karthickgururaj Posted October 6, 2014 Report Share Posted October 6, 2014 This is a result of a fallout from a previous puzzle.. You choose a point "randomly" on the positive side of origin on x-axis. This point is called A. Choose an another point again on x-axis (again, positive). Call this point B. What is the probability that B is closer to the origin than A? Is it 0? Or 0.5? Or something else? If the question is not well-formed, please feel free to qualify it with "reasonable" statements. Quote Link to comment Share on other sites More sharing options...

0 witzar Posted October 7, 2014 Report Share Posted October 7, 2014 Asking about probabilities makes no sense unless there is a probability distribution. Unfortunately there is no reasonable distribution on R^{+}. But an obvious distribution exists on interval [0,x], where x is any positive real number. One thing we can do in such case is to try to solve the problem for [0,x] and then see if the solution has a limit when we go with x to infinity. With this approach it's obvious that the probability we look for equals 0.5 for every x, so obviously it's limit is also 0.5 when x goes to infinity. 1 Quote Link to comment Share on other sites More sharing options...

0 koren Posted October 7, 2014 Report Share Posted October 7, 2014 Depends on the technique of choice. If you choose first the point A and then B, the probability is zero, because A divides the axis on finite and infinite parts. (probability = N/infinity = 0) If you will simultaneously "throw" these points on the axis, then the odds that B is to the left or to the right from A are equal. The answer is 50-50. How you like it? Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted October 7, 2014 Author Report Share Posted October 7, 2014 Depends on the technique of choice. If you choose first the point A and then B, the probability is zero, because A divides the axis on finite and infinite parts. (probability = N/infinity = 0) If you will simultaneously "throw" these points on the axis, then the odds that B is to the left or to the right from A are equal. The answer is 50-50. How you like it? So obviously (at least) one of the approaches is wrong (if not both). Which one and why? Quote Link to comment Share on other sites More sharing options...

0 koren Posted October 8, 2014 Report Share Posted October 8, 2014 (edited) Asking about probabilities makes no sense unless there is a probability distribution. Unfortunately there is no reasonable distribution on R^{+}. But an obvious distribution exists on interval [0,x], where x is any positive real number. One thing we can do in such case is to try to solve the problem for [0,x] and then see if the solution has a limit when we go with x to infinity. With this approach it's obvious that the probability we look for equals 0.5 for every x, so obviously it's limit is also 0.5 when x goes to infinity. For me, it's not obvious. Imagine rain and two plots, one of which is twice larger than the other. Obviously, on a larger plot fall twice drops, i.e. probability for each drop to fall at a larger plot is approximately 0.66. And if the second plot is 1000 times larger, and if it is infinitely large? So, for me it's not obvious. And for you? Edited October 8, 2014 by koren Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted October 8, 2014 Report Share Posted October 8, 2014 (edited) Depends on the technique of choice. If you choose first the point A and then B, the probability is zero, because A divides the axis on finite and infinite parts. (probability = N/infinity = 0) If you will simultaneously "throw" these points on the axis, then the odds that B is to the left or to the right from A are equal. The answer is 50-50. How you like it? The order of choice, by symmetry, cannot matter. The probability that A is finite is the same as the probability that B is finite. Edit: As witzar points out, the probability density function comes into play. For a probability evenly distributed over an infinite domain, that function is zero for every finite region. That is to say the probability of choosing at random any finite number from the infinite positive x-axis is zero. To see this, note that the ratio of the lengths (a) from the origin to any finite number to (b) the entire positive x-axis is zero. This is koren's argument that for an even distribution of probabilities the size of the region gives the answer. That part of his argument is not flawed. The flaw is to assign different domains for the choice of A and B. That is, to assume A is bounded while B is not. Since the premise has zero probability, so does the conclusion. Edited October 8, 2014 by bonanova Add comment Quote Link to comment Share on other sites More sharing options...

