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I am not sure if I am understanding this problem correctly. It says;

Two six-sided dice are rolled. Let X1 and X2 denote the outcomes of the first and second die roll respectively. Define a new random variable Y such that:

Y = max{X1 , X2}

Does this mean I am supposed to roll the two dice and find the largest roll?

From my own understanding, if I am supposed to find the largest roll from the two dice roll, is this example of mine correct?

ex: the outcomes are {2,3} then Y=3

Moving on to what is asked of me:

Find the probability distribution of Y

For this part, do I need to find all the possible outcomes of the two dice? For example;

{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}

etc....

And do I need to find the probability of the sum of two dice? For example;

Sum = 2 , Prob = 1/36

Sum = 3 , Prob = 2/36

Sum = 4 , Prob = 3/36

etc....

After this, what do I do from here to find the probability distribution of Y?

If I am totally off track, explanation from you guys will be great. Sorry if it looks like I'm asking too much. Please feel free to correct me on anything!

David G. Stork
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  • Looks to me like you’ve mostly understand the problem correctly. For the distribution of $Y$, however, you’re not interested in the sums of the dice, but the probabilities that $Y=\max{X_1,X_2}=1$, $Y=2$, etc. So, once you’ve got your table of all $6\times6$ possible rolls, count up how many have a max. of $1$, of $2$, and so one. – amd May 11 '17 at 22:52
  • The distribution has nothing to do with "sum". – David G. Stork May 11 '17 at 23:58
  • Yes, thank you for referring @Max. Requested to be delete! Much help –  May 12 '17 at 01:32

2 Answers2

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Notice that the mirror image of every L-shaped event corresponds to the outcome $Y$

As an example, consider the L-shaped event corresponding to the outcome $Y=3$

$$\begin{align} \{1,3\} \\ \{2,3\} \\ \{3,1\}, \{3,2\}, \{3,3\} \end{align}$$

We can easily see that $$\mathbb{P}(Y=3)=\frac{5}{36}$$

In general, you can generalize that $$\mathbb{P}(Y=y)=\frac{2y-1}{36}$$ where $y=1,2,3,4,5,6$

Roby5
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It may be useful to construct a table, with $X_1$ down the left, and $X_2$ along the top, and each cell gets the maximum of the two values. The table below has been started for you; fill in the rest of the values.

$$ \begin{array}{|c|c|c|c|c|c|c|} \hline & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline \textbf{1} & 1 & 2 & 3 & & & \\ \hline \textbf{2} & 2 & 2 & 3 & & & \\ \hline \textbf{3} & 3 & 3 & 3 & & & \\ \hline \textbf{4} & & & & & & \\ \hline \textbf{5} & & & & & & \\ \hline \textbf{6} & & & & & & \\ \hline \end{array} $$

The probability distribution for $Y = \max(X_1, X_2)$ can then be obtained by counting up the proportion of entries that are $1, 2, 3, \ldots$.

Brian Tung
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