Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 37

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MetaDriver:
I think you're joking.

In this variant, after opening each box (and discovering that it is empty) the probability that the letter is in the next obviously increases.

1 = 1/16

2 = 1/15

3 = 1/14

...

8 = 1/9

9 = 1/8

...

15 = 1/2

16 = 1 (100%)

Taaaaaac.... Exactly.... :)

And if drawers=8 -> ....

And if the initial probability = 1/2 ? ))))

As for petrol, the answer is very simple : you can. (if we know at the beginning how much there is)

 
Manov:

Taaaaaaack.... Exactly.... :)

And if drawers=8 -> ....

And if the initial probability = 1/2 ? ))))

... Then it is equivalent to this:

With probability 1 (100%) a letter was placed in one of the 16 drawers of the table (chosen at random). Then half of the drawers were removed . Then 7 drawers are opened one by one - all are empty. What is the probability that there is a letter in the last drawer?

 
MetaDriver:

... Then it is equivalent to this:

With probability 1 (100%) a letter was placed in one of the 16 drawers of the table (chosen at random). Then half of the drawers were removed . Then 7 drawers are opened one by one - all are empty. What is the probability that there is a letter in the last drawer?

1/2, obviously (the probability of removing the box with the letter = 1/2)
 
Dimitar, you'd better sort it out. The answer is 1/9 is correct. The further you open it, the less likely it is that the letter was laid down at all.
 
TheXpert:
Dimitar, you'd better sort it out. The answer is 1/9 correct. The further you open it, the less likely it is that the letter was even laid down.
Yeah, now it's your turn. I'm off for popcorn.
 

Manov:

MetaDriver:

... Then it is equivalent to this:

With probability 1 (100%) a letter was placed in one of the 16 drawers of the table (chosen at random). Then half of the drawers were removed . Then 7 drawers are opened one by one - all are empty. What is the probability that there is a letter in the last drawer?

1/2, obviously (probability, on having removed the box with the letter = 1/2)

I am going to kill you. Now another problem.

With probability 1 (100%), put a letter into one of the 16 drawers of the table (chosen at random). Then half of the drawers are moved by 1 meter . Then they opened 7 drawers one by one - all of them are empty. What is the probability that there is a letter in the 8th drawer?

 
MetaDriver:

I'll finish you off. Now another problem.

With probability 1 (100%) in one of the 16 drawers of the table (chosen at random) put a letter. Then half of the drawers were moved 1 meter away . Then they opened 7 drawers one by one - all of them are empty. What is the probability that there is a letter in the 8th drawer?

And here is the last one.

Sixteen drawers of the table are randomly placed 16 cards with hexadecimal digits from 0 to F written on them. Then half of the drawers are removed . Then they opened seven drawers one by one. They contain the numbers 3, 5, B, A, 4, 0, E. What is the probability that the number F is in the 8th drawer ?

 
alsu: The essence of the game and the principle of winning are similar to him, so the solution came to mind almost immediately.

There's another one about them, I'll post it now. This one is different.

But I didn't know the rules when I was solving it, so I had to make it up as I went along and justify it.

 
alsu:

Suppose the statement of the theorem is incorrect, i.e. for any grid shifts at least one node is covered by a blot.

Let's fix some position of the grid. Let node 1 of some cell be under the ink. Since the area of the blots is smaller than the area of the cell, there must be an area inside the cell that is not covered by the blot. Consider all possible shifts of the grid such that node 1 moves into a clean region. By our assumption, at least one of the nodes 2,3,4 of the same cell must move under the blot, and necessarily outside the cell (since node 1 has moved inside). Hence, each point of the cell, not filled with ink, corresponds to at least one point outside the cell, filled with ink. Hence, it follows that the area of the ink cannot be smaller than the area of the cell. We come to a contradiction, the theorem is proved.

Well, Alexei came along and blew everyone away. I have almost the same, covering the torus with a plane, I think it's called.

I simply moved all the blots to one cell and moved the origin of coordinates to the blot-free area.

 
TheXpert: Nah, the second interpretation doesn't make sense. Unless you ask the skip.
The correct answer is 1/9. There are no peeks.
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