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While not exactly the same problem this is what cleared it up for me: First pick a positive integer at random. Now pick a second integer at random. What is t
by te_platt 16y ago
While not exactly the same problem this is what cleared it up for me:
First pick a positive integer at random. Now pick a second integer at random. What is the probability that the second number is larger? The answer seems to be 1 because after the first number is chosen there are finitely many numbers smaller but infinitely many larger. But what if two people (A and B) choose two positive integers at random but don't tell anyone just yet. What is the probability that A>B? By symmetry we can see it should be 0.5 . So A tells his number. Now the probability that A>B seems to jump to 0 because there are finitely many numbers smaller than A but infinitely many larger. But why should the probability change when we don't even know what A is, just that it known?
The problem lies with the phrase "choose a positive integer at random". However you choose the number you will be hopelessly biased toward 0 because every number you choose will be finite. Now think about the envelope problem but add the condition that the largest amount that any envelope will have is X. Then if your envelope has N in it there is a chance that 2N is larger than X. That cut's down on the chance that switching gets you more. Magically (because I'm too lazy to actually write it all out) it turns out it cuts it down just so the probability of getting more money by switching works out to .5
- jeromec 16y agoSo A tells his number. Now the probability that A>B seems to jump to 0 because there are finitely many numbers smaller than A but infinitely many larger. This is not correct. You said B chose a positive integer. Therefore, there are not infinitely many larger numbers B's choice can be.
- InclinedPlane 16y agoSure there are. No matter what finite positive integer you choose there are infinitely more finite positive integers greater than that number.
- jeromec 16y agoThat's true, but what te_platt said was "two people (A and B) choose two positive integers at random" If two people, A and B, have chosen -- that's choose in the past tense -- two positive integers, there are not infinitely more positive integers B's number can be greater than A's number. Why? Again, because B has already made a choice.
- InclinedPlane 16y agoActually, it doesn't matter. If you rephrase the problem so that B picks after A and you ask the question "is B likely to pick a larger number than A" before B picks then the answer should be "yes". However, this is another example of being misled in reasoning. But not related to timing. The important revelation is understanding that there's no such thing as a method for generating truly random numbers across the entirety of the positive integers. Since there are infinitely many the probability of picking any one integer is 1/infinity. Any concrete random algorithm for positive integers will be hopelessly biased toward 0 (since it will almost certainly have a maximum value, and thus an infinite number of positive integers above that maximum). As long as both A and B use the same algorithm then the chances are equal that A or B will pick the larger number.
- jeromec 16y agoIf you rephrase the problem so that B picks after A and you ask the question "is B likely to pick a larger number than A" before B picks then the answer should be "yes". I agree, but that wasn't what was stated. You've "moved the goal posts" so to speak. ;) The OP said both A and B had chosen a number (but only one had revealed it). That introduces TWO finite values into the experiment. The timing for when B makes the choice makes all the difference.
- praptak 16y ago"First pick a positive integer at random." Wrong already. There exists no equiprobable probability distribution for integers, so you cannot just pick a positive integer at random, you need to define a distribution and it will definitely be biased - some ints will get chosen with larger probability than others.