This problem explores some fundamental ideas of probability.
Firstly, that the probability of getting a given total when two dice are rolled is equal to the number of possible ways of getting that total, divided by 36 (the number of possible outcomes of two dice).
Secondly, if two totals are equally likely, then the totals must occur equally often.
Lara is annoyed by the dice in the game of Monopoly. "Why is 12 (double 6s) hard to get? It’d be much easier if all totals came up the same number of times."
Can you design two dice so that only the totals 6 and 12 come up? If so, what is the probability of getting a 6 and a 12 with your dice?
Can you design two dice so that only 6 and 12 come up and they come up equally often?
- Introduce the subject of dice with a discussion:
What is the probability of getting a six when you roll a dice once? What about getting a 2?
What do the numbers on pairs of opposite faces of a dice add up to?
Why do you think that opposite faces add up to 7?
- Suppose we wanted to design a dice that only had the numbers 1, 2 and 3 on them.
Could this be done?
In how many ways could it be done?
Could it be done so that we got a 1 as often as we got a 2 but less than a 3?
How about so that 1, 2 and 3 came up equally often?
- Pose Lara’s problem, encouraging students to record their solutions as they work.
- As the groups work on the problem, give assistance where needed. Make the Extension problems available as appropriate.
- Have the students report back to the whole class.
- Give them time to write up their method of solution.
How many different dice can you produce that give totals of only 6 and 12?
How many different dice can you produce that give totals of only 6 and 12 and give the totals equally often?
There are many ways to approach this problem. Your students' solutions may vary from those given here.
We first have to put different numbers onto the two dice so that we get totals of 6 and 12 only. Now we can get a 6 in several ways: 1 + 5, 2 + 4, 3 + 3.
NB. 0 + 6, -1 + 7 and so on could be used, but explore positive numbers to begin with.
Suppose we put a 1 on one dice and a 5 on the other:
If we put another number on dice A, it has to make 6 or 12 with every number of dice B. So it has to make a 6 or a 12 with 5. This means it can only be a 1 or a 7.
On the other hand, any other number on dice B has to make a 6 or 12 with a number from dice A. Hence it could be a 5, a –1 , or –6 but only positive numbers are being used. So the two dice could look like:
Clearly there must be quite a few ways of doing this, given that 6 can be made up in three ways. If 7 were not put on onto dice A, perhaps 12 could be made using two different positive numbers on dice B.
With the two dice above, what is the probability of getting 6 and what is the probability of getting 12?
Each 1 and each 5 give us a 6. There are four 1s and six 5s. This means that we can get 6 every time we roll one of these 1s and one of these 5s. This can be done in 4 x 6 = 24 ways. Since there are 36 possible outcomes the probability of getting a 6 is 24/36 = 2/3. It appears that 6 will come up more often than the 12.
How often will the 12 come up? How can you get a 12 with these dice? We can only get a 12 when a 7 and a 5 are rolled. There are two 7s and six 5s. So 12 will come up in 2 x 6 = 12 ways. This means that the probability of getting a 12 is 12/36 = 1/3.
Another way of finding the probability of getting a 12 is to notice that we can only get a 6 or a 12. So the probability of getting a 12 = Prob (12) = 1 – the probability of getting a 6 = 1 – Prob (6) = 1 – 2/3 = 1/3. Fortunately the two answers are the same.
But it would be nice for Lara if we could make Prob(6) = Prob 12. Can this be done? Well, first of all what would it mean for the number of outcomes? Surely if Prob (6) = Prob (12), then we must get 6 and 12 equally often. Since there are 36 possible outcomes, then 6 and 12 must appear 18 times each.
Suppose that there are b 1s on the dice A. Then we get 6 in b x 6 = 6b ways. We know that 6b must be 18, so b = 3.
Check with the total of 12. Suppose that there are c 7s on dice B. Then we get 12 in c x 6 = 6c ways. And 6c = 18, so c = 3.
The idea is that we put three 1s and three 7s on dice A’ to get the two dice in the diagram.
Solution to the extension:
The first question is challenging. This could be done by listing all the possible dice systematically but it is very long list. It’s therefore quite hard to be sure that we have all possibilities. The first thing that we notice is that we want to get 6 and this can only be done with positive numbers by using 1 + 5, 2 + 4 and 3 + 3.
Case 1: Put 1 and 7 on one dice and 5 on the other. Then we can do this in 5 ways because we can use the 1, either 1, 2, 3, 4, or 5 times.
Case 2: Put 1 only on one dice. Then we have to put 5 and 11 on the other dice. We can do this in 5 ways too.
Case 3: Put 2 and 8 on one dice and 4 on the other. Then we can do this in 5 ways because we can use the 2, either 1, 2, 3, 4, or 5 times.
Case 4: Put 2 only on one dice. Then we have to put 4 and 10 on the other dice. We can do this in 5 ways too.
Case 5: Put 3 and 9 on one dice and 3 on the other. Then we can do this in 5 ways because we can use the 2, either 1, 2, 3, 4, or 5 times.
Case 6: Put 3 only on one dice. Then we have to put 3 and 9 on the other dice. We can do this in 5 ways too. BUT, how will we know the difference between the two dice that we got from case 5 and the two dice we got from Case 6. So case 6 was already taken care of in Case 5.
As there are no more cases to consider, we count the number of possibilities above and see that we get 25 different possible dice that will give only 6s and 12s.
For the second question. We have to sift through the five cases above and take out the ones that give Prob (6) = Prob (12). We get these when there are an equal number of different numbers on a dice. (Now use the case structure from above.)
Case 1: Use three 1s and three 7s.
Case 2: Use three 5s and three 11s.
Case 3: Use three 2s and three 8s.
Case 4: Use three 4s and three 10s.
Case 5. Use three 3s and three 9s.
So we get five pairs of dice that will give a total of 6 and a total of 12 equally often.