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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 o
by 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 agoThe 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).