Colin Foster's Mathematics Education Blog: May 2022

26 May 2022

Are two cars better than one?

So many ideas in probability are really quite unintuitive. How can we help learners make better sense of how simple probabilities combine, without relying on arbitrary rules and mysterious formulae?

I recently heard someone talking about whether they should get a second car for their household (Note 1). They were doubtful that it was a good idea. In addition to the cost and environmental impact, they said, “If you have two cars, there’s twice as much chance that something will go wrong with one of them.” This seemed self-evidently true, and everybody nodded sadly, and the conversation moved on.

I began thinking about how I might respond mathematically. I knew what they meant, and their statement might indeed be approximately true, but it couldn’t be exactly true. Even if all that you know about probabilities is that they are capped at 1 (i.e., 100% is the highest possible probability), it is clear that you cannot just go around doubling probabilities. Doubling any probability greater than 0.5 will give you a total probability greater than 1, which is impossible. And, even if you have super-reliable cars with a very small probability $p$ of failing, you would only need to ensure that you buy more than $\frac{1}{p}$ of them for the total probability to exceed 1.

So, why is such a plausible-sounding statement not right? And under what assumptions might the statement be approximately true?

Suppose that both cars have the same probably $p$ of ‘something going wrong with them’ in a certain time interval. Would these be independent events? If both cars are parked outside the same house, then they are likely to be subject to similar weather conditions and other factors, so it seems unlikely that failure of one would be completely unrelated to failure of the other. But let’s ignore this and suppose that the two events are independent, and also that the two cars are equally likely to fail. This would mean that the probability of both cars failing would be $p^2$. And this means that when we double $p$ we are overcounting by $p^2$ (see Figure 1), because we are counting the same situation of ‘something goes wrong’ when car A fails and counting it again when car B fails, for those occasions when they both fail. In the extreme case, where you had two completely useless cars, you would have a 100% chance of not being able to drive anywhere, but not a 200% chance! Now, if $p$ is very small, then $p^2$ will be very very small, and so we can perhaps ignore the overlap region. But, if $p$ is not very small, then $p^2$ will be non-negligible. This means that the correct probability for either (or both) cars failing is $2p-p^2$.

Figure 1. $p(A∪B)$

Looking at the expression $2p-p^2$, we might wonder whether we can be absolutely sure that it is always less than 1 for all values of $p$. Is the $p^2$ definitely always sufficiently large to bring back $2p$ to below 1 whenever $p>0.5$? One way to see this is by completing the square, to obtain $1-(1-p)^2$, meaning that $2p-p^2$ is equal to '$1-$something that is never negative'. Figure 2 shows the graph $y=2p-p^2=p(2-p)$, and the curve has its maximum value of 1 at $p=1$, and so it never exceeds 1 for any value of $p$ (Note 2).

Figure 2. $y=2p$ and $y=2p-p^2$

We can also see in Figure 2 that the line $y=2p$ is indeed a good approximation to the curve for small values of $p$. So, just doubling the probability is a reasonable approximation if you are considering very reliable cars.

Using the algebra, this reasoning is quite straightforward for anyone comfortable with quadratics and elementary probability. In probability, we can only add the probability of events A and B if they are mutually exclusive (i.e., $p(A∩B)=0$), so that the ‘Venn diagram identity’ $p(A∪B)\equiv p(A)+p(B)-p(A∩B)$ reduces to $p(A∪B)=p(A)+p(B)$. In other cases, we have to subtract the intersection, so as not to double count it.

I was happy with all of this, but I wanted to say something that didn't sound technical or rely on set theory or even Venn diagrams. Could I say in words why doubling was not quite right? I found it hard to come up with a good way to explain to my friend why the possibility of both cars going wrong was relevant to their statement, and why this indeed made their statement technically wrong, even if you were willing to make assumptions about things like independence and so on. If I had said that having two cars that might go wrong is not quite twice as bad as having one car that might go wrong, because, at least some times, both cars will simultaneously go wrong, I think they would be quite surprised! The commonsense response is that there is no consolation in having both cars fail on the same day - that is the worst possible nightmare, and indeed one of the reasons for contemplating having a second car was to try to be sure that they would always have one working car! They would probably respond that “When I said 'either-or', I was including 'both'!”, which misses the point. Yes, we want to include the chance of both cars failing - the point is that we want to include that possibility only once, not twice (Figure 3)! It is still true that $p^2<p$, and possibly dramatically so ($p^2 \ll p$), so the chance of having at least one working car has indeed risen from $1-p$ to $1-p^2$. The point is that if last week Car A failed, say, on Monday, Thursday and Saturday, whereas Car B failed on Tuesday, Saturday and Sunday, our daily frequency of car trouble would have been $\frac{5}{7}$ and not $\frac{6}{7}$, because we don't double count Saturday just because it was a double-failure day.

