While reading the first pages of text book in chapter 2 came up with my own example since we were only introduced to Events, Unions, Intersections of events, and containment within events, one example involved a die and it got me thinking, and I made up my own example using a relation, my question was, what is the probability that if two dice are rolled that the sum of their sides will be odd?

Our sample space, $S =\{a,b \in (\Bbb{N}, \Bbb{N}) \mid a\in \{1,2,3,4,5,6\} \wedge b\in\{1,2,3,4,5,6\}\}$

Then S our sample space is:


Lets define a Relation, $R: A \to B$ to be defined as $R=a+b \vert a+b=2k+1$ for some $k \in Z$

Then our relation is:

$$R=\{(1,2),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5), (5,1),(5,3),(5,5),(6,1),(6,3),(6,5)\}$$

R is not reflexive because $(1,1),(2,2)....(6,6)$ $\notin R$

R is symmetric because $((1,2),(2,1)),((1,3),(3,1)).......((5,6),(6,5))$ $\in R$

This was sort of an unnecessary thing I added in with the relation, in fact $R$ is our event,, $E$, $E(i+j = 2k+1)$ i.e $i+j$ is odd.

As we can see $\vert E \vert=18$

As we can also see $\vert S \vert=36$

From this example it only made sense to me that the probability of two sides of dice adding up to an odd number would be:

$$P(E)=\frac{\vert E \vert}{\vert S \vert} = \frac{18}{36} = \frac{1}{2}$$

Now I am not that far in the book and sort of just made up my own example for fun in fact I am only four pages in out of about 40 pages, so please offer additional insight, like was this in-fact true and is this something that always holds?

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    $\begingroup$ what? the probability is $1/2$ because no matter what the outcome of the first dice is, half the outcomes will make the sum odd. $\endgroup$
    – Asinomás
    Commented Dec 27, 2016 at 22:29
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    $\begingroup$ Yes, the probability that the sum is odd is indeed $\frac{1}{2}$. Yes, your solution is correct. No, you did not need to refer to any relation or even the sample space for this problem. A much cleaner solution is to approach via conditional probability and recognize that regardless what the outcome of the first die is, there is a $\frac{1}{2}$ chance that the second die will be a different parity than the first, implying the result. $\endgroup$
    – JMoravitz
    Commented Dec 27, 2016 at 22:30
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    $\begingroup$ Use \{ and \} to get set braces in MathJax. $\endgroup$ Commented Dec 27, 2016 at 22:35

1 Answer 1


In general if you have a finite sample space $\Omega$ and assign equal probability to each outcome in the sample space and an event $E\subset \Omega$, then indeed, $$ P(E)=\frac{\lvert E \rvert}{\lvert \Omega \rvert} $$ where $\lvert\cdot\rvert$ is cardinality (size). In your case $\Omega$ is the set of all possible dice pairs and $E$ is the set of all pairs which have odd sum. If you directly count as you did you get $1/2$. There are often more efficient methods to solve these problems though.


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