Via Alexander Bogomolny:
At 1 minute to midnight 10 apples fall into a sack. The same happens at half a minute to midnight, then at a quarter minute to midnight, and so on. At each such event you remove an apple randomly from the ones still present in the sack.
What is the probability that at the midnight strike the sack will be empty?
You’re supposed to say “1”. I would say 0.
I’d be just as wrong as the folks who say 1. And in that reason we’d all be wrong lies the answer to solving all probability paradoxes.
All we need remember to solve all probability paradoxes: all probability is conditional.
The conditions include those explicitly stated, such as in the words and numbers of the premises used to state the paradox, and those implied, such as what the words and symbols mean. It cannot be any surprise that two people could look at the same passage and come away with two different implied meanings of the same passage. How many interpretations are there of any one Shakespeare sonnet?
The common error is to assume statements need no implicit premises, that, somehow, statements when directed toward math are immune from multiple interpretations. This is false, and horribly false. But mathematicians try to make it true, by being as careful as possible, and, when they can, lapsing into pure symbolism, where every stroke has an explicitly defined meaning. In rare cases like these, there is only one right answer.
That’s no so in the apple paradox. There are many ambiguities. The trick, then, is to recognize them, define them, then solve the probability for them, while acknowledging that different resolutions can give different answers.
If two people come to a different understanding of what the words, to include the grammar, of a statement means, these people could come to a different probability. Assuming no mistakes in calculation—a large assumption for complex statements—then each person would give correct answers to different problems.
Equally, each person would give the wrong probability—if they were not explicitly clear about their implicit assumptions.
Here’s my take, i.e. my list of implied premises, for the apple problem. All “events” must take place in finite time. However long it takes, it must take some finite time to put apples in and take one out of a sack. This time does not have to be constant; though I believe the words of the problem imply that it is.
Regardless of the length of this discrete event, there will come a time when the clock strikes midnight. Because we could only add a finite amount of apples from 11:59 PM to 12:00 Midnight, and so we could only take out a finite amount, each event, adding and subtracting, takes a discrete tick of the clock, and there are only a finite number of clicks. And since we take only one—and there is no way to take any apple “randomly”: one must take an apple some way—for every ten added, the sack is guaranteed to have apples in it at midnight. Therefore, conditional on these assumptions, the probability the sack is empty is 0.
Further, my implied premises hold (I say) for any event happening in physical (contingent) reality, by which I mean in physical substances. There is no real physical or energetic process that can take an infinite number of steps in a finite time. So whether it’s apples or quarks or strings or whatever, the probability will always be 0.
But physical reality is not all there is to Reality. There is also spiritual, or intellectual reality where the restrictions on time do not apply. With that in mind, here are the list of implied premises for those who answer 1, i.e. the probability almost surely (in math terms) that the sack will be empty.
There are, in this second interpretation, an infinite number of steps at which apples will be added, and an equal number of infinite steps where apples will be removed. In short, a countably infinite number of apples will be added, and a countably infinite number of apples will be removed. Mathematicians will understand that because both sequences are countably infinite, they can be “lined up” against each other in a one-to-one correspondence. (This is the same reason why all fractions of the type p/q, where p and q are integers, are the same “size” as the number of integers.)
Because the cardinality (the size) of both adding and subtracting apples are equal, and because we are in the strange land of Infinity, and therefore because the number of apples added can be put in one-to-one list of apples removed, the probability the sack is empty goes to 1.
That is the right answer to the second set of implied premises. Just as 0 is the right answer to the first set of implied premises. Again, since all probability is conditional on the premises assumed, both answers are correct, but only after the implied premises have been made explicit.
Both sets of implications are possible from English in the passage. Are there other possible implied premises? Maybe. I don’t see any that would change the probability. But this is a relatively simple passage. Others, notoriously, are not. Heated disagreements occur when one set of folks say “It’s obvious these implications hold” while others say, “No, these are!”
Sound familiar?
A version of this post first appeared on 30 October 2017.
