To see how this works, let's consider a toy problem. Imagine that your search space consists of only six items, labeled 1 through 6. Let's say your target is item 6 and that you're going to search this space by rolling a fair die once. If it lands on 6, your search is successful; otherwise, it's unsuccessful. So your probability of success is 1/6. Now let's say you want to increase the probability of success to 1/2. You therefore find a machine that flips a fair coin and delivers item 6 to you if it lands heads and delivers some other item in the search space if it land tails. What a great machine, you think. It significantly boosts the probability of obtaining item 6 (from 1/6 to 1/2).
But then a troubling question crosses your mind: Where did this machine that raises your probability of success come from? A machine that tosses a fair coin and that delivers item 6 if the coin lands heads and some other item in the search space if it lands tails is easily reconfigured. It can just as easily deliver item 5 if it lands heads and some other item if it lands tails. Likewise for all the remaining items in the search space: a machine such as the one described can privilege any one of the six items in the search space, delivering it with probability 1/2 at the expense of the others. So how did you get the machine that privileges item 6? Well, you had to search among all those machines that flip coins and with probability 1/2 deliver a given item, selecting the one that delivers item 6 when it lands heads. And what's the probability of finding such a machine? To keep things simple, let's imagine that our machine delivers item 6 with probability 1/2 and each of items 1 through 5 with equal probability, that is, with probability 1/10. Accordingly, this machine is one of six possible machines configured in essentially the same way. There's another machine that flips a coin, delivers item 1 from the original search space if it lands heads, and delivers any one of 2 through 6 with probability 1/10 each if the coin lands tails. And so on. Thus, of these six machines, one delivers item 6 with probability 1/2 and the remaining five machines deliver item 6 with probability 1/10. Since there are six machines, only one of which delivers item 6 (our target) with high probability, and since only labels and no intrinsic property distinguishes one machine from any other in this setup (the machines are, as mathematicians would say, isomorphic), the principle of indifference applies to these machines and prescribes that the probability of getting the machine that delivers item 6 with probability 1/2 is the same as that of getting any other machine, and is therefore 1/6. But a probability of 1/6 to find a machine that delivers item 6 with probability 1/2 is no better than our original probability of 1/6 of finding the target simply by tossing a die. In fact, once we have this machine, we still have only a 50-50 chance of locating item 6. Finding this machine incurs a probability cost of 1/6, and once this cost is incurred we still have a probability cost of 1/2 of finding item 6. Since probability costs increase as probabilities decrease, we're actually worse off than we were at the start, where we simply had to roll a die that, with probability 1/6, locates item 6. The probability of finding item 6 using this machine, once we factor in the probabilistic cost of securing the machine, therefore ends up being 1/6 x 1/2 = 1/12. So our attempt to increase the probability of finding item 6 by locating a more effective search for that item has actually backfired, making it in the end even more improbable that we'll find item 6. Conservation of information says that this is always a danger when we try to increase the probability of success of a search -- that the search, instead of becoming easier, remains as difficult as before or may even, as in this example, become more difficult once additional underlying information costs, associated with improving the search and often hidden, as in this case by finding a suitable machine, are factored in.Can you spot the error in his calculation? The probability to find the correct machine and then the target is indeed 1/12, but the probability to find the target via chosing a machine at random at first is 1/6, thanks to the symmetry of the problem: The probability for a success is 1/6 * 1/2 + 5/6 * 1/10 = 1/6. So the problem didn't become more difficult.