7 ms·

I may be missing something here, but I'm not sure this is quite right. The paper attached describes a scenario where the player A chooses a number between -inf

I may be missing something here, but I'm not sure this is quite right. The paper attached describes a scenario where the player A chooses a number between -infinite to infinite. The simulation uses ranges -1000000 to 1000000, it's not a surprise that the strategy works in this case. I think if the ranges were truly -infinite to infinite, this strategy would fall apart. No matter where you set C, there are infinite values above and below it.

EvanWard97 7y ago
The idea is that by picking a random real number C, there is a chance it will fall between A and B over time, no matter how the other person picks their numbers A and B. It's the infinite range C can be drawn from which gives you a strictly >50% of guessing the higher of A or B no matter how the other player tries to game the system.
- gentaro 7y ago
  >> The idea is that by picking a random real number C, there is a chance it will fall between A and B over time, no matter how the other person picks their numbers A and B. But for any two finite numbers, A and B, the chance that C will fall between them if selected from an infinite range is 1/infinity (or 0).
  - greenbay20 7y ago
    >> But for any two finite numbers, A and B, the chance that C will fall between them if selected from an infinite range is 1/infinity (or 0). I think you're assuming we pick C uniformly at random from an infinite range, but this is not possible. In general, the claim is not true: consider picking C from a sample of a standard normal, if A = -B = 2 then with >95% C will lie between A and B.
  - lonelappde 7y ago
    It's not 0, it's "an unknown positive number" drawn arbitrarily from a set whose lower bound is 0. It's practically 0, but it's some fixed positive number for every instance of the game. It's equivalent to this much simpler game: "I am going to give you a positive real-number amount of dollars." No matter what I do, you will win money playing this game, guaranteed! Now, how much would you pay to play this game? $0, because for any amount you'd pay to play, I could arbitrarily choose to give you less in winnings. Like most problems involving infinity, it's unintuitive/paradoxical because it pretends to model a physically plausible scenario but actually doesn't.
    - kgwgk 7y ago
      Would you say that the value of the infinite sum 1/2 + 1/4 + 1/8 + ... is 1 or that it is an unknown number? The probability that a random number uniformly drawn from the real line is between 42 and 43 is zero in the same way that “the probability that a random real number selected uniformly from the interval [0 1] is pi/4” is zero.
      - speedplane 7y ago
        > Would you say that the value of the infinite sum 1/2 + 1/4 + 1/8 + ... is 1 or that it is an unknown number? Integral of 1/2^n ... got that no problem. I took AB Calculus in high school, three semesters of calculus in college, and a bunch of calculus heavy linear algebra. I’m certain that I knew how to integrate that thing at some point... it’s pretty sad that now I can only stare at it blankly and at best lean heavily on Wikipedia to find an answer.
        kgwgk 7y ago
        What I wrote is an infinite series and it’s much simpler to solve than the integral: https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_⋯ https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%...
        speedplane 7y ago
        Ha, yup, citing a Wikipedia article too. At least I’m not alone.
dsukhin 7y ago
If you have a range (-inf, inf) and you split it at some random C, even though there are still infinite values on either side, the probability you choose another number on the left side of C is (p) and on the right side is (1-p) regardless of how you choose the next number. That's what makes these probabilities finite and makes the math work out.
- OskarS 7y ago
  But... but... you can't "pick a number uniformly distributed in (-inf, inf)". Like, that's not a thing that is possible [0], so saying anything about the probabilities once you've done so is not reasonable. You can pick a random number in infinite range non-uniformly (e.g. to pick a number in the range [1,inf), pick a number X in (0,1] and use 1/X), but then the probabilities are no longer so neat and tidy. The game only makes sense if you attach limits to the numbers the other person is allowed to pick, and then the only "split" that works is in the middle of the range. At which point the game is sort-of "obviously true". [0] https://math.stackexchange.com/a/14169/641751 https://math.stackexchange.com/a/14169/641751
  - MauranKilom 7y ago
    It is not necessary for the distribution to be uniform, it just has to be nonzero everywhere.
    - kgwgk 7y ago
      If the distribution of A and B is uniform, the probability that C from this proper distribution is between A and B is zero. If the distribution of A and B is also a proper distribution the probability that C is between them is positive. But then the distribution of A and B has a middle point such that 50% of the numbers are above and 50% below and the problem is trivial.
ulucs 7y ago
Pick a random C from R. A and B have equal probabilities of being higher or lower than C. Now with the restriction A>B, there are only three equally likely options: A>B>C, A>C>B, C>A>B. You can guess half the time correctly for A>B>C and C>A>B and all the time for A>C>B, which gives you 66% which is shown with the random trials.
BeetleB 7y ago
>The paper attached describes a scenario where the player A chooses a number between -infinite to infinite. No. It doesn't even require Player A to pick them at random. The strategy for Player B still works over 50% of the time.

