bonanova

You're pulling my leg now, right? I mean, how can there be several ways to do what is impossible?

The number of balls in bin B cannot depend on which ball was removed. If removing the larger number results in a countably infinite set of numbers [the odd integers] at completion, then removing the smaller number cannot result in an empty set at completion. To see this, restate the problem. Step 1. Move two balls from bin A to bin B. Step 2. Move one ball from bin B to bin C. [or discard it] Repeat. Or equivalently, Step 1. Move one ball from bin A to bin B. Step 2. Move one ball from bin A to bin C. [or discard it] Repeat. Please describe how in either case bin B eventually becomes empty. If talking about a finite number of operations and then letting that number approach infinity is fallacious [it isn't] your argument is fallacious. A proper subset of a countably infinite set can also be countably infinite [the set of even integers is a countably infinite proper subset of the set of integers]. Thus, just because you have accounted for a countably infinite number of natural numbers [those removed from bin B] it does not follow that you have accounted for all of them. Consider the sequence {1, 2, 3, ..., N, N+1, N+2, N+3, ... 2N} Remove the first N of the elements. Let N approach infinity. You will have removed a countably infinite subset of integers from the sequence. Does that mean you have removed all the elements of the sequence?

I chose it at random.

The "proof" that bin B is empty is fallacious. Re-order the natural numbers so that the even numbers follow the odd ones. Although you can't specify their place in the sequence by a finite number, the even numbers still exist.

Agree the probability [before the choice was made] of choosing .314 is zero. Impossibility and zero probability are different concepts. Anyway, I asked not about its probability, but its randomness. My question remains for d3k3: prove that .314 was not chosen at random.

The book I read this in presented this problem as an introduction to the idea of cardinality, but I thought it was a very good thought experiment / paradox. There are still some objections to the solution that I can't respond to, such as - the number of balls in bin B increases monotonically with time. At any point in time, I can tell you (a), how many balls are in bin B, and (b), that the number of balls has never been higher. This intuitively contradicts the solution, almost as much as the logic for the solution seems inarguable. My first post shows the answers depend on the difference of two infinities. Suppose set A, B and C contains all the integers, all the even integers and all the odd integers from 1 to 100, respectively. A = {1, 2, 3, ... 100} B = {2, 4, 6, ... 100} C = {1, 3, 5, ... 99} The cardinality of the three sets are 100, 50, and 50 respectively. A is twice as large as B and C. But replace 100 with infinity and you get something very different. Now we have A = {1, 2, 3, ... } B = {2, 4, 6, ... } C = {1, 3, 5, ... } These three sets have the same cardinality. It's called Aleph-null, the cardinality of any countably infinite set. Any set whose members have a 1-1 correspondence with the natural numbers has this cardinality. What's counter-intuitive about this is that set A has members that set B does not - namely the members of set C. Nevertheless they have the same cardinality: the members bi of B correspond 1-1 with the members ai of A: bi = 2ai for all i. Any two countably infinite sets have the same cardinality, even if one has countably infinite more elements than the other. For example, the set of all integers and the set of even integers have the same cardinality.

Interesting statement. Let's pursue it a bit. The set of reals in the interval [0, 1] is infinite. I choose the number 0.314. Prove that it was not chosen at random.

The solution that does not require a computer gives an upper bound: It's simply 362880×86400×216×155520×60000×216×4374×256×1 = 15,284,122,844,890,207,459,737,600,000,000 puzzles. This bound is astronomical and meaningless - it includes duplicates and wrong answers. Are you saying that you can get a meaningful answer by a simple calculation?

