# The missing link for black holes

There has been some recent excitement about the claimed identification of a 400-solar-mass black hole. A team of scientists have recently published a letter in the journal Nature where they show how X-ray measurements of a source in the nearby galaxy M82 can be interpreted as originating from a black hole with mass of around 400 times the mass of the Sun—from now on I’ll use $M_\odot$ as shorthand for the mass of the Sun (one solar mass). This particular X-ray source is peculiarly bright and has long been suspected to potentially be a black hole with a mass around $100 M_\odot$ to $1000 M_\odot$. If the result is confirmed, then it is the first definite detection of an intermediate-mass black hole, or IMBH for short, but why is this exciting?

## Mass of black holes

In principle, a black hole can have any mass. To form a black hole you just need to squeeze mass down into a small enough space. For the something the mass of the Earth, you need to squeeze down to a radius of about 9 mm and for something about the mass of the Sun, you need to squeeze to a radius of about 3 km. Black holes are pretty small! Most of the time, things don’t collapse to form black holes because they materials they are made of are more than strong enough to counterbalance their own gravity.

These innocent-looking marshmallows could collapse down to form black holes if they were squeezed down to a size of about 10−29 m. The only thing stopping this is the incredible strength of marshmallow when compared to gravity.

### Stellar-mass black holes

Only very massive things, where gravitational forces are immense, collapse down to black holes. This happens when the most massive stars reach the end of their lifetimes. Stars are kept puffy because they are hot. They are made of plasma where all their constituent particles are happily whizzing around and bouncing into each other. This can continue to happen while the star is undergoing nuclear fusion which provides the energy to keep things hot. At some point this fuel runs out, and then the core of the star collapses. What happens next depends on the mass of the core. The least massive stars (like our own Sun) will collapse down to become white dwarfs. In white dwarfs, the force of gravity is balanced by electrons. Electrons are rather anti-social and dislike sharing the same space with each other (a concept known as the Pauli exclusion principle, which is a consequence of their exchange symmetry), hence they put up a bit of a fight when squeezed together. The electrons can balance the gravitational force for masses up to about $1.4 M_\odot$, known as the Chandrasekhar mass. After that they get squeezed together with protons and we are left with a neutron star. Neutron stars are much like giant atomic nuclei. The force of gravity is now balanced by the neutrons who, like electrons, don’t like to share space, but are less easy to bully than the electrons. The maximum mass of a neutron star is not exactly known, but we think it’s somewhere between $2 M_\odot$ and $3 M_\odot$. After this, nothing can resist gravity and you end up with a black hole of a few times the mass of the Sun.

Collapsing stars produce the imaginatively named stellar-mass black holes, as they are about the same mass as stars. Stars lose a lot of mass during their lifetime, so the mass of a newly born black hole is less than the original mass of the star that formed it. The maximum mass of stellar-mass black holes is determined by the maximum size of stars. We have good evidence for stellar-mass black holes, for example from looking at X-ray binaries, where we see a hot disc of material swirling around the black hole.

### Massive black holes

We also have evidence for another class of black holes: massive black holes, MBHs to their friends, or, if trying to sound extra cool, supermassive black holes. These may be $10^5 M_\odot$ to $10^9 M_\odot$. The strongest evidence comes from our own galaxy, where we can see stars in the centre of the galaxy orbiting something so small and heavy it can only be a black hole.

We think that there is an MBH at the centre of pretty much every galaxy, like there’s a hazelnut at the centre of a Ferrero Rocher (in this analogy, I guess the Nutella could be delicious dark matter). From the masses we’ve measured, the properties of these black holes is correlated with the properties of their surrounding galaxies: bigger galaxies have bigger MBHs. The most famous of these correlations is the M–sigma relation, between the mass of the black hole ($M$) and the velocity dispersion, the range of orbital speeds, of stars surrounding it (the Greek letter sigma, $\sigma$). These correlations tell us that the evolution of the galaxy and it’s central black hole are linked somehow, this could be just because of their shared history or through some extra feedback too.

