Ways of rolling two dice

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Christian Lawson-Perfect	said	Ready to use	7 years, 8 months ago
Chris Graham	said	Has some problems	7 years, 8 months ago
Elliott Fletcher	said	Needs to be tested	7 years, 8 months ago

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Christian Lawson-Perfect 7 years, 8 months ago

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Christian Lawson-Perfect 7 years, 8 months ago

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Christian Lawson-Perfect 7 years, 8 months ago

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Chris Graham commented 7 years, 8 months ago

I think that students will find this very confusing indeed. You give an answer at the top, 11 possible outcomes, and refer to an "outcome" as the total value of the two dice. But in the question you are expecting the students to interpret the word "outcome" as the possible values $(a,b)$ that the dice could take.

I'm also not certain of the benefit of this sort of question to an incoming student.

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Chris Graham 7 years, 8 months ago

Gave some feedback: Has some problems

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Elliott Fletcher 7 years, 8 months ago

Gave some feedback: Needs to be tested

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Elliott Fletcher 7 years, 8 months ago

Created this.

Name	Status	Author	Last Modified
Ways of rolling two dice	Ready to use	Elliott Fletcher	20/11/2019 14:45
Ways of rolling two dice	draft	Xiaodan Leng	11/07/2019 01:44

Question 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.

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

Method $1$ : Rolling one die twice.

Method $2$ : Rolling two dice consecutively.

Method $3$ : Rolling two identical dice simultaneously.

Give any introductory information the student needs.

			Name	Type	Generated Value

Automatically regenerate variables?

No variables have been defined in this question.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
3. Gap Gap 2. Number entry
  Add
Part b) Choose several from a list

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

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 $1 =$ Gap 0.

Number of outcomes for Method $2 =$ Gap 1.

Number of outcomes for Method $3 =$ Gap 2.

Advice

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 $\displaystyle\frac{2}{36}$ . 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 $\displaystyle\frac{1}{36}$ .

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

$\displaystyle\frac{2}{36} = \displaystyle\frac{1}{18}.$

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Choose several from a list
Choose several from a list		Hasn't run yet

This question is used in the following exams:

Probability by Elliott Fletcher in Elliott's workspace.
Probability by Elliott Fletcher in Elliott's workspace.
Probability by Christian Lawson-Perfect in Transition to university.
Introduction to Probability by Robert Aykroyd in Robert's workspace.
Probability by steve kilgallon in Stats and probability.
Week 7 - Practice Exercises by Bobby Kahlon in Bobby's workspace.
NUMBAS - Probability by Katy Dobson in Katy's workspace.
Probability by Simon Thomas in Stats.
Maria's copy of Introduction to Probability by Maria Aneiros in Maria's workspace.
Week 6 - Practice Exercises - Probability by Bobby Kahlon in Bobby's workspace.
Probability 1 by Jeremie Wenger in Jeremie's workspace.
Maths 1A - Homework Week 10 (Probability) by Maria Aneiros in MY QUESTIONS.
Shaheen's copy of Probability by Shaheen Charlwood in Shaheen's workspace.
Probability Exam by Anna Cartlidge in Glasgow Numbas Question Pool.
Transition Probability by Chris Templet in Chris's workspace.

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Checkpoint description

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Chris Graham commented 7 years, 8 months ago

Chris Graham 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Error in variable testing condition

Parts

Gap-fill

a)

Method 1 - Rolling one die twice

Method 2 - Rolling two dice consecutively

Method 3 - Rolling two identical dice simultaneously

b)