Te Kete Ipurangi
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Monopoly Discussion for Staff Room Seminar

Introduction

The format of this seminar is as follows. In this section we give an overview of the seminar and talk about related staff seminars on probability. In Section 2 we list the equipment that you and the staff might need. Then we pose and answer the question of, when playing Monopoly, what are the chances of landing on Park Lane and Mayfair if you happen to be on Regent Street (Section 3)? This question is very much like one that we posed in the seminar Snakes and Ladders. The difference between this question and the Snakes and Ladders question is that this question involves two dice, so we need to analyse the probability of rolling each of the numbers 2 through 12. We follow this by, in Section 4, the Cross the Stream Game. This requires an analysis of rolling two dice and how this relates to the numbers 2 through 12 coming up over the long haul. Then, as in Snakes and Ladders, we see that there might be a more relevant question if you are actually playing Monopoly (Section 5). This question is, what are the chances of landing on Park Lane on this round? This is a harder question than the corresponding Snakes and Ladders question. Solving the same problem for Mayfair requires even more thought. Finally (Section 6), we provide some OHTs for you to copy if you think that they might help your presentation.

There is no need to go as far as we have done here. Leave something up your sleeve if you want to or add a problem from a different dice game or give your staff some homework. The final seminar is up to you. We have just provided some suggestions.

There are two other staff room seminar suggestions for you on probability. These come under ‘Snakes and Ladders’ and ‘Lotto’. Snakes and Ladders is the easiest, Monopoly is a little more difficult and Lotto is the hardest. If you are going to do all three you may want to do them in that order.

Equipment needed

It will be handy to have the following equipment available:

1. for you:
• a whiteboard or equivalent
• a Monopoly board or copies of the last side of the board
• dice for demonstration purposes
• OHTs (see Section 7)
• a calculator
1. for the staff:
• enough dice for a pair of dice between 2 to 4 teachers (depending on the size of the staff)
• a replica of the Mayfair side of a Monopoly board
• pens and paper
• calculators

Monopoly

Playing Monopoly we have landed on Regent St. There are hotels on Park Lane and Mayfair. What is the probability that we will land on either of these places on our next throw?

To do this you may need to remind the staff of the basic definition of probability. It may help to then cover some simple examples. Ask them to calculate:

• the probability of getting a 3 when you roll a dice (1/6)
• the probability of getting a 9 when you roll two dice (4/36 or 1/9)
• the probability of getting a blue ball when you draw a ball from a box that contains 3 red and 2 blue balls (2/5)
• the probability of throwing two Heads when you toss a 20c piece and a 50c piece (1/4)

We show the business end of the Monopoly board below.  Remember we are on Regent St and moving towards Go.

To land on Park Lane or Mayfair we have to throw a 6 or an 8. How can that be done with two dice? The easiest way to do find that out is to produce a table. The numbers in blue represent the numbers on the dice. Those down the side of the table are for one dice; those across the top are for the other. The elements of the table tell us what has happened when the two dice are rolled. So, for example when we throw a 4 (side die) and a 3 (top die) we get a 7 (where the ‘4’ row and the ‘3’ column intersect).

 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

Now suppose we want to find the probability of landing on Park Lane. That’s the probability of getting a 6. Now we know from Snakes And Ladders that since all of the ‘events’ in the table are equally likely, we can use the basic formula

So.

We can tell from the table that there are 5 ways we can get a 6 (5+1, 4+2, 3+3, 2+4, 1+5). This might be a problem in itself as some people think that 5+1 and 1+5 are one, and not two ways of getting a six. To see that they are indeed two ways, think of the dice as being different colours, red and blue. Then 5 on the red and one on the blue is indeed different from 1 on the red and 5 on the blue.

This should help us to see that there are 36 possible rolls. We get a new possibility for every black number of the table. Since there are six rows and six columns this gives 6x6=36 possible outcomes. Perhaps better, there are six numbers on one dice and six on the other. So for every number from the first dice there are six numbers that we can get from the second. So there are 6 x 6 possible totals (including any repetitions).

That gives us

.

But what about Mayfair? That is 8 steps away from Regent Street. Now we can work out from the table that there are 5 ways of getting 8. So the probability of landing on Mayfair is

Now since the two events of landing on Mayfair or Park Lane are mutually exclusive, they cannot both happen in one roll, the probability of landing on one of these squares is

.

This can be simplified to 5/18 if you like but it means that there is roughly a 1 in 4 chance of landing on either of these two big ones. So if you own these two properties, on average you would expect to collect big money about once in every four times that someone lands on Regent Street and rolls the two dice. Of course there is no cast iron guarantee that someone will land on the big two precisely 10 times out of every 36 that they happen to be on Regent Street, but this is what you can expect over a large number of ‘trials’.