MBHs can grow by accreting matter (swallowing up clouds of gas or stars that stray too close) or by merging with other MBHs (we know galaxies merge). The rather embarrassing problem, however, is that we don’t know what the MBHs have grown from. There are really huge MBHs already present in the early Universe (they power quasars), so MBHs must be able to grow quickly. Did they grow from regular stellar-mass black holes or some form of super black hole that formed from a giant star that doesn’t exist today? Did lots of stellar-mass black holes collide to form a seed or did material just accrete quickly? Did the initial black holes come from somewhere else other than stars, perhaps they are leftovers from the Big Bang? We don’t have the data to tell where MBHs came from yet (gravitational waves could be useful for this).

### Intermediate-mass black holes

However MBHs grew, it is generally agreed that we should be able to find some intermediate-mass black holes: black holes which haven’t grown enough to become IMBHs. These might be found in dwarf galaxies, or maybe in globular clusters (giant collections of stars that formed together), perhaps even in the centre of galaxies orbiting an MBH. Finding some IMBHs will hopefully tell us about how MBHs formed (and so, possibly about how galaxies formed too).

IMBHs have proved elusive. They are difficult to spot compared to their bigger brothers and sisters. Not finding any might mean we’d need to rethink our ideas of how MBHs formed, and try to find a way for them to either be born about a million times the mass of the Sun, or be guaranteed to grow that big. The finding of the first IMBH tells us that things are more like common sense would dictate: black holes can come in the expected range of masses (phew!). We now need to identify some more to learn about their properties as a population.

In conclusion, black holes can come in a range of masses. We know about the smaller stellar-mass ones and the bigger massive black holes. We suspect that the bigger ones grow from smaller ones, and we now have some evidence for the existence of the hypothesised intermediate-mass black holes. Whatever their size though, black holes are awesome, and they shouldn’t worry about their weight.

# An introduction to probability: Great expectations

We use probabilities a lot in science. Previously, I introduced the concept of probabilities, here I’ll explain the concept of expectation and averages. Expectation and average values are one of the most useful statistics that we can construct from a probability distribution. This post contains a traces of calculus, but is peanut free.

## Expectations

Imagine that we have a discrete set of numeric outcomes, such as the number from rolling a die. We’ll label these as $x_1$, $x_2$, etc., or as $x_i$ where the subscript $i$ is used as shorthand to indicate any of the possible outcomes. The probability of the numeric value being a particular $x_i$ is given by $P(x_i)$. For rolling our dice, the outcomes are one to six ($x_1 =1$, $x_2 = 2$, etc.) and the probabilities are

$\displaystyle P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = \frac{1}{6}$.

The expectation value is the sum of all the possible outcomes multiplied by their respective probabilities,

$\displaystyle \langle x \rangle = \sum_i x_i P(x_i)$,

where $\sum_i$ means sum over all values of $i$ (over all outcomes). The expectation value for rolling a die is

$\displaystyle \langle x \rangle = 1 \times P(1) + 2 \times P(2) + 3 \times P(3) + 4 \times P(4) + 5 \times P(5) + 6 \times P(6) = \frac{7}{2}$.

The expectation value of a distribution is its average, the value you’d expect after many (infinite) repetitions. (Of course this is possible in reality—you’d certainly get RSI—but it is useful for keeping statisticians out of mischief).

For a continuous distribution, the expectation value is given by

$\displaystyle \langle x \rangle = \int x p(x) \, \mathrm{d} x$,

where $p(x)$ is the probability density function.

You can use the expectation value to guide predictions for the outcome. You can never predict with complete accuracy (unless there is only one possible outcome), but you can use knowledge of the probabilities of the various outcomes the inform your predictions.

