Calculating Joint Probability Mass Function for Two Dice Toss

[r0bHadz]

Homework Statement

Two dice are tossed. Let X be the smaller number of points. Let Y be the larger number
of points. If both dice show the same number, say, z points, then X = Y = z.

Homework Equations

The Attempt at a Solution

Find the joint probability mass function (X,Y)

For (1,1) = 1/36, I do not understand why every other vector, say (1,2) does not = 1/36

 

[r0bHadz]

I'm pretty sure it has to do with introducing a random variable P = sum of the two rolls but I don't see why this would be necessary. If you roll two dice I don't see why they wouldn't be independent. So the sample space has a size of 36, each of the 36 are equally likely, P(1,1) should = P(1,2)

 

[r0bHadz]

Hmm I think I'm getting closer. Since x is less than or = to y, the probability changes. so for the row of x values, where y = 1, you have 5 0's, This is different from the sample space which has 36 entries

 

[PeroK]

Homework Statement

Two dice are tossed. Let X be the smaller number of points. Let Y be the larger number
of points. If both dice show the same number, say, z points, then X = Y = z.

Homework Equations

The Attempt at a Solution

Find the joint probability mass function (X,Y)

For (1,1) = 1/36, I do not understand why every other vector, say (1,2) does not = 1/36

You mean that $P(1, 2) = P(2, 1) = 1/36$ etc.?

 

[SammyS]

Homework Statement

Two dice are tossed. Let X be the smaller number of points. Let Y be the larger number
of points. If both dice show the same number, say, z points, then X = Y = z.

Homework Equations

The Attempt at a Solution

Find the joint probability mass function (X,Y)

For (1,1) = 1/36, I do not understand why every other vector, say (1,2) does not = 1/36

Hmm I think I'm getting closer. Since x is less than or = to y, the probability changes. so for the row of x values, where y = 1, you have 5 0's, This is different from the sample space which has 36 entries

You have left out enough detail so that we are left to guessing just what difficulties you are having .

You say the probability for (1,2) is not 1/36, but don't say what it is. My guess is that it's 1/18. Is that correct?

 

[r0bHadz]

You mean that $P(1, 2) = P(2, 1) = 1/36$ etc.?

yes. I don't see why not.

 

[r0bHadz]

You have left out enough detail so that we are left to guessing just what difficulties you are having .

You say the probability for (1,2) is not 1/36, but don't say what it is. My guess is that it's 1/18. Is that correct?

Yes its 1/18.. I think I'm starting to understand now. It's not dice X and dice Y, it is high score and low score. I think my difficulty came from my lack of understanding of what a random variable is.

 

[PeroK]

yes. I don't see why not.

$P(2, 1) = 0$, as it is stated in the problem that $X \le Y$. If you roll $1$ and $2$, then $X$ is always counted as $1$.

It doesn't have anything to do with random variables, as such. For example, if you have a table headed $X$ and $Y$, then you would never have an entry $2, 1$. You would always enter that as $1, 2$.

This is quite important, because you often record data in ascending or descending sequence. E.g. the lottery. I.e. you order or sort the data before you record it.

 

[r0bHadz]

$P(2, 1) = 0$, as it is stated in the problem that $X \le Y$. If you roll $1$ and $2$, then $X$ is always counted as $1$.

It doesn't have anything to do with random variables, as such. For example, if you have a table headed $X$ and $Y$, then you would never have an entry $2, 1$. You would always enter that as $1, 2$.

This is quite important, because you often record data in ascending or descending sequence. E.g. the lottery. I.e. you order or sort the data before you record it.

Right, so because there are 2 ways you can get (1,2) you have 2(1/36) = 1/18. So I was wrong in my the first post after my OP, you do not need to introduce a new random variable, and the sample space is of size 36.

 

[PeroK]

Right, so because there are 2 ways you can get (1,2) you have 2(1/36) = 1/18. So I was wrong in my the first post after my OP, you do not need to introduce a new random variable, and the sample space is of size 36.

The sample space is of size 21. This has nothing to do with probabilities or random variables. That's simple the number of pairs $(X, Y)$, where $X \le Y$.

 

[Ray Vickson]

Yes its 1/18.. I think I'm starting to understand now. It's not dice X and dice Y, it is high score and low score. I think my difficulty came from my lack of understanding of what a random variable is.

I think it is more your not acknowledging the importance of understanding what the sample space is. You can have lots of probability problems that do not involve random variables, but require knowing what is the sample space is of major importance. In terms of sample spaces, we have in this problem:
$$
\begin{array}{cc|cc}
\text{die 1}&\text{die 2}& \text{low} & \text{high}\\ \hline
1 & 1 & 1 & 1 \\
1 & 2 & 1 & 2 \\
2 & 1 & 1 & 2 \\
\vdots & \vdots & \vdots & \vdots \\
4 & 6 & 4 & 6 \\
6 & 4 & 4 & 6 \\
\vdots & \vdots & \vdots & \vdots\\
6 & 6 & 6 & 6
\end{array}$$
So, for example, P(low=4,high=6) = P(d1=4,d2=6)+P(d1=6,d2=4) = 2/36.

 

[r0bHadz]

That makes sense. For any probability problem, before I do anything, I think I should completely understand the sample space before even attempting the problem. Your post has answered my question, thanks man.

 

[WWGD]

But to nit pick a bit, you also need to specify what aspect of a situation you are observing. Here you throw dice and observe the face facing you. Pedantic here, but helpful in situations where it is not so obvious what you are keeping track of.