Sample Space Ω is the collection of all possible elementary outcomes of a random experiment.
The key
𝑁𝑁(𝐴𝐴)
here is the term elementary. Probability of an event A is then calculated as 𝑃𝑃(𝐴𝐴) = where N(.) denotes
𝑁𝑁(Ω)
the cardinality or dimension of a set or space. Elementary outcomes are those which can’t be broken down
further into other outcomes with respect to the experiment. For example, let’s consider a random
experiment of throwing a single die. Let’s define:
A: event represents that multiple of 2 turns up
B: event represents that the side ‘1’ turns up
C: event represents that the side ‘3’ turns up
D: event represents that the side ‘5’ turns up
The sample space Ω is NOT written as {A,B,C,D} even though it is the same as {2,4,6,1,3,5}≡{1,2,3,4,5,6}.
This is because A={2,4,6} is not an elementary event as opposed to B={1}, C={3} and D={5} which are
elementary. Not having a sample space with all possible elementary events may create confusion. For
example, just by looking into the sample space {A,B,C,D} and not knowing A, B, C and D are not elementary
events, if we want to find out P(B) we may wrongfully say P(B)=1/4. Here, A, B, C and D are not equally
likely or equally probable outcomes because A is not elementary event and it can be broken down into
simpler events. On the other hand, if we define
A1: the side ‘1’ turns up, A2: the side ‘2’ turns up, A3: the side ‘3’ turns up, A4: the side ‘4’ turns up, A5: the
side ‘5’ turns up, A6: the side ‘6’ turns up.
then A1, A2, A3, A4, A5, and A6 are equally likely elementary events because they have the same chance
of occurrence and none of them can be broken down further into simpler events.
Single die problem:
Let’s consider rolling a die (just one time). The sample space Ω = {1,2,3,4,5,6} represents all possible
elementary outcomes. Let X represents an event that x turns up on the die, hence, x=1,2,3,4,5,6. We want
to find P(X), i.e., we want P(1), P(2), P(3), P(4), P(5) and P(6). It may seem straight forward that P(1) = P(2)
= P(3) = P(4) = P(5) = P(6) =1/6. But is that always true? What happens if the die is biased, or getting a 1
is not as equally likely as getting a 5. So, P(1) = P(2) = P(3) = P(4) = P(5) = P(6) =1/6 is true as long as the
outcomes in the sample space are elementary, and each outcome is equally likely or equally
probable. In that case, one way to measure probability of an elementary outcome or event is by
1
.
number of elements in the sample space
Double (non-identical) dice problem:
Now, let’s roll two unbiased dice simultaneously and consider them non-identical or distinguishable. The
sample space for rolling of each die is {1,2,3,4,5,6} and the sample space for the experiment of rolling two
dice would be:
Table 1 Value on Die 2 (Y)
1 2 3 4 5 6
1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
Value on 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
Die 1 (X) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
The elements in the matrix in Table 1 are all elementary events and equally likely unless the dice are biased
in some way. P(getting a number (X,Y))=1/36 as all outcomes in the sample space are equally likely
because the dices are unbiased. Note that adding up all probabilities leads to 1.
Independence of events:
Let A: event that 5 occurs on die 1, and B be an event that 5 occurs on die 2 then P(A)=1/6, P(B)=1/6 and
P(A∩B)=1/36=1/6×1/6=P(A)×P(B). Hence, A and B are independent statistically.
Double (identical) dice problem:
Now, let’s roll two unbiased dice simultaneously and consider them identical or indistinguishable. The
sample space for rolling of each die is {1,2,3,4,5,6} and the sample space for the experiment of rolling two
dice could be considered as (just for the sake of argument and it’s NOT true):
Table 2 Value on Die 2 (Y)
1 2 3 4 5 6
1 (1,1)
Value on 2 (2,1) (2,2)
Die 1 (X) 3 (3,1) (3,2) (3,3)
4 (4,1) (4,2) (4,3) (4,4)
5 (5,1) (5,2) (5,3) (5,4) (5,5)
6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Since the combination of (1,2) and (2,1) or in general (X,Y) and (Y,X) are the same, so the number of
elements in the sample space is �61� + �62� = 6+15=21 cases. �61� denotes the number of cases
corresponding to choosing 1 element out of 6 for the diagonal elements in Table 2, and �62� corresponds to
choosing 2 numbers out 6 irrespective of their arrangements. The above table is fine as long as we view
this problem from a combinatorial perspective.