Imagine that after buying a large quantity of fudge, for entirely innocent reasons, the owner offers you the chance to play double-or-nothing—you’ll either pay double the bill or nothing, based on some random event—should you play?  Obviously, this depends upon the probability of winning. Let’s say that the probability of winning is $p$ and find out how high it needs to be to be worthwhile. We can use the expectation value to calculate how much we should expect to pay, if this is less than the bill as it stands, it’s worth giving it a go, if the expectation value is larger than the original bill, we should expect to pay more (and so probably shouldn’t play). The expectation value is

$\displaystyle \langle x \rangle = 0 \times (1 - p) + 2 \times p = 2 p$,

where I’m working in terms of unified fudge currency, which, shockingly, is accepted in very few shops, but has the nice property that your fudge bill is always 1. Anyhow, if $\langle x \rangle$ is less than one, so if $p < 0.5$, it’s worth playing. If we were tossing a (fair) coin, we’d expect to come out even, if we had to roll a six, we’d expect to pay more.

The expectation value is the equivalent of the mean. This is the average that people usually think of first. If you have a set of numeric results, you calculate the mean by adding up all or your results and dividing by the total number of results $N$. Imagine each outcome $x_i$ occurs $n_i$ times, then the mean is

$\displaystyle \bar{x} = \sum_i x_i \frac{n_i}{N}$.

We can estimate the probability of each outcome as $P(x_i) = n_i/N$ so that $\bar{x} = \langle x \rangle$.

## Median and mode

Aside from the mean there are two other commonly used averages, the median and the mode. These aren’t quite as commonly used, despite sounding friendlier. With a set of numeric results, the median is the middle result and the mode is the most common result. We can define equivalents for both when dealing with probability distributions.

To calculate the median we find the value where the total probability of being smaller (or bigger) than it is a half: $P(x < x_\mathrm{med}) = 0.5$. This can be done by adding up probabilities until you get a half

$\displaystyle \sum_{x_i \, \leq \, x_\mathrm{med} } P(x_i) = 0.5$.

For a continuous distribution this becomes

$\displaystyle \int_{x_\mathrm{low}}^{x_\mathrm{med}} p(x) \, \mathrm{d}x = 0.5$,

where $x_\mathrm{low}$ is the lower limit of the distribution. (That’s all the calculus out of the way now, so if you’re not a fan you can relax). The LD50 lethal dose is a median. The median is effectively the middle of the distribution, the point at which you’re equally likely to be higher or lower.

The median is often used as it is not as sensitive as the mean to a few outlying results which are far from the typical values.

The mode is the value with the largest probability, the most probable outcome

$\displaystyle P(x_\mathrm{mode}) = \max P(x_i)$.

For a continuous distribution, it is the point which maximises the probability density function

$\displaystyle p(x_\mathrm{mode}) = \max p(x)$.

The modal value is the most probable outcome, the most likely result, the one to bet on if you only have one chance.

## Education matters

Every so often, some official, usually an education minister, says something about wanting more than half of students to be above average. This results in much mocking, although seemingly little rise in the standards for education ministers. Having discussed averages ourselves, we can now see if it’s entirely fair to pick on these poor MPs.

The first line of defence, is that we should probably specify the distribution we’ve averaging. It may well be that they actually meant the average bear. It’s a sad truth that bears perform badly in formal education. Many blame the unfortunate stereotyping resulting from Winnie the Pooh. It might make sense to compare with performance in the past to see if standards are improving. We could imagine that taking the average from the 1400s would indeed show some improvement. For argument’s sake, let’s say that we are indeed talking about the average over this year’s students.

If the average we were talking about was the median, then it would be impossible for more (or fewer) than half of students to do better than average. In the case, it is entirely fair to mock the minister, and possibly to introduce them to the average bear. In this case, a mean bear.

If we were referring to the mode, then it is quite simple for more than half of the students to do better than this. To achieve this we just need a bottom-heavy distribution, a set of results where the most likely outcome is low, but most students do better than this. We might want to question an education system where the aim is to have a large number of students doing poorly though!

