Probability
by Joe McDonald
Let us roll 2 die one at a time...
Recall, there are 6 sides to a die
so there are six elements in the sample space
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An experiment (craps) consists of throwing two
dice, one red and one green, and observing the uppermost face. Notice, there are 6 sides
of each die. The possible different rolls would 6
× 6 = 6² = 36 outcomes.
The reason for using 2 different color dice is to emphasize
that (1,2)
and (2,1)
are 2 different throws. Here is the sample space.
| occurrences | sum | sum | occurrences | ||||||
| 1 | 2 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) | ||
| 2 | 3 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) | 8 | 5 |
| 3 | 4 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) | 9 | 4 |
| 4 | 5 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) | 10 | 3 |
| 5 | 6 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) | 11 | 2 |
| 6 | 7 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) | 12 | 1 |
Notice, there are exactly 1 way 2 throw a 2, 2 ways to throw a 3, 3 ways to throw a 4, etc...
Pr( 7 ) = 6 / 36 = 1 / 6 since there are exactly 6 ways to throw a 7 and there are 36 different outcomes in the sample space.
How does this relate to the individual throws?
You could roll a (6,1) or (5,2) or (4,3) or (3,4) or (2,5) or (1,6)
Each outcome here has an equal chance of being thrown.
Pr (6,1) = Pr( 6 ) × Pr( 1 ) = 1 / 6 × 1 / 6 = 1 / 36
Therefore, we have 1 / 36 + 1 / 36 + 1 / 36 +1 / 36 + 1 / 36 + 1 /
36 = 6 / 36
(6 times)
Let us roll 2 die one at a time...
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