3 ms·

I’m really struggling to see how this is counter intuitive? You pick one of the numbers, and if it’s positive you say “this is likely the largest” and if negati

by nabdab 7y ago

I’m really struggling to see how this is counter intuitive? You pick one of the numbers, and if it’s positive you say “this is likely the largest” and if negative “this is likely the smallest”. Since there’s a 50/50 chance for each sign, and all positive numbers are larger than all negative numbers. That already gives you the minimum 1/2 probability they are asking for. But you’ve got an improvement because not only does the other number have to match the sign for you to lose, it also has to be larger in magnitude, and that happens in half the cases where they have the same sign, so that gives you 1/4 extra. You end up being correct 75% of the time. The thing with a random number in between is just an odd way of reducing your probability of winning by adding randomness to the split, what that’s supposed to show I’m really confused by. It feels like the “puzzlement” you are supposed to feel that you can beat 50% comes from people ignoring the fact that you can look at the number before making your call.

EvanWard97 7y ago
They don't have to pick numbers equally from below and above zero. They could try to make your odds as difficult as possible by picking 10^20 and 10^20-1. The power of picking a random C is that there is a nonzero probability it will fall between their two numbers.
com2kid 7y ago
It is confusing (and I am still confused) because my understanding is: 1. Person A picks 2 random #s 2. Person B picks a random split #. 3. Person B looks at one of the random #s. Based on its relationship to the arbitrary split # he choose, he then decides it the other random # is larger or smaller than the one he has in hand. I am confused as to how choosing a random split # has any impact on the probability of the other paper in hand. I'd normally think that each random # choice is an independent event and that my "split #" has no impact on the system as a whole. Like, you choose the numbers -500 and 250. I choose to split at 1000. Or 200. Or ten billion. Obviously the math works, but why does my picking another random # make a difference in the system as a whole? The relationship of all the #s to each other is still presumably completely random. Would this work if I am instead handed two numbers, and I have to guess if a third # is greater than the first number, and I make a choice based on the relationship of the 1st number to the 2nd number? Since the 2nd number is random, it shouldn't matter who provides it, right? So stating it that way, "Here is 200, 750, and some third number, is the third number higher or lower than 200?" Does that still work?
- MauranKilom 7y ago
  The important part in the original is that every choice of splitting point has nonzero probability. This means you will _eventually_ land on a splitting number between the two where you will then have 100% chance to be right (instead of 50% chance for all the other times), pushing your average above 50%. If the splitting point is given to you (by an adversary, not an actual random process), this nonzero probability for any given number is not necessarily fulfilled (at least you didn't indicate so).
BeetleB 7y ago
>But you’ve got an improvement because not only does the other number have to match the sign for you to lose, it also has to be larger in magnitude, and that happens in half the cases where they have the same sign, so that gives you 1/4 extra. This assumes that Player 1 allows for negative numbers. What if Player 1 always writes down positive numbers? Let's take it a step further. What if Player 1 only writes A and B whenever he plays the game (but we keep changing Player 2 so they never realize this)? The claim in the paper is that even then they'll guess correct over 50% of the time. I'm probably wrong about the statement below, but will state it anyway: I suspect (but cannot quickly prove in my mind) that the other flaw in your argument is utilizing rules like "If the distribution is uniform, you have a 50% chance of picking a positive number, and 50% chance of picking a negative number". The reason I have trouble disproving it is that classical probability theory doesn't allow for such statements: The probability of picking a single point in a continuous distribution is 0. Edit: I figured out why it works. See this comment and its parent: https://news.ycombinator.com/item?id=21160811 https://news.ycombinator.com/item?id=21160811

by nabdab 7y ago

EvanWard97 7y ago

They don't have to pick numbers equally from below and above zero. They could try to make your odds as difficult as possible by picking 10^20 and 10^20-1. The power of picking a random C is that there is a nonzero probability it will fall between their two numbers.

com2kid 7y ago

It is confusing (and I am still confused) because my understanding is: 1. Person A picks 2 random #s 2. Person B picks a random split #. 3. Person B looks at one of the random #s. Based on its relationship to the arbitrary split # he choose, he then decides it the other random # is larger or smaller than the one he has in hand. I am confused as to how choosing a random split # has any impact on the probability of the other paper in hand. I'd normally think that each random # choice is an independent event and that my "split #" has no impact on the system as a whole. Like, you choose the numbers -500 and 250. I choose to split at 1000. Or 200. Or ten billion. Obviously the math works, but why does my picking another random # make a difference in the system as a whole? The relationship of all the #s to each other is still presumably completely random. Would this work if I am instead handed two numbers, and I have to guess if a third # is greater than the first number, and I make a choice based on the relationship of the 1st number to the 2nd number? Since the 2nd number is random, it shouldn't matter who provides it, right? So stating it that way, "Here is 200, 750, and some third number, is the third number higher or lower than 200?" Does that still work?

MauranKilom 7y ago
The important part in the original is that every choice of splitting point has nonzero probability. This means you will _eventually_ land on a splitting number between the two where you will then have 100% chance to be right (instead of 50% chance for all the other times), pushing your average above 50%. If the splitting point is given to you (by an adversary, not an actual random process), this nonzero probability for any given number is not necessarily fulfilled (at least you didn't indicate so).

BeetleB 7y ago

>But you’ve got an improvement because not only does the other number have to match the sign for you to lose, it also has to be larger in magnitude, and that happens in half the cases where they have the same sign, so that gives you 1/4 extra. This assumes that Player 1 allows for negative numbers. What if Player 1 always writes down positive numbers? Let's take it a step further. What if Player 1 only writes A and B whenever he plays the game (but we keep changing Player 2 so they never realize this)? The claim in the paper is that even then they'll guess correct over 50% of the time. I'm probably wrong about the statement below, but will state it anyway: I suspect (but cannot quickly prove in my mind) that the other flaw in your argument is utilizing rules like "If the distribution is uniform, you have a 50% chance of picking a positive number, and 50% chance of picking a negative number". The reason I have trouble disproving it is that classical probability theory doesn't allow for such statements: The probability of picking a single point in a continuous distribution is 0. Edit: I figured out why it works. See this comment and its parent: https://news.ycombinator.com/item?id=21160811 https://news.ycombinator.com/item?id=21160811