Finally, there is the mean; to use the mean, we first have to decide if we have a sensible if we are averaging a sensible quantity. For education performance this normally means exam results. Let’s sidestep the issue of if we want to reduce the output of the education system down to a single number, and consider the properties we want in order to take a sensible average. We want the results to be numeric (check); to be ordered, such that high is good and low is bad (or vice versa) so 42 is better than 41 but not as good as 43 and so on (check), and to be on a linear scale. The last criterion means that performance is directly proportional to the mark: a mark twice as big is twice as good. Most exams I’ve encountered are not like this, but I can imagine that it is possible to define a mark scheme this way. Let’s keep imagining, and assume things are sensible (and perhaps think about kittens and rainbows too… ).

We can construct a distribution where over half of students perform better than the mean. In this case we’d really need a long tail: a few students doing really very poorly. In this case, these few outliers are enough to skew the mean and make everyone else look better by comparison. This might be better than the modal case where we had a glut of students doing badly, as now we can have lots doing nicely. However, it also means that there are a few students who are totally failed by the system (perhaps growing up to become a minister for education), which is sad.

In summary, it is possible to have more than 50% of students performing above average, assuming that we are not using the median. It’s therefore unfair to heckle officials with claims of innumeracy. However, for these targets to be met requires lots of students to do badly. This seems like a poor goal. It’s probably better to try to aim for a more sensible distribution with about half of students performing above average, just like you’d expect.

# On symmetry

Dave Green only combs half of his beard, the rest follows by symmetry. — Dave Green Facts

Physicists love symmetry! Using symmetry can dramatically simplify a problem. The concept of symmetry is at the heart of modern theoretical physics and some of the most beautiful of scientific results.

In this post, I’ll give a brief introduction to how physicists think about symmetry. Symmetry can be employed in a number of ways when tackling a problem; we’ll have a look at how they can help you ask the right question and then check that your answer makes sense. In a future post I hope to talk about Noether’s Theorem, my all-time favourite result in theoretical physics, which is deeply entwined with the concept of symmetry. First, we shall discuss what we mean when we talk about symmetry.

## What is symmetry?

We say something is symmetric with respect to a particular operation if it is unchanged after that operation. That might sound rather generic, but that’s because the operation can be practically anything. Let’s consider a few examples:

• Possibly the most familiar symmetry would be reflection symmetry, when something is identical to its mirror image. Something has reflection symmetry if it is invariant under switching left and right. Squares have reflection symmetry along lines in the middle of their sides and along their diagonals, rectangles only have reflection symmetry along the lines in the middle of their sides, and circles have reflection symmetry through any line that goes through their centre.
The Star Trek Mirror Universe actually does not have reflection symmetry with our own Universe. First, they switch good and evil, rather than left and right, and second, after this transformation, we can tell the two universes apart by checking Spock’s beard.
• Rotational symmetry is when an object is identical after being rotated. Squares are the same after a 90° rotation, rectangles are the same after a 180° rotation, and circles are the same after a rotation by any angle. There is a link between the rotational symmetry of these shapes and their mirror symmetry: you can combine two reflections to make a rotation. With rotations we have seen that symmetries can either be discrete, as for a square when we have to rotate by multiples of 90°, or continuous, as for the circle where we can pick any angle we like.
• Translational symmetry is similar to rotational symmetry, but is when an object is the same when shifted along a particular direction. This could be a spatial direction, so shifting everything to the left, or in time. This are a little more difficult to apply to the real world than the simplified models that physicists like to imagine.
For translational invariance, imagine an infinite, flat plane, the same in all directions. This would be translational invariant in any direction parallel to the ground. It would be a terrible place to lose your keys. If you can imagine an infinite blob of tangerine jelly, that is entirely the same in all directions, we can translate in any direction we like. We think the Universe is pretty much like this on the largest scales (where details like galaxies are no longer important), except, it’s not quite as delicious.
The backgrounds in some Scooby-Doo cartoons show periodic translational invariance: they repeat on a loop, so if you translate by the right amount they are the same. This is a discrete symmetry, just like rotating my a fixed angle. Similarly, if you have a rigid daily routine, such that you do the same thing at the same time every day, then your schedule is symmetric with respect to a time translation of 24 hours.
• Exchange symmetry is when you can swap two (or more) things. If you are building a LEGO model, you can switch two bricks of the same size and colour and end up with the same result, hence it is symmetric under the exchange of those bricks. The idea that we have the same physical system when we swap two particles, like two electrons, is important in quantum mechanics. In my description of translational symmetry, I could have equally well have used lime jelly instead of tangerine, or even strawberry, hence the argument is symmetric under exchange of flavours. The symmetry is destroyed should we eat the infinite jelly Universe (we might also get stomach ache).
Mario and Luigi are not symmetric under exchange, as anyone who has tried to play multiplayer Super Mario Bros. will know, as Luigi is the better jumper and has the better moustache.

