[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 609

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C-4:

So? Who's arguing? If the numbers are even, the money goes to the megabrain's opponent. In other cases, it'll be exactly 50-50. Now explain how the megamozg will make money on this?

Here is an example of the cubes:

1: 111444

2: 222666

3: 555333

there are no equal numbers here.

In games of 1st and 2nd dice, the probability of winning the 2nd is 0.75

In games of 1st and 3rd dice the probability of winning the 3rd is 0.75

In games of dice 2 and 3, the probability of winning is the same: 0.5

If the megabrain's opponent is smart, he will choose the 2nd or 3rd die, if not, he will choose the 1st die as well. If the number of games is high, the megabrain will win statistically.

 
Mathemat:
How do you understand - in your own words - the term "polbita"?

I wrote two pages ago. One bit is conventionally a system capable of accepting two values. The task requires that the size of the information transmitted should not exceed one bit (only one question, the answer can be either "yes" or "no"). However, in order to process this one bit, we need to know all its 2 possible states, and we do not know this. If the system returned "0" (one of the two states of one bit or 0.5 bits), for example, we do not know whether the answer is true or false, because there is no agreement about what is true or false.
 
PapaYozh:

Here is an example of the cubes:

1: 111444

2: 222666

3: 555333

there are no equal numbers here.

In the 1st and 2nd dice games, the probability of winning the 2nd is 0.75

In the 1st and 3rd dice games, the probability of winning the 3rd is 0.75

In games of 2nd and 3rd dice, the odds of winning are the same: 0.5

If your opponent is very clever, he will choose the 2nd or 3rd die, and if not, he will choose the 1st die as well. If the number of games is high, the megabrain will statistically win.


I agree. There really is no overlap. However, you admit that at least from time to time a megabrain will come across a moron as an opponent, who will pick a die with a lower probability. Then and only then will the odds mix up in his favour. But then it would be easier to introduce another condition: the opponent of the megabrain is always a moron and must always pick the die with the lowest odds. Or here's another one: the megabrain's opponent is blindfolded, and at the first opportunity the megabrain will take out the die with the higher odds.
 
Mathemat:

You're making that up again: no simplification, just a clarification of the term.

I get the impression that somewhere inside the C-4 nickname sits a worm that is constantly forcing its owner to twist any phrase or task. No offence, OK?


Why are you making it up? The terms are written in black and white: city. A city is always a plurality. Have you seen a city of two people? When they say "city" they mean a lot of people, in this case some of whom are telling the truth and some of whom are lying. This is the crux of the problem, and it doesn't seem to have a solution in this formulation.
 
C-4:
... in this formulation it has no solution.
I agree. if only because we don't know which "townspeople" are in front of us.
 

Here is a counter-function returning exactly 1 bit. Call it 1 000 000 times, you will not be able to draw any conclusions about its results from the 1 000 000 bits it returns:

bool Expert(bool ThisLeftTrue)
{
   bool INowThisLeft = true;
   bool AmILiar;
   if(Rand()%2 == 0)AmILiar = true;
   else AmILiar = false;
   if(AmILiar)return !INowThisLeft;
   else return INowThisLeft;
}
 
moskitman:
C-4:
How about this question: "(Are you a liar and you don't have a coloured body) or (Are you a truth-teller and you have a coloured body)?"
 
GaryKa:
How about this question: "(You are a liar and you don't have a coloured body) or (You are a truth-teller and you have a coloured body)?"

and I don't care. I have a colour TV and I'm a liar at times.

Well, on the subject, it really doesn't make any difference. The truth, by going through one YES and one NO is inverted, that's the solution to the problem.

 

Well, C-4 and moskitman in their little circle have already decided that the problem has no solution. That's good, that's good.

But there is a state in the USA, which also decided that pi equals exactly 3, without any tenths.

 
GaryKa:
How about this question: "(Are you a liar and you don't have a coloured body) or (Are you a truth-teller and you have a coloured body)?"


That is, after all, two questions wrapped in the same shell if. This is a softer version of such an interrogation:

if(Вы_лжец)
{
   if(А_Телевизор_у_Вас_есть)
   {
       if(А_может_быть_и_домашняя_кошечка_найдется)
       {
           if(Ну_тогда_вы_просто_чудо)
           {
               Чудо = true;
           }
       }
   }
}
You can, of course, wrap it in one if box. Who likes it:)
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