The unit looks at, analyses, and extends, a game of chance in which three coins are tossed. A player wins if two heads and a tail come up.
In this unit we are essentially looking at five ideas. These are how to calculate theoretical probabilities, how to test probabilities by trialling, how to generalise to arbitrary numbers of objects, that theoretical probability and experimental estimates of probability may not be equal, and how to make games fair.
We show that probabilities can be calculated directly from the definition by determining appropriate outcomes, by using the product of two probabilities, and by tree diagrams. We encourage students to use more than one method here as it deepens their understanding and gives them more than one strategy to use in new situations. Sometimes it is not possible to provide a theoretical probability so it is important to be able to determine an estimate of probability by using trials (experimental estimates of probability).
Generalisation is an important skill in all mathematics. We encourage it here to extend students' experience of finding patterns. The more able students may be able to justify the generalisation. Students should know when games that they are playing are fair as they may otherwise lose a lot of money in later life. Accordingly, we experiment here with some simple games and see more than one way to make a game fair. All of the above skills will be useful in later statistical and mathematical work.
For a similar unit to this see The Coloured Cube Question, Level 4.
Sufficient coins so that each group of students can have 4
theoretical probability, experimental estimates of probability, generalise, trialling, estimate, chance, combinations, outcomes, fairness, consecutive, systematic, mathematical argument
The basic game here is for the student to toss three coins. A student wins if they get two heads and one tail. For when a head is showing we write H; for a tail T. (You might note here that coin tossing can be disruptive as coins may end up all over the place. This problem can be reduced a little by having the students spin the coins about a vertical axis. Better still they can take even and odd numbers from a roll of a die as heads and tails, respectively, or red and black suits from a pack of cards, or they can make spinners with two equal colours.)
Now Jo says that there are four equal outcomes: HHH; HHT; HTT; and TTT. So the chances of winning are one in four.
On the other hand, Brian says that the chance of getting a head or a tail is 1 in 2. So the chance of getting two heads and a tail is ½ by ½ by ½ = 1/8 and so the chance of winning is 1 in 8.
(Incidentally this will be more interesting if you use names of people that are known to the students in your class. Maybe you can use a rap star’s name and name of the Prime Minister.)
The unit is based on the following 10 questions.
Question 1: Is Jo right? Why?
Question 2: Is Brian right? Why?
Question 3: Are they both wrong? If so, what is the probability of getting HHT?
Question 4: What argument can you give to convince the class that one of the answers above is wrong? Are there any other arguments that you can use?
Question 5: What experiment can you do to support the verbal argument(s)?
Question 6: If the above questions are too hard, can you see a simpler situation that will give you some insight into what is happening with this problem?
If you can do the questions above look at more difficult situations with more coins.
Question 7: If you toss two coins many times you can record the result on the diagram below. Here the angle between the lines marked HH, and HT is 120º; between HT and TT is 120º; and between TT and HH is 120º. To record the results assume you are anywhere, P, on the diagram after several tosses of the coins. Then on the next throw, move one unit parallel to the HH line if you throw two heads; or one unit parallel to the HT line if you throw a head and a tail; or one unit parallel to the TT line if you throw two heads. Where will you be after 1000 tosses of the coin?
Question 8: Jo and Brian play a game. If there an even number of heads thrown Jo wins; if not, Brian wins. Is this a fair game? (Does this depend on the number of coins used?)
We now look at the answers to all of the questions and then look at a possible teaching sequence.
Answer 1: Jo is wrong. The situation is more complicated than she thinks but she has the bones of an idea that will work. If she can nail the correct outcomes she’ll be in business. We’ll see how to do this as we go further.
Answer 2: Brian is wrong. His argument could be applied just as well to each of the outcomes HHH, HHT, HTT and TTT. That would mean that all of these events have the same probability. But 4 x 1/8 = ½ and since these are the only things that can happen, the sum of the probabilities of all of the events should be 1.
Answer 3: Yes they are both wrong. The probability should be 3/8.
