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 

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...

Pr() =  1/6

Pr() =  1/6

 What is the probability of throwing +   =

    

What number(s) has the best chance of being thrown?

What number(s) has the least chance of being thrown?

What is the probability of throwing craps ( 7 or 11 ) on a single roll?

Tutorials and Applets by
Joe McDonald
Community College of Southern Nevada