... forcing us, right in the middle of thinking about infinity, to think in finite terms. Not nice! But Ok, I guess that's fair. Let's see what we have. By the OP, after a full minute has passed, there have been an infinity of operations on the bins. Those operations result in an infinity of integers [or balls, but let's think of integers because we've been asked about specific numbers] moving from bin A to bin B, and the lower half of these integers leaving bin B. We're all kind of thinking that the bins, even after the full minute, still have balls in them. Now we're being asked: if bin A and bin B are NOT EMPTY, name some of the the integers they contain. Seems impossible, but ... maybe not. Read on. What integers will have been left in bin A? [1] The answer is: bin A will have all the integers it initially had, minus the infinity of integers that were taken from it. What integers will remain in bin B? [2] The answer is: bin B will contain the upper half of the infinity of integers that were taken from bin A. What specific integers are in the A and B sets? [3] The answer is: none that you can specify. If you specify an integer, 13 is as good as any, it will eventually leave bin A, and later it will leave bin B. So let's take these three statements, which seem all to be unavoidably true, and see where they leave us. And since I've forgotten how to write the symbol, let me use 8 to represent infinity: Bin A has 8-8 balls remaining. Bin B has 8/2 balls remaining. Bins A and B contain no identifiable integers. We can wriggle out of [1] by saying 8-8 is undetermined, and therefore could be taken to be 8. We can wriggle out of [2] by saying that 8/2 is still 8. Okay so far. Now [3] seems like a proof [by induction - it's true for 1, and if it's true for n, it's also true for n+1] that the bins are empty. But [3] can be restated in the same terms as [1] and [2]. Let 8a be the count of the integers originally in bin A. Let 8b be the count of operations performed during the full minute described in the OP. Then, the above 3 points can be stated this way: Initially, bin A contains 8a integers [or balls] bin B contains 0 integers All the integers - 13, for example - are in bin A; no integers are in bin B. After the full minute,bin A contains 8a - 8b integers bin B contains the upper half of the first 8b integers bin A contains all of the integers that lie between 8b and 8a; bin B contains all the integers between 8b/2 and 8b. Now we see that all the answers contain 8-8 in some form, and you can Google "infinity minus infinity" to proceed further. Anyway, you may now resume the part of your life that is not indeterminate, assuming it exists.

It may be time for us to simply agree with Enigma. Enigma states that s/he disagrees. In all likelihood that is a true assertion. We can all agree with it. Remember: You can lead a horse to drink, but you can't make him water.

In [1] and [2] I read we pick positive or non-negative integers. In [3] I pick a random real number - rational or irrational - from the interval [0, 1]. In [4] and [5] it's clear.

August has it.

Romeo and Juliet Locked.

octopuppy and esjohnson are correct in their comments about too many odd parity nodes. jimmyt's suggestion of adding a line can reduce their number to two. The spoiler contains one way to do this. Still, that's not what the OP asks us to do, so this doesn't seem to be a solution.

You're using seven numbers.

I'm back... Locked.

The problem you're describing is this: I have two children with androgenous names of Taylor and Alison. I tell you that Taylor is a girl. I ask you the probability that Alison is a girl. You reply 1/2. You are correct. That's not really much of a puzzle now, is it? And it's not what the OP asks. Here's a hint: The statements "one of the children is a girl" and "child number one is a girl" are different. [1] Try to figure out why they are different. [2] Then try to figure out which statement is in the OP. Here's another hint: If you believe the answer is 1/2, then you must believe that two-girl families and mixed-gender families occur with equal probability. Why? Because if one is a girl and the other is a girl has a 1/2 probability, then the only other case one is a girl and the other is a boy must have the remaining 1/2 probability. That's a tough position to defend. But something tells me you might try.

I remember that day. I also remember April 5, 1967 at 1:23 am. I was studying for an exam, and my mind wandered to the clock in time to enjoy the moment. Question - how many like moments will occur this century?

You are not told that child one is a girl, and then asked whether child two is a girl. The OP does not single out either child. The statement 'one of them is a girl' is true for three disjoint and equally likely cases: GG BG GB. In one of those cases the other is a girl. The probability is 1/3. It's not all that hard. The OP rules out 1/2 as an answer. Read it.

And it seems someone else got this before me, so I won't claim credit. You have it .. nice going.

Since the word only is not in the OP 0% does not work.

Only one answer fits.

You have the right formula - just use it. N = 3 [GG GB BG] - if you know one is a girl, [you don't know which one is certainly a girl] these are the possible outcomes n = 1 [GG] - this is the favorable outcome. p = 1/3

Nice chestnut.

Asked and answered.