All of JackOfAllTrades's Comments + Replies

Nothing complex is a black box , because it has components, which can potentially be understood. Nothing artificial is a black box to the person who built it. An LLM is , of course, complex and artificial. What justifies that claim? I wasn't arguing on that basis.

Here's the general calculation. Take any probability distribution defined on the set of all values (x,y) where x and y are non-negative reals and y=2x. It can be discrete, continuous, or a mixture. Let p(x) be the marginal distribution over x. This method of defining p avoids the distinction between choosing x and then doubling it, or choosing y and then halving it, or any other method of choosing (x,y) such that y=2x. Assume that p has an expected value, denoted by E(p). The expected value of switching when the amount in the first envelope is in the range [x−dx/2,x+dx/2] consists of two parts: (i) The first envelope contains the smaller amount. This has probability p(x)dx/2+O(dx2). The division by 2 comes from the 50% chance of choosing the envelope with the smaller amount. (ii) The first envelope contains the larger amount. This has probability p(x/2)dx/4+O(dx2). The extra factor of 2 comes from the fact that when the contents are in an interval of length dx, half of that (the amount chosen by the envelope-filler) is in an interval of length dx/2. In the two cases the gain from switching is respectively x+O(dx) or −x/2+O(dx). The expected gain given the contents is therefore xp(x)/2−(x/2)p(x/2)/4+O(dx). Multiply this by dx, let dx tend to 0 (eliminating the term in O(dx2)) and integrate over the real line: ∫∞0(xp(x)/2−(x/2)p(x/2)/4)dx =∫∞0xp(x)/2dx−∫∞0(x/2)p(x/2)/4dx The first integral is E(p)/2. In the second, substitute y=x/2 (therefore dx=2dy), giving ∫∞0yp(y)/2dy=E(p)/2. The two integrals cancel.

To follow a maxim of Edwin Jaynes, when a paradox arises in matters of probability, one must consider the generating process from which the probabilities were derived. How does the envelope-filler choose the amount to put in either envelope? He cannot pick an "arbitrary" real number. Almost all real numbers are so gigantic as to be beyond human comprehension. Let us suppose that he has a probability distribution over the non-negative reals from which he draws a single value x, and puts x into one envelope and 2x into the other. (One could also imagine that he puts x/2 into the other, or tosses a coin to decide between 2x and x/2, but I'll stick with this method.) Any such probability distribution must tail off to zero as x becomes large. Suppose the envelope-chooser is allowed to open the first envelope, and then is allowed to switch to the other one if they think it's worth switching. The larger the value they find in the first envelope, the less likely it is that the other envelope has twice as much. Similarly, if they find a very small value in the first envelope (i.e. well into the lower tail of the distribution), then they can expect to profit by switching. In the original version, of course, they do not see what is in the envelope before deciding whether to switch. So we must consider the expected value of switching conditional on the value in the first envelope, summed or integrated over the probability distribution of what is in that envelope. I shall work this through with an example probability distribution. Suppose that the probability of the chosen value being xn=2n is pn=2⋅3−n for all positive integers n, and no other value of x is possible. (Taking pn=2−n would be simpler, but that distribution has infinite expected value, which introduces its own paradoxes.) I shall list all the possible ways the game can play out. 1. $2 in the envelope in your hand, $4 in the other. Probability p1=2/3 for selecting the value x1=2, and 1/2 for picking up the env