2)\;(6,3)\;(6,4)\;(6,5)\;(6,6)$

We can now count the favorable outcomes and compute the probability as $\frac{favorable\; outcomes}{total\; outcomes}\;\;$, as we are dealing with an equiprobable sample space

Answer 3

Just to echo @trueblueanil and @DanielC..

The experiment is about tossing a pair of dice so we really have to look at the results of two dice rolls first before calculating the sum. Because going directly to the sum makes you lose valuable information about the experiment ,that is the number of ways an outcome can occur because the mistake in your second approach is you assumed all sums have same number of possibilities to occur.

The first table below shows all possibilities, while the second table shows you how many times a sum can occur.

First	Second	Sum
1	1	2
2	1	3
3	1	4
4	1	5
5	1	6
6	1	7
1	2	3
2	2	4
3	2	5
4	2	6
5	2	7
6	2	8
1	3	4
2	3	5
3	3	6
4	3	7
5	3	8
6	3	9
1	4	5
2	4	6
3	4	7
4	4	8
5	4	9
6	4	10
1	5	6
2	5	7
3	5	8
4	5	9
5	5	10
6	5	11
1	6	7
2	6	8
3	6	9
4	6	10
5	6	11
6	6	12

Sum	Number of Possibilities it occurs
2	1
3	2
4	3
5	4
6	5
7	6
8	5
9	4
10	3
11	2
12	1

Answer 4

The reason your method does not work is because the eleven possible sums of the dice, between $2$ and $12$, are not equally likely. Instead, the $36$ possible ordered pairs of numbers rolled by the dice are equally likely. The reason that each pair of numbers is equally likely is a consequence of the fact that the two dice are independent. Namely, if we let $X$ be the result of the first die, and $Y$ of the second, then for any numbers $x,y\in \{1,\dots,6\}$, $$ P(X=x\text{ and }Y=y)\stackrel{\text{indep.}}=P(X=x)P(Y=y)=(1/6)\cdot(1/6)=1/36. $$ Above, used the fact that $X$ and $Y$ are independent, along with the fact that each value of $X$ is equally likely, and the same for $Y$ (the dice are fair).