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This is why I hate probability. But here’s my take, and I’ll go with P=0. Each action, addition and removal, is paired. One addition, one removal. Therefore, although both are countably infinite, one is countable infinite as adding 10, and the other is countably infinite as removing 1. Therefore at each countably infinite step, 9 apples are added. The only way to get to zero apples is to have more removal steps, but the problem specifically says “At each such event you remove an apple”. I don’t see how you get around a net of 9 apples added per clock strike, no matter how close together they occur.
Or you could argue that it will never be midnight, therefore the concept of a probability happening at some non-existent time will be non-existent, I suppose.
There are apples in the sack at midnight. I offer three distinct proofs.
1) The countably infinite steps are (+10 -1) = +9, therefore there are apples left in the bag at midnight.
?|x=0..?| (10-1)x = ?
2) At midnight, the series stops. The final step was either adding ten apples, or removing one apple from the (much larger) number of apples previously added. Either way, there must still be apples in the bag at midnight.
3) The apples explicitly FALL into the sack. There must be a pause between layers, however minor, for you to remove one. Assuming a sack with a mouth large enough for all 10 apples to fall in at once, and with the average diameter of a small apple being around 2.5″ (10cm), then only a finite number of apples can fall into the bag given gravitational acceleration at earth’s surface of 9.8m/s^2. So let us assume that many flat layers of 10 apples were arranged prior to the experiment just above the mouth of the bag, then dropped all at once. (This continuous flow of apples results in MANY more apples than specified in the thought experiment.) This gives time for an object to fall 17,652 meters (ignoring air resistance). At 10 apples per meter height, and 10 apples per layer, gives 100 apples per meter fallen. So a maximum of 1,765,200 apples could fall into the bag in one minute. A total of 176,520 apples could (somehow) be removed. That leaves 1,588,680 apples in the bag at midnight, given a continuous flow of free falling apples.
The answer is 0. This is just a geometric series but it’s INFINITE. No way will there ever be no apples.
I’ve always hated this problem. It’s nonsensical, but it’s presented as though there is a single way to interpret it and that this interpretations leads to a single correct probability.
What is described could of course never happen in the real world. This is why the problem is nonsensical. So we have to rephrase it in some sort of mathematical limit sense. If we do that the most obvious thing to do is just keep track of the number of apples at each iteration: there’s a 100% chance that there will be 10 apples at the 0th, 19 on the 1st, 28 on the 2nd and in general 10+9n on the n-th. So if we take the limit of 10+9n as n goes to infinity, infinitely many apples! And if infinitely many, at least one, so the probability is 0.
But of course there is no guarantee in a limiting process like this that measures of the set are preserved as we go to the limit. For example, we could approximate a line of length 1 by zig zags that each have a total length of more than 1, but which converge pointwise to the line. So the length is not preserved by the limiting process. The trouble here is that no one told us what precisely is being used to take the limit! And honestly I’m not sure what it’s supposed to be.
This is a variant of the idea taking a set of numbers S_0 ={1,2,…,10} and then getting a set S_n by taking S_{n-1} removing the first number, and adding the next 10 integers to the end. Then you define the limit as the intersection of all S_n sets. This is empty, since for every integer n we have n not in S_{n+1}. Fine. Everything is explicit, and we can verify that this works out by definition of intersections. (Whether this corresponds to any “real world” process is another matter, but mathematically it works out.)
But adding in probabilities changes things quickly. If we randomly remove numbers, then it is possible that any individual number remains. For example, maybe we never remove 1, so it is there at the end. Now of course the response is “I never said that the bag would be empty, but that it would be empty with probability 1. We could still have events where it is not empty, as long as they have collective probability 0.” Okay, fine. But how do you know that?
Whenever I’ve seen this explained the explanation has always been “the probability of any individual apple being in the bag is 0, so therefore there is a probability of 1 that the bag contains no apples.” But this is obviously fallacious. For example, say that I use a uniform probability distribution on the interval [0,1] to select numbers 10 times. The probability that any individual number would be in my selection is 0, but the probability that my selection would contain at least one number is 1.
