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I think it can be more intuitively explained without getting into math: if you know someone wrote down two random numbers between 1 and 10. You look at one and

I think it can be more intuitively explained without getting into math: if you know someone wrote down two random numbers between 1 and 10. You look at one and have to guess if the other one is higher or lower. Say you flip over a 2: what do you think is the right guess?

ChristianBundy 7y ago
Of course, but I'm having trouble understanding why this works on an infinite number line. Say you flip over a 2,387,290,723,013,172,348,238,987: what do you think is the right guess?
- tobbykop 7y ago
  Its the largest because its positive, 75% I’m right. And if we are only looking at positive numbers in an infinite axis I’d need to borrow your generator for a test to approximate its midpoint (which obviously it doesn’t have), which I then weigh against the original number with the logic “if it’s larger than what I got, so is your other number. If it’s smaller so is the other number.” Now since the chance for my number being the middle of the three drawn is 1/3(only case I lose), that gives me 66% chance of winning that bet.
godelski 7y ago
I think this distracts from the actual problem. They have a more generalized case, (-inf, inf). In your case, if we don't allow double selecting, 2 is the lowest number between 1 and 10. So you have a 100% chance of guessing right. You're using the bound to help you. Being unbounded you still have an infinite set of numbers smaller than the number you see and an infinite number higher. To be more analogous, that would be like flipping over 5 every single time (on [0,10]). You have {1,2,3,4} below and {6,7,8,9} above. EVERY TIME.
amalcon 7y ago
The reason it's counterintuitive is because, in the game, you don't know that the two numbers are random. In fact, they are chosen by an adversary. You flip a 2: do you think this is most likely the lower number, or do you think the other player picked 1 and 2 just to mess with you? That's the point of picking a random pivot: it reduces your opponent's ability to influence the situation to only placing their numbers as close to each other as possible.

ChristianBundy 7y ago

Of course, but I'm having trouble understanding why this works on an infinite number line. Say you flip over a 2,387,290,723,013,172,348,238,987: what do you think is the right guess?

tobbykop 7y ago
Its the largest because its positive, 75% I’m right. And if we are only looking at positive numbers in an infinite axis I’d need to borrow your generator for a test to approximate its midpoint (which obviously it doesn’t have), which I then weigh against the original number with the logic “if it’s larger than what I got, so is your other number. If it’s smaller so is the other number.” Now since the chance for my number being the middle of the three drawn is 1/3(only case I lose), that gives me 66% chance of winning that bet.

godelski 7y ago

I think this distracts from the actual problem. They have a more generalized case, (-inf, inf). In your case, if we don't allow double selecting, 2 is the lowest number between 1 and 10. So you have a 100% chance of guessing right. You're using the bound to help you. Being unbounded you still have an infinite set of numbers smaller than the number you see and an infinite number higher. To be more analogous, that would be like flipping over 5 every single time (on [0,10]). You have {1,2,3,4} below and {6,7,8,9} above. EVERY TIME.

amalcon 7y ago

The reason it's counterintuitive is because, in the game, you don't know that the two numbers are random. In fact, they are chosen by an adversary. You flip a 2: do you think this is most likely the lower number, or do you think the other player picked 1 and 2 just to mess with you? That's the point of picking a random pivot: it reduces your opponent's ability to influence the situation to only placing their numbers as close to each other as possible.