This question asks you to think about a very subtle property of probability. Don't worry if it seems complicated - it is!
a)
You're asked to consider what can happen when you roll dice following each of the given methods, before adding up the scores on the dice. So you need to think about how many different outcomes you can observe.
Method 1 - Rolling one die twice
There are a total of 36 different outcomes when we roll one die twice; these are shown in the table below. for each outcome, the value of the first throw is shown before the outcome of the second throw.
|
1 |
2 |
3 |
4 |
5 |
6 |
| 1 |
(1,1) |
(1,2) |
(1,3) |
(1,4) |
(1,5) |
(1,6) |
| 2 |
(2,1) |
(2,2) |
(2,3) |
(2,4) |
(2,5) |
(2,6) |
| 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) |
When we roll one die twice, we know the order in which the rolls happened.
This means that we can differentiate between rolling a 1 followed by a 2 - that's written (1,2) in the table above - and rolling a 2 followed by a 1 - that's written (2,1).
Although the sum of the numbers in these outcomes is the same, these two outcomes are different because we are able to distinguish between the two rolls of the die.
Method 2 - Rolling two dice consecutively
Similarly to Method 1, there are a total of 36 different outcomes when we roll two dice consecutively (one after the other); these are the same outcomes as in the table for Method 1.
As in Method 1, we know that the dice were rolled in a certain order so we can distinguish between them.
Method 3 - Rolling two identical dice simultaneously
If you roll two indistinguishable dice simultaneously (at the same time), you can't tell the difference between (a 1 and a 2) or (a 2 and a 1), because there is nothing different about the dice and they weren't rolled in any order. From your point of view, they are the same outcome with probability 236. The probabilities of the underlying events (each die's score) haven't changed, but the outcomes you observe have.
In this case you have 21 outcomes:
- (1,1)
- (1,2) or (2,1)
- (1,3) or (3,1)
- (1,4) or (4,1)
- (1,5) or (5,1)
- (2,2)
- (2,3) or (2,3)
- (2,4) or (4,2)
- (2,5) or (5,2)
- (2,6) or (6,2)
- (3,3)
- (3,4) or (4,3)
- (3,5) or (5,3)
- (3,6) or (6,3)
- (4,4)
- (4,5) or (5,4)
- (4,6) or (6,4)
- (5,5)
- (5,6) or (6,5)
- (6,6)
b)
In methods 1 and 2 we have 36 outcomes, which are all different. Hence any one of these 36 outcomes is equally likely to occur and the probability of each outcome is 136.
Contrastingly in Method 3, we have 21 outcomes but these outcomes are not all equally likely.
For example, there's only one way of obtaining two 1s, while there are two ways of obtaining a 1 and a 2, corresponding to two squares in the table in part a).
Therefore, the probability of obtaining two 1s would be 136 but the probability of obtaining a 1 and a 2 would be
236=118.