Suppose I choose box A, and on opening it I find that it contains an amount

*x*. I am asked if I wish to switch. What is the expected outcome? It is equally likely that I have chosen the smaller or the larger amount. If I stick with my choice, I gain

*x*. If I switch, my expected gain is

*1/2.(x/2) + 1/2(2x)*, which is

*5x/4*, a higher number, so my rational choice is to switch. But that applies whatever the value of

*x*, so regardless of what I find when I open the box, I am going to switch. So why didn't I just choose box B in the first place? Well, because the same argument would have required me to switch to A on opening that box.

(A variation could allow the sums to be negative, with the same rule. In that case logic tells me I stick if I find that

*x*is negative and switch if

*x*is positive.)

Apparently this paradox was attributed to Schrodinger by the mathematician J.E. Littlewood, who mentions it in his

*Miscellany*..

Professor Barrow addressed the apparent paradox by saying that it assumes that money is infinitely divisible. But that is not the case: if we are dealing in notes and I find that box A contains £1, I know that must be the smaller amount because there is no smaller value of money that box A could contain. That is a valid point, but it doesn't seem to me to be a wholly satisfying resolution.

Suppose the boxes contains not money but quantities of gold, so that there is no limit to how far it can be divided. Then there is no lowest possible value. Of course, in our physical world, matter cannot be divided infinitely, but in an infinite universe with different laws of physics one could postulate that gold is infinitely divisible and then how do we resolve the paradox? I don't quite believe that the Two-Box Paradox is proof that the laws of the Universe must be based on an atomic theory, so that infinite divisibility is impossible.

Coincidentally I had asked students an apparently very different problem (taken from Chris Maslanka's column in the Guardian a few weeks ago). If I choose three consecutive integers randomly, what is the probability that their product is divisible by 48? I'll leave that for you to solve: the answer is in one sense quite straightforward, but in another it is problematic.

What is the probability that a randomly chosen integer is divisible by 7? You might think it is obviously one in 7. One of my mathematical heroes, the group theorist William Burnside, got involved in a bitter argument with R.A. Fisher over exactly this question. If you choose an integer randomly from the set {1,2,3,4,5,6,7,8, 9,10,11,12,13,14,15,...} it looks as if there is a one in seven chance that it is a multiple of 7. But we can also write the set of all integers as {1,7,2,14,3,21,4,28,5,35,6,42,8,49,9,56,10,63,11...}, alternating numbers not divisible by 7 with those which are. This is again the set of positive integers: we can see that every positive integer is listed. But every second integer in the list is divisible by 7, so it looks like the probability that a random integer is a multiple of 7 is as high as 1/2.

Choosing from infinite sets is problematic. It's not that an event with probability 0 cannot happen - it can. Spin a bicycle wheel and let it come to rest: the bottom-most point is a random point of the circumference. There are infinitely many points on the circumference (in a non-atomic universe, anyway) so the probability that the wheel came to rest on that point was 0. So either an event of probability 0 has happened or the wheel cannot stop spinning. And there is no such thing as perpetual motion!

But choosing a random integer from the infinite possibilities presents problems. If all integers are equally likely, then the possibility that the chosen integer

*n*is less than, say, one million, is infinitesimally small. The probability that it is less than one googol, or one googolplex, is still infinitesimally small. It's inconceivable that a random integer could be identified by name (or in any other way) in the lifetime of the universe so far. With probability one,

*n*has more digits than one could write using all the atoms in the universe. So finding out whether

*n*is divisible by 7 is not really a sensible question. The "divisible by 48?" problem I mentioned above needs to be specified much more carefully (and then the intuitive answer is correct).

To work out the true odds in the case of the Two-Box Paradox one needs to know how the contents of the boxes were chosen. The paradox assumes the values are chosen from an infinite range, and that's where the problems lie.

Here's another one. I have chosen a random number in the range [0,1]. What is the probability that it is rational? A pure mathematician would say 0 - there are many more irrational numbers than rational. But I've just tried this with my computer. I'm using Excel: that surely has high enough precision for every practical purpose. In ten million trials I have yet to generate an irrational number. So the theoretical probability is 0 but my estimate from my experiments is 1. Can computer simulation really be that inaccurate?

That may seem a frivolous example but I think it gets to the heart of the Two-Box paradox. Random selection from infinite possibilities needs to be handled very carefully.

A fascinating problem. As you say, I think the paradox arises because the values are chosen from an infinite range, which may not lead to a consistent definition of what choosing at random from this range means. But given the actual value does not need to be chosen randomly, can this give any information about whether it is likely to be the smaller or larger?

ReplyDeleteI've been thinking about similar issues as I'm due to give a short talk about the history of the Secretary Problem at the BSHM Christmas meeting - one of the ongoing debates is whether the game of Googol is in fact the same as the Secretary problem, i.e. if someone is given a free choice of writing positive numbers can you infer from some of the values that are revealed to you anything about their size compared with ones you haven't yet seen?

The puzzle sounds something similar to what Randall Munroe posted a while ago https://plus.google.com/111588569124648292310/posts/dv9Fi45h91T

ReplyDelete>I'm using Excel: that surely has high enough precision for every practical purpose. In ten million trials I have yet to generate an irrational number.

ReplyDeleteThere is probably a more fundamental reason why that is so, but from a very practical standpoint, Excel, with all its randomness is unable to generate an irrational number. In fact, there are many rational numbers it can't generate, either, because there is no binary, non-repeating versions of them to store in whatever format it uses.

Yes, I was making a joke about computers' inability to deal with irrational numbers. It's paradoxical that, even although these devices are so valuable in numerical computation, almost all numbers cannot be represented by a computer!

ReplyDeleteIsn't your assumption that your expected gain is:

ReplyDelete1/2.(x/2) + 1/2(2x) or 5x/4

wrong?

Shouldn't it really be:

1/2*( x+n )+1/2*( x-n ), where n is the difference between the higher and the lower amount.

Which can be simplified to:

x

Which in turn is exactly what we expected?

In other words:

The thought of "doubling up and halving down" feels wrong from the start.

But "gaining or loosing" the difference feels much more natural.

(And the result doesn't create a paradox.)

I think as specified the calculation is correct: the gain from getting the higher amount is twice the loss from getting the lower amount.

ReplyDeletethanks for share..

ReplyDelete