Recent content by FragileFred

I think ELM is right about the backwards thing. There are two numbers in Al's hands, one low and one high. There is exactly a 50% chance of either being uncovered first and therefore exactly a 50% chance Bob will be correct when he guesses higher or lower. Surely, the act of Bob "thinking of...

So you are saying the original solution looks backwards in the same way?

I see what you are saying now, but I still think there is something wrong with it. It clashes with the following way of looking at the situation:There are two numbers less than n+.5 in Al's hands, one of which is less than the other. Surely the chances of the lower one being uncovered first is...

There are an infinite numbers of integers larger than -1000 that would make him right, not 989.5.

Okay, so after some quick internet review of calculus, I have been reminded that 1/infinity only makes sense in terms of limits. The limit of 1/x as x approaches infinity = 0. I think this brings us back to my first question. If Bob could truly choose a random integer from an infinite number...

mytagid = Math.floor( Math.random() * 100 );document.write("This solution raises an interesting question. Is it even possible to produce a random integer of any value? That is, you can produce a random integer between 1 and 6 by rolling a die. But how do you produce a random integer between...

I would say yes based on how it is written. I copied it directly off this website: http://philosophy.hku.hk/think/logic/hardest.php

Here's one that has been called the the hardest recreational logical puzzle ever created. Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter...

But not an infinite number that can fit on a piece of paper of finite size. :jawdrop: That doesn't have anything to do with the answer though does it?

But they can be integers of any value, not just 1-10.

Yeah, you're right. Sorry about that. Original post is now corrected.

Let me try to explain this one. The solution requires each dragon to believe the other dragons have perfect logic. Let's start with 3 dragons since some of you seem to think this is where the riddle breaks down. Dragon A sees that Dragon B and Dragon C have red eyes. He first considers what...

Miami