Figure 3. $(A∪B)-(A∩B)$

Perhaps this is part of a broader theme in mathematics of situations in which things can’t be simplistically added up. Other examples could include vectors that are not in the same direction, numerators of fractions that have different denominators, and dimensionally-incompatible quantities, such as distance and time (Foster, 2019). However, there is something particular about probability for me. I enjoy probability very much, but I think it's the area of mathematics in which I'm most likely to struggle to truly make sense or to be able to explain concepts clearly to others without using technical language and symbols (see Foster, 2021). Rarely when I'm doing a probability calculation do I have a rough ballpark estimate of the answer I should be getting, and, if I obtain an answer like $\frac{127}{351}$, it would be scarcely worth my while converting it to a decimal to see if it looked a 'reasonable' size, as I would have no idea how to tell reasonable from unreasonable. Do other people share this sense?

Questions to reflect on

1. Do you have a better way of explaining why doubling is invalid here?

2. Do you have other examples of 'similar' situations to this?

3. Do you share my sense that probability is often 'harder to explain' than other areas of mathematics?

Notes

1. With apologies for the highly 'middle-class' nature of this 'first-world problem'!

2. The expression $1-(1-p)^2$ can also be obtained intuitively by saying that the required probability is the complement of the probability that both cars are working properly. Since the probability that either car is working properly is $1-p$, the probability that both (assumed independent) work properly is $(1-p)^2$, and so the probability that this is not the case must be $1-(1-p)^2$.

References

Foster, C. (2019). Questions pupils ask: Why can’t it be distance plus time? Mathematics in School, 48(1), 15–17. https://www.foster77.co.uk/Foster,%20Mathematics%20in%20School,%20Why%20can't%20it%20be%20distance%20plus%20time.pdf

Foster, C. (2021). In a spin. Teach Secondary, 10(1), 11. https://www.foster77.co.uk/Foster,%20Teach%20Secondary,%20In%20a%20spin.pdf

12 May 2022

Learning times tables efficiently

Times tables can be a controversial subject. Can we help students to learn their tables in ways that promote conceptual understanding? This is my take on teaching times tables. I imagine there will be some strong opinions...

For many children, learning the times tables feels like a huge mountain to climb. And for those who have tried and feel that they have failed, going back and trying again fills them with dread. Perhaps all seems to go well in the beginning, with the 2s, 5s and 10s, say, but before long we reach the 6s, 7s and 8s, and it feels like every new fact that is mastered displaces an old fact that then becomes lost. As more and more facts are covered, the potential for muddling them up increases (e.g., $7 \times 8$: Is it $54$, or maybe $48$?), until the student really doesn’t have much idea which things they know and which they don’t. In the worst-case scenario, the only thing the child really trusts is skip-counting up from zero every time. And with skip-counting you only have to make one mistake for all of your remaining numbers to be wrong.

Teachers are highly strategic in the order in which they teach the tables: often 2s, 5s, 10s, 4s, … etc. But the effect of this is that the ‘hard stuff’ (6s, 7s, etc.) is delayed, so that when it arrives it can feel overwhelming and as though it is coming at learners far too quickly. I am not sure that learning one table after another like this – however carefully planned the sequence – is ideal (Note 1).

Here is a different way, that tries to build up from the multiplicative connections between the facts and deliberately avoids any addition/subtraction/skip-counting approaches, so as to build on the multiplicative structure of the tables and work more in harmony with that.

At first sight, there are 144 facts to learn:

But, of course, this is highly deceptive, and it is nowhere near as bad as this. Because of the commutativity of multiplication ($a \times b \equiv b \times a$, see Foster, 2022), we can immediately delete nearly half of these facts:

Everyone knows their 1-times table (which is almost as easy as the 0-times table, Note 2), so we can grey those out. And I will assume that the 2s (the even numbers), 5s, 10s and 11s (at least up to $9 \times 11$) are also known, or easily learned, and so I’ve marked those in green below:

So, from the original 144, this now leaves just 30 which need some teaching. And these are the tougher ones. Because of how the picture looks at this point, the best way to tackle this, I think, is not to go table by table (Note 3), but to exploit the structure a bit more strategically. In particular, we want to begin with the highest-leverage multiplication facts – the ones that help most with getting others. When students arrive, say, at secondary school and clearly ‘do not know their tables’, it is basically these 30 that are the problem. Convincing them that their difficulty is not a functionally infinite number of unknown facts but a relatively small number can be helpful. (It really is not like having to memorise a telephone directory!) And starting with the ones most likely to be of immediate help seems to make sense.

Big claim: The most useful of these remaining products to know are the eight squares in red below:

In desperate circumstances, where students have repeatedly tried without success to master tables, I have been known to (reluctantly) settle for just knowing the squares. The beauty of the squares is that they march diagonally through the table, and so they really take you deep in amongst all the difficult facts. If you know the squares, the difficult products you don’t know are often only a step away.

For example, if you know that $8 \times 8 = 64$, then $7 \times 8$ must be $64-8 = 56$.
Or, if you know that $7 \times 7 = 49$, then $7 \times 8$ must be $49+7 = 56$.

So, the squares are really high-leverage facts to know, and I wouldn’t do anything else on the multiplication facts until the student knows these 8 squares. However, I am not really advocating pushing things like $7 \times 8 = 8 \times 8-8$, because students find this reasoning hard (Do I subtract 8 or 7?), and it breaks with the multiplicative theme.