So how do you actually calculate the probability that the bag is empty? The only way that I can see to do this is to create a probability space where we take every possible choice of removal at every stage. Some of these, like removing 1, 2, 3, etc. in numerical order, will eventually remove every apple. Others, like removing the even numbers in numerical order, will leave apples. You then assign some “natural” probability to each event, and show that the events that don’t remove all the apples have probability 1.
But what is the probability distribution? We have an infinite discrete set. So “apply 1/n, where n is the number of possibilities” doesn’t work. Nor can we do things easily by assigning some continuous function we integrate, as with the uniform distribution on the interval [0,1]. So what do we do? We could try to do something like saying that if S_n is the set of selections that remove n from {1,2,…,10} first, then this should have a collective probability of 10. Then for example if we remove 1 first and let S_{1,n} indicate the selections that remove n from {2,3,…,20} next, that it should have collective probability of (1/10)(1/19) (i.e. 1/10 chance to remove 1 and 1/19 chance to remove n next) but it’s not obvious to me how how then take infinite sums to determine what the probability of not removing every number is. Probably this can be calculated but it’s certainly NOT obvious how to do it, and this is only one way of assigning probabilities. Maybe there is another that is just as “natural” but leads to a different answer.
Such a garbage problem.
I don’t understand your cardinality approach. If I remove an infinite set from another of the same cardinality, the result is not necessarily the empty set. For example, the even integers and all integers have the same cardinality, but removing the evens from the integers leaves the odds. And we can have that happen in this apple problem: if we label the apples it is possible that we remove only the evens, in which case there is a one to one correspondence between the apples added and removed, but some apples are never removed.
Now there is a question of how LIKELY these sorts of selections are, but I don’t think that cardinality by itself answers that question.
Mr Harrier is on the right lines: see previous blog where I was berating sundry for worrying about using probability theory to test dice when you can actually measure the dice to determine correctness and, in any case, how it performs on the craps table for the casino matters not whether it’s loaded or not. Long live Rosencrantz and Guildenstern.
I think Rudolf comes closest except for this being a garbage problem. My guess is we will see similar arguments come around when Uncle Briggs (almost certainly) presents us with Bertrand’s chord problem. I’ll take as given that Briggs is correct about the pragmatic answer with real apples and only touch on the abstract mathematical component, changing apples to integers.
I’m pretty sure I heard my favorite definition of random here on this site – that is random is just some definite process about which we are ignorant. So with that definition we need to consider all possible ways of removing an integer from a set of integers, not the limit of one given process.
First we note that any number of points remaining is possible. For example, if we want N integers left in the set we can at each step remove the largest integer from the set until there are at least N integers left, then at each step remove smallest integer in the set that is greater than the Nth integer in the bag.
So now we see that the number of integers left is the sum of a different arrangement of the same alternating series. I don’t have a good way to put it in nice notation here but some form of ( +9/10 – 1/10). By the Reimann rearrangement theorem we have if an infinite series of real numbers is conditionally convergent, then its terms can be rearranged in a permutation such that the new series converges to any arbitrary real number, or even diverges. This implies that a series of real numbers is absolutely convergent if and only if it is unconditionally convergent. Because of the if and only if that means if we are able to get any number by rearranging terms we have a conditionally convergent series.
Conditionally convergent series are well studied but in our case here we just need to note that the number of integers left in the bag is any number from 0 to infinity. Also we have no reason to prefer one over any other leaving the probability as 1 / (number of integers). Also the probability that there will be N integers left is 1/ (number of integers). I hate to say that means the probability is 0 for other reasons but at least it is close enough to 0 that there isn’t a positive real number you can explicitly state that is closer to 0.
In simpler terms we can think of the number of integers left as one of the integer values of the harmonic alternating series. The one particular value the problem asks about is 0 so the probability is 1/(number of integers).
Infinity is neither a number nor something measurable.