# Fair four sided die is rolled twice, what is the possibility of the sum to be 4 or less?

A fair four sided die is rolled twice, what is the possibility of the sum of the 2 die rolls to be 4 or less? could you please explain in detail

-
What is the sample space? Hint: there are 16 = 4 x 4 elements in it. Which of the elements in the sample space have "sum 4 or less"? How many? – The Chaz 2.0 Mar 1 '12 at 3:57
If you scroll down a bit, you can see the sample for two SIX-sided dice on this unrelated site. You can literally dispense with the right two columns and bottom two rows and you'd have your sample space. – The Chaz 2.0 Mar 1 '12 at 4:00
@TheChaz there are 6 elements that meet the requirement, so does that mean that the possibility of 4 or less would be 4/16 = 1/4?? – Ahoura Ghotbi Mar 1 '12 at 4:01
If there are six elements that meet the requirement, why is your numerator four? – Brian M. Scott Mar 1 '12 at 4:11
The probability of an event $E$ is the number of elements in $E$ divided by the number of elements in $S$ (the sample space). – The Chaz 2.0 Mar 1 '12 at 4:16

Record the result of the tossing as an ordered pair $(a,b)$, where $a$ and $b$ are integers between $1$ and $4$. Here the first coordinate records the result of the first toss, and the second coordinate records the result of the second toss.

(a) Let the number of such ordered pairs be $N$. What is the value of $N$? There is a fast way of finding $N$. But one can do it slowly, by listing and counting. Here is a start: $(1,1)$, $(1,2)$, $(1,3)$, $(1,4)$.

(b) All these ordered pairs are equally likely. How many ordered pairs give you a sum of $4$ or less? List and count them carefully. Let the number be $S$.

(c) Your probability is $\dfrac{S}{N}$.

-
So Mr. Ahoura, what was the final answer that you got? – The Chaz 2.0 Mar 1 '12 at 4:34
Thanks alot, that explained everything, do you mind taking a look at my other question related to probability : math.stackexchange.com/questions/115151/… ?? – Ahoura Ghotbi Mar 1 '12 at 4:38
@Ahoura Ghotbi: Done. I hope my writeup of that question is useful to you. – André Nicolas Mar 1 '12 at 4:56

The accepted answer didn't provide an actual numeric answer so I wanted to attempt to answer it myself to ensure I got it right.

$$\mathbb{P}(Sum \le 4) = \frac{\text{Number of Winning Pairs}}{\text{Number of Pairs}}$$

I'm pretty sure $\textit{Number of Pairs} = {4 \choose 1} ^ 2 = 16$

What I'm not sure about is if there's a smarter way to determine the number of winning pairs than just listing them: (1, 1), (1, 2), (1, 3), (2, 2) and their inverses so 8 in all.

So, $\mathbb{P}(Sum \le 4) = 8 / 16 = .5$

Is that what everybody else came up with?

-