There are lots more potential symmetries. Some used by physicists seem quite obscure, such as Lorentz symmetry, but the important thing to remember is that a symmetry of a system means we get the same thing back after a transformation.

Sometimes we consider approximate symmetries, when something is almost the same under a transformation. Coke and Pepsi are approximately exchange symmetric: try switching them for yourself. They are similar, but it is possible to tell them apart. The Earth has approximate rotational symmetry, but it is not exact as it is lumpy. The spaceship at the start of Spaceballs has approximate translational invariance: it just keeps going and going, but the symmetry is not exact as it does end eventually, so the symmetry only applies to the middle.

## How to use symmetry

When studying for an undergraduate degree in physics, one of the first things you come to appreciate is that some coordinate systems make problems much easier than others. Coordinates are the set of numbers that describe a position in some space. The most familiar are Cartesian coordinates, when you use $x$ and $y$ to describe horizontal and vertical position respectively. Cartesian coordinates give you a nice grid with everything at right-angles. Undergrad students often like to stick with Cartesian coordinates as they are straight-forward and familiar. However, they can be a pain when describing a circle. If we want to plot a line five units from the origin of of coordinate system $(0,\,0)$, we have to solve $\sqrt{x^2 + y^2} = 5$. However, if we used a polar coordinate system, it would simply be $r = 5$. By using coordinates that match the symmetry of our system we greatly simplify the problem!

Pirates are trying to figure out where they buried their treasure. They know it’s 5 yarrrds from the doughnut. Calculating positions using Cartesian coordinates is difficult, but they are good for specifying specific locations, like of the palm tree.

Using polar coordinates, it is easy to specify the location of points 5 yarrrds from the doughnut. Pirates prefer using the polar coordinates, they really like using r.

Picking a coordinate system for a problem should depend on the symmetries of the system. If we had a system that was translation invariant, Cartesian coordinates are the best to use. If the system was invariant with respect to translation in the horizontal direction, then we know that our answer should not depend on $x$. If we have a system that is rotation invariant, polar coordinates are the best, as we should get an answer that doesn’t depend on the rotation angle $\varphi$. By understanding symmetries, we can formulate our analysis of the problem such that we ask the best questions.

At the end of my undergrad degree, my friends and I went along to an awards ceremony. I think we were hoping they’d have the miniature éclairs they normally had for special occasions. There was a chap from an evil corporation™ giving away branded clocks, that apparently ran on water. We were fairly convinced there was more to it than that, so, as now fully qualified physicists, we though we should able to figure it out. We quickly came up with two ideas: that there was some powder inside the water tank that reacted with the water to produce energy, or that the electrodes reacted in a similar way to in a potato clock. We then started to argue about how to figure this out. At this point, Peter Littlewood, then head of the Cavendish Laboratory, wandered over. We explained the problem, but not our ideas. Immediately, he said that it must be to do with the electrodes due to symmetry. Current flows to power the clock. It’ll either flow left to right through the tank, or right to left. It doesn’t matter which, but the important thing is the clock can’t have reflection symmetry. If it did, there would be no preferred direction for the current to flow. To break the symmetry, the two (similar looking) electrodes must actually be different (and hence the potato clock theory is along the right lines). My friends and I all felt appropriately impressed and humbled, but it served as a good reminder that a simple concept like symmetry can be a powerful tool.

