Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 160

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When I was a vip and had this problem in the credits, I saw the benchmark solution. Eh, too bad I don't have a phenomenal memory, I would have given it a pass this time, and I could have shown the solution here too. All I can say is I remember one thing. The solution is nonstandard, non-school, such that not only an ordinary person cannot solve it, but also many teachers of mathematics in universities do not get to it (and it even they eventually remain unsolved, because the problem probably has no other solutions at all). In general, it can be solved if you have additional knowledge of mathematics, in short, the problem for special knowledge, in fact. There is a special series of numbers (I forget the name of the mathematician, like Fourier series, Fibbonacci, etc.), which has a certain regularity, which has been proved. And so the solution to this problem is built on the basis of that regularity. It is necessary to compare them, to show that the essence of the problem is equivalent to this series (which is already proved and patented), by which it becomes unambiguously obvious, what the probability is (i.e. the number of successful outcomes of all possible ones) for any number of pairs of buyers. And it is equal to 1/(N+1). That's the way it is. If this series of numbers is not known, the problem can not be solved, i.e. logically it is possible to come to the answer, but the justification is not strict, which in general nobody will count.
 
Road_king:
That's the way it is. If this series of numbers is not known, the problem cannot be solved, i.e. logically it is possible to come to an answer, but the justification is not rigorous, which in general nobody will give any credit to.
Is it not possible to solve it recurrenceally?
 
In any case I'm not aware of a solution that assumes school knowledge and nothing more, while being sufficiently rigorous to make it clear for sure that the answer is like that. However, I don't pretend that there is no such solution at all. Maybe you will find one.
 
Heroix: I disagree with the moderators. Otherwise, correct the terms of the problem.

Whatever you like.

The rules of the game on braingames.ru are not set by you.

I gave the problem as worded, as it is on this site. Usually, by reading the comments along with the moderators notes, you can find additional information valuable for clarifying the condition.

Your modification of the condition simplifies the problem too much, after which it becomes uninteresting and completely without a twist. It's a five-point problem!

Road_king: There is some special series of numbers (I forget the name of this mathematician, like Fourier series, Fibbonacci, etc.) that has some kind of regularity that has been proved. And so the solution to this problem is based on that regularity.

It's not as bad as it looks.
 

I can't think of any suitable formula...

But if you play with numbers on paper, then:

for one pair of buyers options:

10 (+) 01

(where 1 is the buyer with a 50 kopeck coin, 0 is the buyer with a ruble coin + - the variant in which everyone buys matches)

the probability that they both buy matches is 1/2.


For two pairs:

1100 (+) 0110

1010 (+) 0101

1001 0011

We get probability 2/6 or 1/3.


For three pairs

111000 (+) 101100 (+) 100101 011001 001110

110100 (+) 101010 (+) 100011 010110 001101

110010 (+) 101001 011100 010101 001011

110001 100110 011010 010011 000111

We get probability 5/20 or 1/4.

I.e. a pattern emerges: p=1/(n+1), where n is the number of pairs of buyers. Then for 50 pairs the probability p=1/51.

 
Heroix:

I disagree with the moderators. Otherwise, correct the terms of the problem.

Those moderators have probably never queued :)
 
Contender:
Those moderators have probably never queued :)
Did you mean those megamos?
 
Mathemat:
Did you mean those megamosks?

Those that:

Clarification from the forum moderators: this is not allowed.

For an experiment, I set this task in the department. Some of the staff hesitated, some started calculating percentages (i.e. as those moderators want), the smartest (well, I know the people I work with) immediately gave a probability of 1.0, because "if there is no change, the customer who doesn't need change will be allowed to use the counter".

This is a normal problem for intelligence. Such problems should not be complicated by artificial conditions.

 
Contender:

This is a normal problem of cleverness. Such problems should not be complicated by artificial conditions.

There you have it, homo prakticus-no-desiraus-thinking :) (not about you, but about "the most intelligent").

Well, consider that it happens with a hundred horses in a spherical vacuum, which will crash but will strictly follow the condition: no one ever gets ahead, and as soon as the seller runs out of change, the rest of them will go away unattended.

Guys, where are we - in the pure maths branch or something!

 

Here's a rather interesting challenge straight out of real life that happened to me yesterday in the office. So, there is a standard small drinking water boiler installed in the office. A standard 19 litre water bottle is placed in the boiler, with the neck down, as shown in the diagram below. Some of the water in the bottle is no longer there. However, the bottle itself is defective and there is a very small crack in the bottle neck, which is sufficient for the water to flow out of it (see black dash in the diagram).


This raises the question: What happens to the water in the bottle? There are two obvious possibilities:

a) Water will flow out of the bottle through the crack (sub-options: with or without pressure).

b) water will not flow out of the bottle through the crack (sub-options: all the time, only when water is being poured into the glass, etc.).

Perhaps someone will suggest other options. In general, I suggest speculation.

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