Monday, 5 April 2010


When we say that something will probably happen, we may not be aware that this is a vague statement lacking in mathematical precision. In other words, how probable is ‘probable’? Clearly, the minimum standard required for this adjective to be applicable is that an event be more likely to happen than not to happen, and this may be the criterion that most people instinctively use, but a good case can be made for reserving ‘probably’ for events that have a far greater likelihood of occurring than 50 percent.

In fact, most people have only the sketchiest notion of probability: for example, anyone who gambles in a casino will probably lose money. And everyone will lose eventually if they play long enough. Yet gamblers tend to think that a long losing streak will be compensated for by a long winning streak. Some may even believe that they have a ‘system’, especially for roulette, wantonly disregarding the built-in bias in favour of the house, which for roulette in an American casino gives a long-term return of $94.74 for every $100 gambled. It is worth pointing out, although it tends to be forgotten nowadays, that the original purpose of a casino, notably the one in Monte Carlo, was not as a place where one could win money, even if some did, but as a place where one could be seen to lose money (if a player lost, say, $20,000, it demonstrated to those who witnessed the loss that here was someone who could afford to lose such an amount—an early if extreme example of conspicuous consumption).

Probability theory throws up some interesting cases, particularly with regard to inherently improbable events. For example, the universe consists predominantly of hydrogen and helium, heavier elements making up a very small fraction of the total mass. There is only one place where these heavier elements are formed: in the cores of super-massive stars such as red giants. Anyway, to cut a complicated technical story down to more manageable proportions, carbon is formed when three helium nuclei collide within a timescale that is measurable in fractions of a nanosecond (one nanosecond equals one million millionth of a second). If the third nucleus arrives after the window of opportunity has closed, then the result of the fusion of the first two nuclei, a highly unstable isotope of beryllium, will have decayed again to helium. However, despite the intrinsic unlikelihood of such an event, the opportunities for successful three-way fusion are so vast that given sufficient time carbon will form in significant quantities. It is thus possible to say with complete confidence that every carbon atom in every protein molecule in my body, or your body, and every carbon atom in the billions of tons of carbon dioxide that have been carelessly added to our atmosphere, there to contribute to the greenhouse effect and global warming, was formed in the centre of a star in the improbable process just described.

Mention of unstable (i.e. radioactive) isotopes recalls an interesting application of probability theory involving quantum mechanics. Every radioactive isotope has a fundamental property known as its half-life. This is the time taken for precisely half the atoms in a given sample to decay, either to another radioactive isotope or to a stable one. In other words, it is possible to say exactly how many atoms will decay in a given time; however—and this is the odd part—it is not possible to tap a given atom on the shoulder and say: “okay, you’re next.” This is the uncertainty that is built into quantum mechanics.

Back in the real world, a good test of how well you understand probability is the following problem, which elicited a great deal of discussion on BBC Radio 4’s flagship news and current affairs program Today a few years ago and as far as I can recall was never satisfactorily explained.

Imagine that you are a participant in a television game show in which you are shown three locked boxes. You are told that one of the three contains $10,000, while the other two are empty. You are allowed to select one box, and if you choose correctly the money is yours. Let us refer to your selection as box #1. Now, before you are allowed to open your chosen box, the show’s compere, who knows in advance which box contains the money, opens one of the other boxes (call this box #2) to show that it is empty. Now comes the offer: do you want to stick with box #1, or would you prefer to change your mind and choose box #3?

At this time, I do not propose to provide a detailed analysis of the problem, but I will point out in passing that if you know what you’re doing, you will now choose box #3. Why?


  1. i guess i don't know what i'm doing..don't you have a 50/50 chance with either one of them?

  2. Sarah, I’m afraid you're wrong. I plan to provide a detailed analysis in due course, but I’d like to see what other readers make of the problem first.

  3. You switch box. When you first chose, you had a 33% chance of choosing the correct box. If you don't switch box, you're still working with the same probability, I'd assume.
    However, if you switch box, you have a 50% chance of choosing the right box.

  4. Correct E.Y. Care to attempt a more detailed analysis?

  5. Your first choice has a one-third chance of being correct, so one-third of the time the host has two boxes to choose from. However, two-thirds of the time, the host has no choice regarding which box to open, because the other unselected box contains the money. Therefore, two-thirds of the time, the box that the host doesn't choose will contain the money.


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