A concept I now try to impress upon my students, is to use symmetry to guide their answers. Most are happy enough to use symmetry for error checking: if the solution is meant to have rotational symmetry and their answer depends on $\varphi$ they know they’ve made a mistake. However, symmetry can sometimes directly tell you the answer.

Lets imagine that you’ve baked a perfectly doughnut, such that it has rotational symmetry. For some reason you sprinkled it with an even coating of electrons instead of hundreds and thousands. We now want to calculate the electric field surrounding the doughnut (for obvious reasons). The electric field tells us which way charges are pushed/pulled. We’d expect positive charges to be attracted towards our negatively charged doughnut. There should be a radial electric field to pull positive charges in, but since it has rotational symmetry, there shouldn’t be any field in the $\varphi$ direction, as there’s now reason for charges to be pulled clockwise or anticlockwise round our doughnut. Therefore, we should be able to write down immediately that the electric field in the $\varphi$ direction is zero, by symmetry.

Most undergrads, though, will feel that this is cheating, and will often attempt to do all the algebra (hopefully using polar coordinates). Some will get this wrong, although there might be a few who are smart enough to note that their answer must be incorrect because of the symmetry. If symmetry tells you the answer, use it! Although it is good to practise your algebra (you get better by training), you can’t learn anything more than you already knew by symmetry. Working efficiently isn’t cheating, it’s smart.

Symmetry is a useful tool for problem solving, and something that everyone should make use of.

# An introduction to probability: Leaving nothing to chance

## Probabilities and science

Understanding probabilities is important in science. Once you’ve done an experiment, you need to be able to extract from your data information about your theory. Only rarely do you get a simple yes or no: most of the time you have to work with probabilities to quantify your degree of certainty. I’ll (probably) be writing about probabilities in connection with my research, so I thought it would be useful to introduce some of the concepts.

I’ll be writing a series of posts, hopefully going through from the basics to the limits of my understanding. We’ll begin with introducing the concept of probability. There’s a little bit of calculus, but you can skip that without effecting the rest, just remember you can’t grow up to be big and strong if you don’t finish your calculus.

## What is a probability?

A probability describes the degree of belief that you have in a proposition. We talk about probabilities quite intuitively: there are some angry-looking, dark clouds overhead and I’ve just lit the barbecue, so it’s probably going to rain; it’s more likely that United will win this year’s sportsball league than Rovers, or it’s more credible that Ted is exaggerating in his anecdote than he actually ate that much fudge…

We formalise the concept of a probability, so that it can be used in calculations, by assigning them numerical values (not by making them wear a bow-tie, although that is obviously cool). Conventionally, we use 0 for impossible, 1 for certain and the range in between for intermediate probabilities. For example, if we were tossing a coin, we might expect it to be heads half the time, hence the probability of heads is $P(\mathrm{head}) = 1/2$, or if rolling a die, the probability of getting a six is $P(6) = 1/6$.

For both the coin and the die we have a number of equally probable outcomes: two for the coin (heads and tails) and six for the die (1, 2, 3, 4, 5 and 6). This does not have to be the case: imagine picking a letter at random from a sample of English text. Some letters are more common than others—this is why different letters have different values in Scrabble and why hangman can be tricky. The most frequent letter is “e”, the probability of picking it is about 0.12, and the least frequent is “z”, the probability of picking that is just 0.0007.

