How many different outcomes are there in each of the three methods, **before** adding up the scores on the dice?

Number of outcomes for Method $1 =$ [[0]]

\nNumber of outcomes for Method $2 =$ [[1]]

\nNumber of outcomes for Method $3 =$ [[2]]

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Method 1

", "Method 2

", "Method 3

"]}], "preamble": {"css": "", "js": ""}, "type": "question", "functions": {}, "extensions": [], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "A game asks for two dice scores to be added together, but doesn't specify how. You want to work out the probability of each possible outcome occurring.

\nYou come up with three different methods for obtaining two dice scores:

\nMethod $1$: Rolling one die twice.

\nMethod $2$: Rolling two dice **consecutively**.

Method $3$: Rolling two **identical** dice **simultaneously**.

This question asks you to think about a very subtle property of probability. Don't worry if it seems complicated - it is!

\nYou'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**.

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.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 | 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.

\nThis 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).

\nAlthough 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.

\nSimilarly 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.

\nAs in Method 1, we know that the dice were rolled in a certain order so we can distinguish between them.

\nIf 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 $\\displaystyle\\frac{2}{36}$. The probabilities of the underlying events (each die's score) haven't changed, but the outcomes you observe have.

\nIn this case you have $21$ outcomes:

\n- \n
- (1,1) \n
- (1,2) or (2,1) \n
- (1,3) or (3,1) \n
- (1,4) or (4,1) \n
- (1,5) or (5,1) \n
- (2,2) \n
- (2,3) or (2,3) \n
- (2,4) or (4,2) \n
- (2,5) or (5,2) \n
- (2,6) or (6,2) \n
- (3,3) \n
- (3,4) or (4,3) \n
- (3,5) or (5,3) \n
- (3,6) or (6,3) \n
- (4,4) \n
- (4,5) or (5,4) \n
- (4,6) or (6,4) \n
- (5,5) \n
- (5,6) or (6,5) \n
- (6,6) \n

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 $\\displaystyle\\frac{1}{36}$.

\nContrastingly 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 $\\displaystyle\\frac{1}{36}$ but the probability of obtaining a 1 and a 2 would be

\n\\[\\displaystyle\\frac{2}{36} = \\displaystyle\\frac{1}{18}.\\]

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