Question 1196729
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Here is what the addition table looks like when we add values from two 7-sided dice<table border=1 cellpadding=10><tr><td>+</td><td><font color=red>1</font></td><td><font color=red>2</font></td><td><font color=red>3</font></td><td><font color=red>4</font></td><td><font color=red>5</font></td><td><font color=red>6</font></td><td><font color=red>7</font></td></tr><tr><td><font color=blue>1</font></td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td><font color=blue>2</font></td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td></tr><tr><td><font color=blue>3</font></td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td><font color=blue>4</font></td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td></tr><tr><td><font color=blue>5</font></td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td></tr><tr><td><font color=blue>6</font></td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td></tr><tr><td><font color=blue>7</font></td><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td></tr></table>For example, if we roll a <font color=blue>1 on a blue die</font> and a <font color=red>7 on a red die</font>, then we get <font color=blue>1</font>+<font color=red>7</font> = 8 as shown in the upper right corner of that table.


The possible sums are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14


X = sum of the two 7-sided dice
P(X) = probability of that sum showing up


The X values range from 2 to 14 inclusive.
The sum "2" shows up exactly once out of 7*7 = 49 possible outcomes. Therefore P(X) = 1/49 when X = 2
<table border = "1" cellpadding = "5"><tr><td>X</td><td>P(X)</td></tr><tr><td>2</td><td>1/49</td></tr><tr><td>3</td><td></td></tr><tr><td>4</td><td></td></tr><tr><td>5</td><td></td></tr><tr><td>6</td><td></td></tr><tr><td>7</td><td></td></tr><tr><td>8</td><td></td></tr><tr><td>9</td><td></td></tr><tr><td>10</td><td></td></tr><tr><td>11</td><td></td></tr><tr><td>12</td><td></td></tr><tr><td>13</td><td></td></tr><tr><td>14</td><td></td></tr></table>


Then we have the sum "3" show up twice out of 49 possible outcomes. We write 2/49 next to 3 like so
<table border = "1" cellpadding = "5"><tr><td>X</td><td>P(X)</td></tr><tr><td>2</td><td>1/49</td></tr><tr><td>3</td><td>2/49</td></tr><tr><td>4</td><td></td></tr><tr><td>5</td><td></td></tr><tr><td>6</td><td></td></tr><tr><td>7</td><td></td></tr><tr><td>8</td><td></td></tr><tr><td>9</td><td></td></tr><tr><td>10</td><td></td></tr><tr><td>11</td><td></td></tr><tr><td>12</td><td></td></tr><tr><td>13</td><td></td></tr><tr><td>14</td><td></td></tr></table>


Keep this process going until you have this completed table.<table border = "1" cellpadding = "5"><tr><td>X</td><td>P(X)</td></tr><tr><td>2</td><td>1/49</td></tr><tr><td>3</td><td>2/49</td></tr><tr><td>4</td><td>3/49</td></tr><tr><td>5</td><td>4/49</td></tr><tr><td>6</td><td>5/49</td></tr><tr><td>7</td><td>6/49</td></tr><tr><td>8</td><td>7/49</td></tr><tr><td>9</td><td>6/49</td></tr><tr><td>10</td><td>5/49</td></tr><tr><td>11</td><td>4/49</td></tr><tr><td>12</td><td>3/49</td></tr><tr><td>13</td><td>2/49</td></tr><tr><td>14</td><td>1/49</td></tr></table>


You could write the table out like this if you prefer<table border = "1" cellpadding = "5"><tr><td>X</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td></tr><tr><td>P(X)</td><td>1/49</td><td>2/49</td><td>3/49</td><td>4/49</td><td>5/49</td><td>6/49</td><td>7/49</td><td>6/49</td><td>5/49</td><td>4/49</td><td>3/49</td><td>2/49</td><td>1/49</td></tr></table>A pattern to notice is the numerators start to increase {1,2,3,4,5,6,7} when going from X = 2 to X = 8; afterward we have a decreasing set of numerators {6,5,4,3,2,1}
Optionally you could reduce 7/49 to get 1/7, but I find it's better to have all the denominators be the same.


Side note: all of the P(X) values are between 0 and 1. Also, the P(X) values sum to 1.
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