Often we consider a parameter that has a continuous range, rather than discrete values (as in the previous examples). For example, I might be interested in the mass of a black hole, which can have any positive value. We then use a probability density function $p(x)$ such that the probability for the parameter lies in the range $a \leq x \leq b$ is given by the integral

$\displaystyle P(a \leq x \leq b) = \int_a^b p(x)\, \mathrm{d}x$.

Performing an integral is just calculating the area under a curve, it can be thought of a the equivalent of adding up an infinite number of infinitely closely spaced slices. Returning to how much fudge Ted actually ate, we might to find the probability that he a mass of fudge $m$ that was larger than zero, but smaller than the fatal dose $M$. If we a had probability density function $p(m)$, we would calculate

$\displaystyle P(0 < m \leq M) = \int_0^{M} p(m)\, \mathrm{d}m$.

The probability density is largest where the probability is greatest and smallest where the probability is smallest, as you’d expect. Calculating probabilities and probability distributions is, in general, a difficult problem, it’s actually what I spend a lot of my time doing. We’ll return to calculating probabilities later.

## Combining probabilities

There are several recipes for combining probabilities to construct other probabilities, just like there are recipes to combine sugar and dairy to make fudge. Admittedly, probabilities are less delicious than fudge, but they are also less likely to give you cavities. If we have a set of of disjoint outcomes, we can work out the probability of that set by adding up the probabilities of the individual outcomes. For example, when rolling our die, the probability of getting an even number is

$\displaystyle P(\mathrm{even}) = P(2) + P(4) + P(6) = \frac{1}{6} +\frac{1}{6} +\frac{1}{6} = \frac{1}{2}$.

(This is similar to what we’re doing when integrating up the probability density function for continuous distributions: there we’re adding up the probability that the variable $x$ is in each infinitesimal range $\mathrm{d}x$).

If we have two independent events, then the probability of both of them occurring is calculated by multiplying the two individual probabilities together. For example, we could consider the probability of rolling a six and the probability of Ted surviving eating the lethal dose of fudge, then

$\displaystyle P(\mathrm{6\: and\: survive}) = P(6) \times P(\mathrm{survive})$.

The most commonly quoted quantity for a lethal dose is the median lethal dose or LD50, which is the dose that kills half the population, so we can take the probability of surviving to be 0.5. Thus,

$\displaystyle P(\mathrm{6\: and\: survive}) = P(6) \times P(\mathrm{survive}) = \frac{1}{12}$.

Events are independent if they don’t influence each other. Rolling a six shouldn’t influence Ted’s medical condition, and Ted’s survival shouldn’t influence the roll of a die, so these events are independent.

Things are more interesting when events are not independent. We then have to deal with conditional probabilities: the conditional probability $P(\mathrm{A}|\mathrm{B})$ is the probability of $\mathrm{A}$ given that $B$ is true. For example, if I told you that I rolled an even number, the probability of me having rolled a six is $P(6|\mathrm{even}) = 1/3$. If I told you that I have rolled a six, then the probability of me having rolled an even number is $P(\mathrm{even}|6) = 1$—it’s a dead cert, so bet all your fudge on that! When combining probabilities from dependent events, we chain probabilities together in a logical chain. The probability of rolling a six and an even number is the probability of rolling an even number multiplied by the probability of rolling a six given that I rolled an even number

$\displaystyle P(\mathrm{6\: and\: even}) = P(6|\mathrm{even}) \times P(\mathrm{even})= \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$,

or equivalently the probability of rolling six multplied by the probability of rolling an even number given that I rolled a six

$\displaystyle P(\mathrm{6\: and\: even}) = P(\mathrm{even} | 6) \times P(6) = 1 \times \frac{1}{6} = \frac{1}{6}$.

Reassuringly, we do get the same answer. This is a bit of a silly example, as we know that if we’ve rolled a six we have rolled an even number, so all we are doing if calculating the probability of rolling a six.

We can use conditional probabilities for independent events: this is really easy as the conditional probability is just the straight probability. The probability of Ted surviving his surfeit of fudge given that I rolled a six is just the probability of him surviving, $P(\mathrm{survive}|6) = P(\mathrm{survive})$.