0 dgreening Posted October 8, 2014 Report Share Posted October 8, 2014 (edited) 'Could be over thinking this' for any 2 positive numbers A and B, there are just 3 possible relationships A = B A > B A < B If we assume that the range of numbers is really large then the P [A = B] is very, very low. If we say that P [A=B] = 0, then there are just the 2 choices [A>B or A<B] the relative magnitude of the numbers doesn't matter and it is simply a matter of which number was chosen first. A random process is just as likely to pick "small then large" as it is to pick "large then small" and the P [A>B] = P[A<B] = 0.5 this holds whether the 2 numbers are (1 and 2) or (1 and a trillion) Could be over thinking this! for any 2 positive numbers A and B, there are just 3 possible relationships A = B A > B A < B If we assume that the range of numbers is really large then the P [A = B] is very, very low. If we say that P [A=B] = 0, then there are just the 2 choices [A>B or A<B] the relative magnitude of the numbers doesn't matter and it is simply a matter of which number was chosen first. A random process is just as likely to pick "small then large" as it is to pick "large then small" and the P [A>B] = P[A<B] = 0.5 this holds whether the 2 numbers are (1 and 2) or (1 and a trillion) Edited October 8, 2014 by dgreening Quote Link to comment Share on other sites More sharing options...

0 witzar Posted October 8, 2014 Report Share Posted October 8, 2014 For me, it's not obvious. Which part? If you don't understand the first statement, then just look at the definition of probability. Let me quote from Wikipedia: "Probability is the measure of the likeliness that an event will occur." The key word here is measure. Measure is a non-negative value you assign to each measurable set of your space (the assignment should have some special properties). No measure - no probability, as simple as that, just by the mere definition. So let me state it once again: 1) we need to agree upon which subsets of our space are measurable, 2) we need do assign them measure, and only then 3) we can ask about probabilities of different events. Before steps 1) and 2) probability is not even defined, so questions like 3) are meaningless. PS Your example problem with rain and plots could be easily resolved by choosing a finite number of sets that should me measurable and assigning them proper measures. Same procedure cannot be done to solve the original problem. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted October 8, 2014 Report Share Posted October 8, 2014 Given that both random points lie at infinity and cannot be compared, the argument of symmetry that holds on any finite portion of the real axis can be taken to hold in the infinite case. P[b<A] = 0.5 Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted October 9, 2014 Report Share Posted October 9, 2014 Let's start by considering the finite case for a uniform distribution. Let's consider A and B on an interval [0,S]. Probability of picking point A at the value x is 1/S. Probability that point B has a value greater than x is (S-x)/S To calculate the probability for all x, we integrate P(A is at x)*P(B has a value greater than x) over the interval [0,S]. which is INT[(S-x)/S^2] = (Sx-x^2/2)/S^2 from 0 to S. Plugging in the boundaries, we get (S^2-S^2/2)/S^2 For finite S, we can simply divide out S^2 from top and bottom and we see it simplifies easily to 1/2. To take the limit as S approaches infinity, we can use L'Hopitals rule twice and we also get 1/2. Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted October 9, 2014 Author Report Share Posted October 9, 2014 Let's start by considering the finite case for a uniform distribution. Let's consider A and B on an interval [0,S]. Probability of picking point A at the value x is 1/S. Probability that point B has a value greater than x is (S-x)/S To calculate the probability for all x, we integrate P(A is at x)*P(B has a value greater than x) over the interval [0,S]. which is INT[(S-x)/S^2] = (Sx-x^2/2)/S^2 from 0 to S. Plugging in the boundaries, we get (S^2-S^2/2)/S^2 For finite S, we can simply divide out S^2 from top and bottom and we see it simplifies easily to 1/2. To take the limit as S approaches infinity, we can use L'Hopitals rule twice and we also get 1/2. While the final answer doesn't change.. Probability of picking point A at x is 0. I think you meant to say, probability of picking point A less than x. And that would be x/S (not 1/S). Note that A and B can be any points on x-axis, not necessarily at integral distance from origin. Quote Link to comment Share on other sites More sharing options...

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## karthickgururaj

This is a result of a fallout from a previous puzzle..

You choose a point "randomly" on the positive side of origin on x-axis. This point is called A.

Choose an another point again on x-axis (again, positive). Call this point B.

What is the probability that B is closer to the origin than A? Is it 0? Or 0.5? Or something else?

If the question is not well-formed, please feel free to qualify it with "reasonable" statements.

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