# Representing a die roll in a form of a table

I am trying to understand this question but not having much luck:

Suppose two fair dice are tossed one time. Let X be the number of 1's observed and Y be the number of 4's observed. Represent this in the form of a joint pf table.

This is what I tried but the sum of rows/cols is not 1 so it can't be right


X
0      1      2
_________________________
0 |  0    1/36  2/36  |
Y   1 | 1/36  1/36   0    |
2 | 2/36   0     0    |
__|___________________|_
| not 1


-
The sum can add up to 1, but the table is incorrect. Give me a second I am writing up an answer. – mathguy Nov 12 '12 at 8:58
Would it be possible to know how to correct it? – Kay Nov 12 '12 at 8:59

                 X
0      1      2
_________________________
0 | 16/36 8/36  1/36  |
Y   1 | 8/36  2/36   0    |
2 | 1/36   0     0    |
__|___________________|_
|


This is the table I would have made. If X = 2, then that means that both the dice had a 1, and there is only one scenario out of 36 where both 1's show up. If X=1 and Y=0, then that means only one 1 showed up, and the other dice was not a 4 or a 1. There are 8 unique cases of such. (You can mirror these two for Y=2 and Y=1,X=0. For X=1 and Y=1, that means one 1 showed up and one 4 showed up. There are two possible scenarios for that. For X=0 and Y=0, that means no 1's showed up and no 4's showed up. There are only 4 other sides to each of the die, so 4*4 = 16.

Below are the 36 possible die rolls: 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

-
This answer has helped me understand this a lot better. Thank you sir! – Kay Nov 12 '12 at 9:04