Cross the Stream

The board below represents a stream that has to be crossed. Each player is given 10 counters that they may place anywhere on their side of the stream.  The players take it in turns to roll two dice and the scores are added.  The players may then move one counter to the other side of the stream if they have a counter on that number.  The winner is the first player to move all their counters to the other side of the stream.

 2 3 4 5 6 7 8 9 10 11 12 2 3 4 5 6 7 8 9 10 11 12

The question is, what is the best strategy for placing the counters?

To answer this, you can either play, and learn by experience, or you can use the table in the last section. That table will tell you what numbers are most likely to come up and so which numbered squares it is best to put your pieces on.

We note that the probability of getting a 7 is 6/36 or 1 in 6. So what is your strategy going to be? It seems that there are three approaches we might take. Put all of the pieces on 7 because that has the biggest probability of happening, spread all the pieces equally over the board, or spread them with the most on 7, fewer on 6 and 8 and so on symmetrically down to 2 and 12.

To make things easier, let’s suppose that we have 3600 pieces to put on the board. So do we put:

1. 3600 pieces on the 7
2. as close as possible to 3600/11 pieces on each of the eleven numbers
3. 100 on 2 and 12, 200 on 3 and 11, 300 on 4 and 10, 400 on 5 and 9, 500 on 6 and 8 and 600 on 7?

Remember the probability of getting a 2 is 1/36. This means that in 36 throws we’d expect to get 2 about once. So in 3600 rolls of the dice we’d get a 2 about 100 times. You can work the rest out for yourself.

Now we’ll talk about each of these in turn. Suppose that we try (i) first. Because of the probabilities in the table, in any 3600 rolls of the dice we would expect 7 to come up roughly 600 times. So if we put all of the pieces on 7, we’d only have crossed about 3000 of them over after 3600 rolls.

What about (ii)? Here we put as close as possible, 3600/11 (or about 327) pieces on each number. (We divide by 11 here because there are 11 possible totals when we roll two dice.) After 3600 rolls, we’d expect to get 2 (and 12) about 100 times, so there would still be about 200 pieces left on 2 (and 12). After 3600 rolls, we’d expect to get 3 (and 11) about 200 times, so there would still be about 100 pieces left on 3 (and 11). After 3600 rolls, we’d expect to get 4 (and 10) about 300 times, so there would still be a few pieces left on 4 (and 10). You can work the rest out for yourself.

So go to strategy (iii). Since we’d expect to throw 2 about 100 times, we’d expect to have lost most of the pieces from 2 after 3600 throws. Since we’d expect to throw 3 about 200 times, we’d expect to have lost most of the pieces from 3 after 3600 throws. Since we’d expect to throw 4 about 300 times, we’d expect to have lost most of the pieces from 4 after 3600 throws. And so on.

Now we couldn’t expect everything to work quite like clockwork here, that’s not the way that chance works, but we can expect to have just a few pieces left after 3600 rolls if we adopt strategy (iii). We’d certainly have less than by the other two strategies.

But you only have 10 pieces to put on the numbers. That means that you will have to do some approximating. But a variant of (iii) ought to be the best way to go in the long run.

Park Lane Again

(You may need to look at Snakes and Ladders to see why the probabilities of this section are what they are.)

Now knowing what the probability of hitting Mayfair or Park Lane in one go from Regent Street is, is only part of the game. What we really want to know is what are the chances of missing Park Lane or Mayfair sometime this time through, with one roll or many? To find that out we are going to have to look at bit more closely. In what ways are we likely to land on either of these two giants? To find this out we are going to have to be systematic.

Let’s concentrate on Park Lane. Now we can get to there as follows:

1. in one roll by throwing 6;
2. in two rolls by throwing 2 and then 4; 3 and then 3; or 4 and then 2;
3. in three rolls by throwing 2 and then 2 and then 2.

(After four rolls we have to be past Park Lane.  Why?)

The chances of getting (i) we know. They are 5/36.
The chances of getting (ii) are 1/36 x 3/36 = 3/1296; and 2/36 x 2/36 = 4/1296; and 3/36 x 1/36 = 3/1296. This is a total of 10/1296.
The chances of getting (iii) are 1/36 x 1/36 x 1/36 = 1/46656.

Altogether the chances of hitting Park Lane are the sum of these. That is,

5/36 + 10/1296 + 1/46656 = 6841/46656.

If you do the arithmetic, you’ll see that this is only marginally bigger (less than 1 in a 100 more) than landing straight on Park lane in one throw. That ought to be a comfort to those of you who don’t own Mayfair and Park Lane.

We’ll leave you to analyse the situation for Mayfair by itself. You might also like to think about the worst case of all – landing on both Park Lane and Mayfair!

The Monopoly Board.

Probability table for rolling two dice

 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

Cross the Stream Board

 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12