But the inherent problem with the matrix in Table 2 is that the elements are NOT equally likely and they are
not elementary events. (1,1) means the 1st dice gets 1 and the 2nd dice gets 1 and this can happen in only
ONE possible way. (1,1) is elementary. But (1,2) means the 1st dice gets 1 and the second dice gets 2.
Being indistinguishable this outcome is same as getting 2 on the 1st dice and 1 on the 2nd dice. So, we will
notice that getting (1,2) is more likely than (1,1) because now since we can’t distinguish between (1,2) and
(2,1), (2,1) will be masked in (1,2). In this case, if we define M to be the event that the 1st die gets 1 and the
2nd die gets 2, i.e. (1,2) turns up in the simultaneous throw of two identical dice, then M is NOT an elementary
event as it can be broken down to (1,2) and (2,1) which are elementary individually. Therefore, we can’t
use Table 2 to calculate P(1,2) and we can’t say P(1,2)=1/21, because (1,2) is NOT as equally likely as
(1,1) or (2,2). Now, we don’t know how to calculate P(1,2) unless we consider the matrix in Table 1.
Since we are no longer able to distinguish between (1,2) and (2,1), we consider both to be say (2,1). Then
based on Table 1 which represents the true sample space, P(2,1)=2/36=1/18. In short:
𝟏𝟏
⎧ , 𝒙𝒙 = 𝒚𝒚
⎪ 𝟑𝟑𝟑𝟑
𝑷𝑷(𝒙𝒙, 𝒚𝒚) = 𝟏𝟏
⎨ , 𝒙𝒙 > 𝒚𝒚
⎪𝟏𝟏𝟏𝟏
⎩ 𝟎𝟎, 𝒙𝒙 < 𝒚𝒚
So, why is this confusion? Let me provide a counter example. Suppose we denote (re-write) 4, 5,
and 6 on a die to be 1,2 and 3, respectively. And let’s denote (re-write) 4 and 5 on the second die to
be 1 and 2, respectively. So, the total elements in the sample space would still be 36 corresponding
to each side that turns up, but the mapped numbers we observe are the following combinations.
{(𝟏𝟏, 𝟏𝟏), (𝟏𝟏, 𝟐𝟐), (𝟏𝟏, 𝟑𝟑), (𝟏𝟏, 𝟔𝟔), (𝟐𝟐, 𝟏𝟏), (𝟐𝟐, 𝟐𝟐), (𝟐𝟐, 𝟑𝟑), (𝟐𝟐, 𝟔𝟔), (𝟑𝟑, 𝟏𝟏), (𝟑𝟑, 𝟐𝟐), (𝟑𝟑, 𝟑𝟑), (𝟑𝟑, 𝟔𝟔)}
So, we may be compelled to think that P(1,1)=1/12, but really it isn’t because chance of getting (1,1)
is not same as the chance of getting (1,6) because they are not equally likely or elementary. The
correct matrix would be:
Table 3 Value on Die 2 (Y)
1 2 3 1 2 6
1 (1,1) (1,2) (1,3) (1,1) (1,2) (1,6)
Value on 2 (2,1) (2,2) (2,3) (2,1) (2,2) (2,6)
Die 1 (X) 3 (3,1) (3,2) (3,3) (3,1) (3,2) (3,6)
1 (1,1) (1,2) (1,3) (1,1) (1,2) (1,6)
2 (2,1) (2,2) (2,3) (2,1) (2,2) (2,6)
3 (3,1) (3,2) (3,3) (3,1) (3,2) (3,6)
P(1,1)=4/36=1/9 whereas P(1,6)=1/18
Finally, throwing two identical dice is same as throwing a single die two times one after the other. In the
latter scenario, the above confusion doesn’t arise because dice are made non-identical w.r.t to the time of
throwing.
Take away: Probability is always calculated based on the elementary outcomes of a random
experiment rather than the mapping of each outcome. Even though the sample space elements’
mapping is correct (as given in Table 2) from the point of view of a combinatorial problem, however,
it’s not appropriate for calculating probability of events as it won’t give us the correct sample space.