EvanWard97 7y ago

The idea is that by picking a random real number C, there is a chance it will fall between A and B over time, no matter how the other person picks their numbers A and B. It's the infinite range C can be drawn from which gives you a strictly >50% of guessing the higher of A or B no matter how the other player tries to game the system.

gentaro 7y ago
>> The idea is that by picking a random real number C, there is a chance it will fall between A and B over time, no matter how the other person picks their numbers A and B. But for any two finite numbers, A and B, the chance that C will fall between them if selected from an infinite range is 1/infinity (or 0).
- greenbay20 7y ago
  >> But for any two finite numbers, A and B, the chance that C will fall between them if selected from an infinite range is 1/infinity (or 0). I think you're assuming we pick C uniformly at random from an infinite range, but this is not possible. In general, the claim is not true: consider picking C from a sample of a standard normal, if A = -B = 2 then with >95% C will lie between A and B.
- lonelappde 7y ago
  It's not 0, it's "an unknown positive number" drawn arbitrarily from a set whose lower bound is 0. It's practically 0, but it's some fixed positive number for every instance of the game. It's equivalent to this much simpler game: "I am going to give you a positive real-number amount of dollars." No matter what I do, you will win money playing this game, guaranteed! Now, how much would you pay to play this game? $0, because for any amount you'd pay to play, I could arbitrarily choose to give you less in winnings. Like most problems involving infinity, it's unintuitive/paradoxical because it pretends to model a physically plausible scenario but actually doesn't.
  - kgwgk 7y ago
    Would you say that the value of the infinite sum 1/2 + 1/4 + 1/8 + ... is 1 or that it is an unknown number? The probability that a random number uniformly drawn from the real line is between 42 and 43 is zero in the same way that “the probability that a random real number selected uniformly from the interval [0 1] is pi/4” is zero.
    - speedplane 7y ago
      > Would you say that the value of the infinite sum 1/2 + 1/4 + 1/8 + ... is 1 or that it is an unknown number? Integral of 1/2^n ... got that no problem. I took AB Calculus in high school, three semesters of calculus in college, and a bunch of calculus heavy linear algebra. I’m certain that I knew how to integrate that thing at some point... it’s pretty sad that now I can only stare at it blankly and at best lean heavily on Wikipedia to find an answer.
      - kgwgk 7y ago
        What I wrote is an infinite series and it’s much simpler to solve than the integral: https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_⋯ https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%...
        speedplane 7y ago
        Ha, yup, citing a Wikipedia article too. At least I’m not alone.

dsukhin 7y ago

If you have a range (-inf, inf) and you split it at some random C, even though there are still infinite values on either side, the probability you choose another number on the left side of C is (p) and on the right side is (1-p) regardless of how you choose the next number. That's what makes these probabilities finite and makes the math work out.

OskarS 7y ago
But... but... you can't "pick a number uniformly distributed in (-inf, inf)". Like, that's not a thing that is possible [0], so saying anything about the probabilities once you've done so is not reasonable. You can pick a random number in infinite range non-uniformly (e.g. to pick a number in the range [1,inf), pick a number X in (0,1] and use 1/X), but then the probabilities are no longer so neat and tidy. The game only makes sense if you attach limits to the numbers the other person is allowed to pick, and then the only "split" that works is in the middle of the range. At which point the game is sort-of "obviously true". [0] https://math.stackexchange.com/a/14169/641751 https://math.stackexchange.com/a/14169/641751
- MauranKilom 7y ago
  It is not necessary for the distribution to be uniform, it just has to be nonzero everywhere.
  - kgwgk 7y ago
    If the distribution of A and B is uniform, the probability that C from this proper distribution is between A and B is zero. If the distribution of A and B is also a proper distribution the probability that C is between them is positive. But then the distribution of A and B has a middle point such that 50% of the numbers are above and 50% below and the problem is trivial.

ulucs 7y ago

Pick a random C from R. A and B have equal probabilities of being higher or lower than C. Now with the restriction A>B, there are only three equally likely options: A>B>C, A>C>B, C>A>B. You can guess half the time correctly for A>B>C and C>A>B and all the time for A>C>B, which gives you 66% which is shown with the random trials.

BeetleB 7y ago

>The paper attached describes a scenario where the player A chooses a number between -infinite to infinite. No. It doesn't even require Player A to pick them at random. The strategy for Player B still works over 50% of the time.