Answer 4: Investigate Jo’s argument a little more deeply. What are the possible outcomes? The problem is that the coins act as individuals even though we are only looking at the heads or tails they provide. If we think of the coins as being 1, 2, and 3, then the outcomes are H1H2H3, H1H2T3, H1T2H3, T1H2H3, H1T2T3, T1H2T3, T1T2H3, T1T2T3. (However, the outcome T1H2H3 gives HHT in the same way that H1T2H3 does.) So this says that there are 8 equally likely outcomes. So the chance of getting HHT is 3/8.
On the other hand Brian wasn’t so far off but he forgot that there are three coins being used. The thing is that each one of the coins could come up H or T, so there are 3 ways that you can get just one tail. We showed this in the last paragraph. To correct his argument you need to must multiply his 1/8 by 3.
You can also do this using a tree diagram if you know how to construct such a device. The tree diagram can be based on either of the arguments above. If you use separate branches for each coin you will have 8 endpoints. We have shown this below.
Answer 5: Just do it! Get students to toss three coins and record what happens. If each pair does this 30 times the data can be grouped to give a whole class data set. This won’t show that HHT appears 3/8th of the time but it should be closer to 3/8 than to 1/4 or 1/8.
Note that this gives evidence. It doesn’t give a proof!
Answer 6: Try with two coins. Analyse that to see what happens.
Answer 7: This is very hard to predict but let’s talk about it a bit to see if we can make any progress. Now we first need to analyse what happens with two coins. Theoretically we’ll get HH ¼ of the times; HT ½ of the times; and TT ¼ of the times. If this happens over a 1000 tosses of the coins, you should expect to get HH 250 times, HT 500 times and TT 250 times. Naturally you won’t get this but bear with us for a minute. In that purely theoretical case, the HHs and the TTs will ‘cancel’ each other out in the sense that for every move parallel to the HH line you’ll have a move parallel to the TT line and those two moves will put you on the central (HT) axis and down a unit (you could check this by using trigonometry or just drawing a scale diagram. So the net effect of all of the HH and TT moves will be to put the point P on the central axis but below the horizontal axis by 250 units.
On the other hand, the 500 HTs will move P vertically 500 times and so be 500 units above the horizontal axis through the intersection of the HH, HT and TT lines. So, theoretically P should end up 250 units up the HT axis.
Well that’s theory. But what will happen in practice? You should expect P to end up well above the horizontal axis.
Note that it is difficult to plot 1000 moves as you will need to have a large piece of paper. So start with fewer moves and get them to see that it doesn’t matter what the order of the moves is, it’s only the final list of throws that matters. That way they can plot P after all the coin tossing has been done.
Answer 8: The answer is that this is a fair game. This can be seen easily if there is only one coin, or two, or three. It is even the case if 789 coins are used. They might not be able to justify this but they should appreciate it. One way of looking at this is to think about the way the tossing of 4 coins is related to the tossing of 3 coins. We show this below.
Many of you will recognise this as Pascal’s Triangle. To get the number of ways of obtaining three heads and a tail, for instance, you could see that as coming from the HHH with a T on the fourth coin; or from THH with a T on the fourth coin; or from THH with a T on the fourth coin; or from HTH with a T on the fourth coin; or from HHT with a T on the fourth coin. The THH, HTH, HHT are represented on the third horizontal line by HHT(3). Add the 1 from HHH(1) to the 3 from HHT(3) and you get HHHT(4). The other numbers of occurrences follow in the same way. What’s more the pattern continues for ever down the Triangle.
To get the probability of each occurrence, you just divide the number against it by the total number of occurrences in that row. So the probability of getting HHTT is 6 divided by 1 + 4 + 6 + 4 + 1 = 16, i.e., 6/16 = 3/8.
In this session we look at the coins game played with three coins.
In this session, we go further with the problem by experimenting to see what might be the exact probability. Then we consider four ways to get the right answer.
In this session, we build on the arguments of the last two sessions and extend finding probabilities to other situations.
In this session we plot the outcomes of experimental trials on a diagram.
In the final session we look at making the game fair.