Heads and Tails


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. 

Achievement Objectives
Specific Learning Outcomes
  • find a theoretical probability
  • use more than one way to find a theoretical probability
  • check theoretical probabilities by trials
Description of Mathematics



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.

Required Resource Materials

Sufficient coins so that each group of students can have 4

Key Vocabulary

 theoretical probability, experimental estimates of probability, generalise, trialling, estimate, chance, combinations, outcomes, fairness, consecutive, systematic, mathematical argument


Teachers’ Notes

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)?


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?)



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.

tree diagram.

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.
tree diagram.

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.

Teaching Sequence

Session 1

In this session we look at the coins game played with three coins.

  1. Introduce the basic game of the three coins.
  2. Tell the class about Jo’s solution and Brian’s solution.
  3. Ask them who they think is right and who is wrong. Take a class vote. (There are three possibilities here: Jo right; Brian right; neither right.) Check that everyone has voted.
  4. Now go through Jo’s argument.
    How does it sound?
    Is it OK?
    If it’s not, what’s wrong with it?
  5. Now go through Brian’s argument.
    How does it sound?
    Is it OK?
    If it’s not, what’s wrong with it?
    What do you think the probability is of getting HHT?

    (Don’t ask for an explanation at this stage.)
    What is the probability of getting two heads and a tail? Why?
    Ask them to write down their arguments.
  9. Let different students say what they think the probability is. Get them to argue their cases. Do not resolve any differences.
  10. Repeat the vote of step 3.

Session 2


  1. Recall the problem from the last session. We want to know what the probability of getting two heads and a tail when we toss three coins.
  2. What do you think the probability of getting two heads and a tail is?
    Take suggestions.
    Make sure that they realise that if they take an experimental approach, the answer they get is unlikely to be exactly the same as theoretical probability. However, it should be closer to the correct probability than to the others.
  5. Let the students work in pairs to collect evidence by tossing three coins. They should do about 30 trials; record the number of successful trials; and divide the successful trials by the number of trials. Any pair that finishes more quickly than the rest of the group can do more trials. Walk round to check that everyone has understood the task correctly.
  6. Bring the students together and list their results separately in three columns (successful trials, number of trials, and quotient of these two) on the board. Combine the results to give a class data set. Divide the number of successful trials by the number of trials.
    Which suggested theoretical probability is this number closest to?
    Why isn’t it exactly the same as any of the theoretical probabilities we thought of?
    Why would you expect that?
    Is it possible that you could toss three coins 13 times and each time you get 2 heads and a tail?
    If that had happened, what would you expect to happen the next time you tossed three coins?
  7. Now return to the arguments for the different theoretical probabilities. Discuss the various arguments (see Answer 4 above) for the correct probability.
    Which of these arguments do you find most easy to understand?
    What aspects of the others do you find hard to see?
  8. Let the students write down their favourite argument in their books. Check this at some stage to make sure that what they have written is correct.

Session 3


  1. Recall the game of the three coins and get the students to recall the arguments that were used to find the probability of finding two heads and a tail.
  2. Send the students into their groups to find the probability that you get three heads and that you get three tails. Ask the students to find at least two arguments for each of these probabilities. What are the chances of a student winning this game?
  3. Bring them back to discuss their results.
  4. Get them to work in pairs again with four coins. Suppose you win if you get two heads and two tails. Ask them (i) what the possible outcomes are; (ii) what the probabilities are of each possible event; (iii) what is the probability of winning the game and (iv) to provide two arguments to justify their arguments.
  6. This time send the pairs off to work together with five coins. Ask them to do the same two things that they did in step 4.
  8. Ask what the outcomes and probabilities would be for tossing 100 coins. (You might look for patterns here by showing them the Pascal Triangle approach.)
    Can you justify these results?
    Can you generalise to n coins?


Session 4

In this session we plot the outcomes of experimental trials on a diagram.

  1. Explain to them the diagram of Question 7. Send them into their groups to plot the results of 30 trials.
  2. Check that they are on track.
  3. Discuss the results. Let each group say where P ended up. Get them to see that it is easier to put in P at the end than plotting it after each throw of the coins.
    What was the longest consecutive run of HH, HT, TT that they got?
    If you had tossed 20 times and got 20 HH each time, what would you expect to get next/Why?
  4. Let them go back to their groups and think about where P would end up after 1000 tosses using the theoretical probabilities.
  5. Discuss their ideas.
  6. Let them work in groups to devise and use a diagram for tossing four coins 30 times.
    Where should P end up if theoretical probabilities are used?
  7. Have a reporting back session.

Session 5

In the final session we look at making the game fair.

  1. Recall the original game with three coins. A player wins if they get two heads and a tail. They lose otherwise.
    Is this game fair? Why?
  2. Let the students work in groups on a four coin game where you win if you get an even number of heads. Get them to play the game 30 times.
    Is the game fair?
    If so, why? If not, why not?
  3. Discuss their results.
  4. Let the students work in their groups to make their own game with four coins. They can make any rules for a player to win. Get them to play the game. Is their game fair? How can it be made fair?
  5. Discuss their results.
  6. Suppose you have 100 coins. Toss them all. If you get an even number of heads you win; if you don’t you lose.
    Is the game fair?
    Why? Why not?
    (You might want to use Pascal’s Triangle here.)
    What if you did this with 234 coins?
  7. All of these games seem fair.
    Can you make a game that is not fair?
    Why is it not fair?
    What would you have to do to make it fair?