Look at a Sample Space
Return to Probability
One common introduction to samples spaces is to look at the possible outcomes
of rolling two dice, and, in particular, to look at the sum of the values "showing"
(i.e., on the top of the dice when then come to rest) on the roll.
This sample space is displayed in a chart along the lines of Figure 1.
Figure 1
Sum of faces | |
F I R S T D I E |
|
1 | 2 | 3 |
4 | 5 | 6 |
| | |
S E C O
N D D I E |
1 | | 2 | 3 | 4 |
5 | 6 | 7 |
2 | | 3 | 4 | 5 |
6 | 7 | 8 |
3 | | 4 | 5 | 6 |
7 | 8 | 9 |
4 | | 5 | 6 | 7 |
8 | 9 | 10 |
5 | | 6 | 7 | 8 |
9 | 10 | 11 |
6 | | 7 | 8 | 9 |
10 | 11 | 12 |
From that table we can see that there are 36 possible outcomes.
We can see that there are five outcomes that have the value 6 in them.
And we can see that there is just one outcome that has the value 2 in it.
We could look at slightly more complex events
and find the number of outcomes that meet the requirements of those evens. For example,
we coud ask how many outcomes are evenly divisible by 4? The answer is 9. Or we could ask
how many outcomes are greater than 5? That answer is 26.
There are many ways to change the problem. For one thing, the dice need not
have the traditional faces. Consider a pair of dice that each have
the values 3, 4, 6, 8, 9, and 13 on them.
Our new table appears in Figure 2.
Figure 2
Sum of faces | |
F I R S T D I E |
|
3 | 4 | 6 |
8 | 9 | 13 |
| | |
S E C O
N D D I E |
3 | | 6 | 7 | 9 |
11 | 12 | 16 |
4 | | 7 | 8 | 10 |
12 | 13 | 17 |
6 | | 9 | 10 | 12 |
14 | 15 | 19 |
8 | | 11 | 12 | 14 |
16 | 17 | 21 |
9 | | 12 | 13 | 15 |
17 | 18 | 22 |
13 | | 16 | 17 | 19 |
21 | 22 | 26 |
Now, for Figure 2, we could ask how many of the outcomes are divisble by 4?
The answer is still 9, but the pattern of successes is much different
from the pattern found in Figure 1.
We do not even need to have the same faces on the two dice. One could
have the faces from Figure 2 while the other has the traditional faces of
Figure 1. That would produce Figure 3.
Figure 3
Sum of faces | |
F I R S T D I E |
|
3 | 4 | 6 |
8 | 9 | 13 |
| | |
S E C O
N D D I E |
1 | | 4 | 5 | 7 |
9 | 10 | 14 |
2 | | 5 | 6 | 8 |
10 | 11 | 15 |
3 | | 6 | 7 | 9 |
11 | 12 | 16 |
4 | | 7 | 8 | 10 |
12 | 13 | 17 |
5 | | 8 | 9 | 11 |
13 | 14 | 18 |
6 | | 9 | 10 | 12 |
14 | 15 | 19 |
Again, we use the table to help answer questions.
We examine the table and count the outcomes hat satisfy the
constraint of the question.
In Figure 3 there are 8 outcomes that are evenly divisible by 4.
There is nothing special about the rule used to make the
table. In the first three examples we used the rule that finds the sum of the two dice.
Let us change the rule. In Figure 4 the new rule is to find the remainder
when we divide the number from the first die by
the number from the second. This operation is called "modulo" and is often abbreviated as "mod".
Thus 11 mod 4 is 3 because 11 divided by 4
leaves a remainder of 3. Also note that Figure 4 not only uses two differently marked dice,
it also illustrates that the values need not be in ascending order.
Figure 4
1st mod 2nd | |
F I R S T D I E |
|
5 | 7 | 12 |
19 | 23 | 28 |
| | |
S E C O
N D D I E |
6 | | 5 | 1 | 0 |
1 | 5 | 4 |
9 | | 5 | 7 | 3 |
1 | 5 | 1 |
3 | | 2 | 1 | 0 |
1 | 2 | 1 |
7 | | 5 | 0 | 5 |
5 | 2 | 0 |
11 | | 5 | 7 | 1 |
8 | 1 | 6 |
4 | | 1 | 3 | 0 |
3 | 3 | 0 |
Again, we can count the number of outcomes that meet any particular criteria.
For example, we can count and find that there are 10 outcomes that are the number 1.
There is nothing special about having sice with 6 sides. We might have five sides on one die and
eight on another. If the first has the numbers 2, 3, 5, 7, and 11 on its sides and the second
has the numbers 1, 1, 2, 3, 5, 8, 13, and 21 on its sides, and if our rule is to look at the
product of the two values, then our sample space will have the 40 items shown in
Figure 5. Note that Figure 5 illustrates that we can even have repeated numbers
on the die.
Figure 5
Product of the faces | |
F I R S T D I E |
|
2 | 3 | 5 |
7 | 11 |
| | |
S E C O
N D D I E |
1 | | 2 | 3 | 5 |
7 | 11 |
1 | | 2 | 3 | 5 |
7 | 11 |
2 | | 4 | 6 | 10 |
14 | 22 |
3 | | 6 | 9 | 15 |
21 | 33 |
5 | | 10 | 15 | 25 |
35 | 55 |
8 | | 16 | 24 | 40 |
56 | 88 |
13 | | 26 | 39 | 65 |
91 | 143 |
21 | | 42 | 63 | 105 |
147 | 231 |
The event that we seek might be having the digit 4 in the value. Then we would have to
count the number of outcomes in Figure 5 that have the digit 4 in them. That
answer is 7.
Finally, there is not even a requirement that we have a verbalized rule for
determining the make-up of the values in the table. That is, we could roll the dice and
then use the table to determine the outcome.
Consider some board game where you need to move your playing piece (or pieces) a certain number
of places depending upon the roll of the dice.
Figure 6 gives such rules. Furthermore, the rules for Figure 6 change
each time you reload this page, producing a different game defined by each new table.
Return to Probability
©Roger M. Palay
Saline, MI 48176 Febbruary, 2016