So, instead, I would build differently from the squares:

$6 \times 3$ is half of $6 \times 6$ or double $3 \times 3$
$4 \times 8$ is half of $8 \times 8$ or double $4 \times 4$
$6 \times 12$ is half of $12 \times 12$ or double $6 \times 6$

This is really powerful. Mental doubling and halving may need some work, but that is very important anyway, so I am happy to be dependent on that (see Francome, 2020).

So, now we have three more facts, in orange below:

Knowing that $6 \times 6 = 36$ is the single most powerful fact in the entire tables square, so long as you are able to mentally break down the 6s into 2s and 3s. Students who haven’t had much practice doing this ‘prime decomposition’ find it initially difficult, but this is at the heart of how multiplication works, so is an important awareness, and, with practice, it allows students to see why all the 36s in the table are equal (there are no 'coincidences' in the multiplication table):

$4 \times 9 = (2 \times 2) \times (3 \times 3) = (2 \times 3) \times (2 \times 3) = 6 \times 6 = 36$
So, $8 \times 9 = 2 \times 4 \times 9 = 2 \times 36 = 72$ and
$3 \times 12 = 6 \times 6$ (double the 3, halve the 12) $= 36$

These are in gold below:

Next, I would do 12, 24, 48 and 96. If you learn that $3 \times 4 = 12$ (which most students will know), then $3 \times 8$ (double the 4), $6 \times 4$ (double the 3), $12 \times 4$ (double the 3 twice), $6 \times 8$ (double the 3 and double the 4) and $12 \times 8$ (double three times) all come along without too much trouble if students are fluent doublers - and only the the last one of these involves any 'carrying' when doubling.

This means that when students are stuck on $6 \times 8$, the prompt would not be to count up in 6s or work from the nearest multiple of 6 or 8 they can think of (e.g., $6 \times 10$). It would be: Do you know $ \textit 3 \times \textit 4$? (Both numbers are doubled, so the answer must be 12 double-doubled, which can be done easily mentally, without any 'carrying'.)

These six are in blue below, so, by this point, we have dealt with 20 of the tricky ones and there are just 10 left.

The remaining ones are all ‘hard’, and we need to take time and care over these. I think I would spend 50% of my total energies on these 10.

There is 21, 42, 84 and 63 (in purple below), which all come from $3 \times 7$, which therefore needs to be learned. Then, given $3 \times 7$, we can do $6 \times 7$ (double), $12 \times 7$ (double twice) and $9 \times 7$ (triple). (None of these scalings is hard to do quickly mentally, as none involves any 'carrying'.)

Then there is 28 and 56 (in yellow below), where $8 \times 7$ is just double $4 \times 7$ (which is just double $2 \times 7$).

And then we have 27, 54 and 108 (in pink below), which come from $3 \times 9$, which needs to be learned (perhaps as $3^3$). We have $6 \times 9$ (double) and $12 \times 9$ (double twice).

Which just leaves 132 to remember (or know as $11^2 + 11$, which is possibly easier than double $6 \times 11$). I think this is probably the least connected of all of the multiplication facts, and so perhaps the hardest to remember.

So, in conclusion, this means that the only ones that potentially 'need' memorising are these 12:

$3^2, 4^2, 6^2, 7^2, 8^2, 9^2, 11^2, 12^2, 3 \times 4, 3 \times 7, 3 \times 9$ and $11 \times 12.$

And, if you have the 3-times table, then that reduces this list to just the other 7 squares and $11 \times 12$, which really feels manageable. It does show the power you get from knowing the squares (Note 4).

I think the key to supporting all of this is in the kind of prompts that you provide when a student is stuck. Rather than asking them to figure it out from ‘anything relevant that you know’, or waiting patiently while they skip-count up from zero, with this approach you have a clear plan for how they might be getting from known things to unknown. With practice, figuring out something like $7 \times 8$ by saying ‘double $7 \times 4$, which is 28, so that's 56’ can be extremely quick, and the more you do this the more you are incidentally practising doubling. (And this is one of the trickiest ones, because doubling 28 involves a mental 'carry'.) Of course, nothing will be as fast as ‘just knowing’, but, where that has repeatedly failed for a student, then this kind of approach may help. And I would teach it to everyone for the sake of understanding the multiplicative connections (Note 5). I certainly prefer to spend energy on this than on those one-off mnemonics, like ‘5-6-7-8’ for $56 = 7 \times 8$, which are flukes that don't generalise.

Here is my attempt at a (rather messy) summary of where everything comes from (pdf version):

In conclusion, I am not suggesting that any of this is easy, especially with students who have experienced repeated failure with tables or have developed ‘tables anxiety’. There is no quick, easy fix. And I’m not saying that I think this approach is definitely the best (e.g., I don’t make anything much of the 9s, which can be fairly easy to learn). But, I think that if you work through in this order you at least get the highest-leverage facts (e.g., the squares) before the lower-leverage ones (e.g., $11 \times 12$). However, if you have a better order - or entire approach - please put it in the comments below!