Probability: Rolling Two Dice

Rolling Two Dice

When rolling two dice, distinguish between them in some way: a first
one and second one, a left and a right, a red and a green, etc. Let
(a,b) denote a possible outcome of rolling the two die, with a the
number on the top of the first die and b the number on the top of the second
die. Note that each of a and b can be any of the integers from 1 through 6.
Here is a listing of all the joint possibilities for (a,b):

(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)

Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So,
the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of
all possible outcomes for (a,b) is called the sample space
of this probability experiment.

With the sample space now identified, formal probability theory requires
that we identify the possible events.
These are always subsets of the
sample space, and must form a sigma-algebra. In an example such as this,
where the sample space is finite because it has only 36 different outcomes,
it is perhaps easiest to simply declare ALL subsets of the sample space to
be possible events. That will be a sigma-algebra and avoids what might
otherwise be an annoying technical difficulty. We make that declaration
with this example of two dice.

With the above declaration, the outcomes where the sum of the two
dice is equal to 5 form an event.
If we call this event E, we have

E={(1,4),(2,3),(3,2),(4,1)}.

Note that we have listed all the ways a first die and second die add
up to 5 when we look at their top faces.

Consider next the probability of E, P(E). Here we need more information.
If the two dice are fair and independent
, each possibility (a,b) is equally likely. Because there are
36 possibilities in all, and the sum of their probabilities must equal
1, each singleton event {(a,b)} is assigned probability equal to 1/36.
Because E is composed of 4 such distinct singleton events, P(E)=4/36=
1/9.

In general, when the two dice are fair and independent, the probability
of any event is the number of elements in the event divided by 36.

What if the dice aren’t fair, or aren’t independent of each other?
Then each outcome {(a,b)} is assigned a probability (a number in [0,1])
whose sum over all 36 outcomes is equal to 1. These probabilities aren’t
all equal, and must be estimated by experiment or inferred from other
hypotheses about how the dice are related and and how likely each number
is on each of the dice. Then the probability of an event such as E
is the sum of the probabilities of the singleton events {(a,b)} that make
up E.

Go to the home page for Tom Ramsey
Go to the home page for the UHM Department of Mathemati
cs

Your comments and questions are welcome. Please email them
[email protected]

Trả lời

Email của bạn sẽ không được hiển thị công khai.