Let’s try a more complicated example, let’s imagine that Ted is playing fudge roulette. This is like Russian roulette, except you roll a die and if it comes up six, then you have to eat the lethal dose of fudge. His survival probability now depends on the roll of the die. We want to calculate the probability that Ted will live to tomorrow. If Ted doesn’t roll a six, we’ll assume that he has a 100% survive rate (based on that one anecdote where he claims to have created a philosopher’s stone by soaking duct tape in holy water), this isn’t quite right, but is good enough. The probability of Ted surviving given he didn’t roll a six is

$\displaystyle P(\mathrm{not\: 6\: and\: survive}) = P(\mathrm{survive} | \mathrm{not\: 6}) \times P(\mathrm{not\: 6}) = 1 \times \frac{5}{6} = \frac{5}{6}$.

The probability of Ted rolling a six (and eating the fudge) and then surviving is

$\displaystyle P(\mathrm{6\: and\: survive}) = P(\mathrm{survive} | \mathrm{6}) \times P(\mathrm{6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.

We have two disjoint outcomes (rolling a six and survivng, and not rolling a six and surving), so the total probability of surviving is given by the sum

$\displaystyle P(\mathrm{survive}) =P(\mathrm{not\: 6\: and\: survive}) +P(\mathrm{6\: and\: survive}) = \frac{5}{6} +\frac{1}{12} =\frac{11}{12}$.

It seems likely that he’ll make it, although fudge roulette is still a dangerous game!

There’s actually an easier way of calculating the probability that Ted survives. There are only two possible outcomes: Ted survives or he doesn’t. Since one of these must happen, their probabilities must add to one: the survive probability is

$P(\mathrm{survive}) = 1 - P(\mathrm{not\: survive})$.

We’ve already seen this, as we’ve used the probability of not rolling a six is$P(\mathrm{not\: 6}) = 1 - P(6) = 5/6$. The probability of not surviving is much easier to work out as there’s only one way that can happen: rolling a six and then overdosing on fudge. The probability is

$\displaystyle P(\mathrm{not\: surviving}) = P(\mathrm{fudge\: overdose}|6) \times P(6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$,

and so the survival probability is $P(\mathrm{survive}) = 1 - 1/12 = 11/12$, exactly as before, but in fewer steps.

In a future post we’ll try working out the probability that Ted did eat a lethal dose of fudge given that he is alive to tell the anecdote. This is known as an inverse problem, and is similar to what scientists do all the time. We do experiments and get data, then we need to work out the probability of our theory (that Ted ate the fudge) being correct given the data (that he’s still alive).

## Interpreting probabilities

We have now discussed what a probability is and how we can combine them. We should now think about how to interpret them. It’s easy enough to understand that a probability of 0.05 means that we expect something should happen on average once in 20 times, and that it is more probable than something with a probability of 0.01, but less likely than something with a probability of 0.10. However, we are not good at having an intuitive understanding of probabilities.

Consider the case that a scientist announces a result with 95% confidence. That sounds pretty good. Think how surprised you would be (assuming that their statistics are all correct) that the result was wrong. I feel like I would be pretty surprised. Now consider rolling tow dice, how surprised would you be if you rolled two sixes? The probability of the result being wrong is $1 - 0.95 = 0.05$, or one in twenty. The probability of rolling two sixes is $1/6 \times 1/6 = 1/36$ or about one in forty. Hence, you should be almost twice as surprised by rolling double six as for a 95% confidence-level result being incorrect.

When dealing with probabilities, I find it useful to make a comparison to something familiar. While Ted is more likely than not to survive fudge roulette, there is a one is twelve chance of dying. That’s three times as likely as rolling a double six, or equally probable as rolling a six and getting heads. That’s riskier than I’d like, so I’m going to stick to consuming fudge in moderation.