Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 36

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MetaDriver:

but listen, right.

you first need to scatter two barrels. for two, the algorithm is clear (for one, even more so)

we need to correctly work out an algorithm for moving to the number of barrels n+1

It is necessary to remember the volume of petrol in the tank after each refill and dance from it. What happens, if while driving from a new arbitrary barrel to the nearest one (in the same direction) there is enough petrol to the previous amount (then the route is passed because of the matinduction assumption, as all missing quantities from every barrel were already poured into the new one and so they are already in the tank) and what happens, if there is not enough (there are couple more cases to consider).
 

In general, I have a suspicion that there is always a solution both ways.

There was also the idea of distributing barrels equivalently to a pile of litre barrels. There is also the suspicion that it is possible to prove the opposite.

 
TheXpert:

In general, I have a suspicion that there is always a solution both ways.

I have a physico-geometrical solution in my head. Take a ring (preferably a weightless one) and place flat weights on its inner side, proportional to the volumes of barrels. put it on the table, wait until it balances out. Then count down the barrels from the bottom point (separately to the left and to the right), counting gasoline in them so that there is enough when we move to the bottom edge (towards the counting). The counting is interrupted if you meet a barrel with not enough petrol to reach the previous one. Then we see where (on the left or on the right) the chain is bigger (according to the amount of petrol). From this edge we start, in the direction of the lower edge of the ring.

The algorithm obviously works, I don't know how to prove it.

Moreover, it is possible that you are right, and it is possible to start from the opposite side, though it is not so obvious.

But there's bound to be a one way solution, unequivocally.

--

if the ring rolls freely (balances in any position) - then you can start from any barrel and move towards the nearest one.

 
alsu:
that's why such probabilities are called a posteriori, Bayes formula has been invented for them, which gives the same answer.

)))))

Let's do a little quiz and you will probably see where you made a mistake:

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

My guess is that the probability of the letter being in the last drawer is 1 (100%) ! According to you it's 1/8 ( 12.5% ) ?!?

p.s. i wonder what Mathemat.... has to say

 
Manov:

)))))

Let's do a little quiz and you will probably see where you made a mistake:

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

My guess is that the probability of the letter being in the last drawer is 1 (100%) ! You think it's 1/8 ( 12.5% ) ?! ?!?

I propose to simplify it even further.

With probability 1 (100%) a letter is placed in one (1) drawer of the table. Then one by one opened 7 drawers...............

Is that better? :)

 
Manov:

)))))

Let's do a little quiz and you will probably see where you made a mistake:

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


Seriously, it seems to me that the original problem is equivalent to this:

With probability 1 (100%), a letter is placed in one of the 16 desk drawers (chosen at random). Then 7 drawers are opened one by one - all are empty. What is the probability that there is a letter in the 8th drawer?

And with it everything becomes clear at once, or not?

 
MetaDriver:

In all seriousness, it seems to me that the original problem is equivalent to this:

With probability 1 (100%) a letter was placed in one of the 16 desk drawers (chosen at random). Then 7 drawers were opened one by one - all are empty. What is the probability that there is a letter in the 8th drawer?

And with it everything becomes clear at once, or not?

The probability increases with each open box, and I've shown how. If the initial probability is 1, then with probability 1 the letter is in the last drawer. If 0.5, then 0.5. I don't know what probability theory says about the existence of an intertemporal interdimensional letter carrier, but the letter lies in the last box with a probability equal to the initial probability for all boxes.

->

joo : Since 7 boxes are empty, the probability is 0.5, either there is or there isn't.

 
MetaDriver:

In all seriousness, it seems to me that the original problem is equivalent to this:

With probability 1 (100%) a letter was put in one of the 16 desk drawers (chosen at random). Then 7 drawers were opened in turn - all empty. What is the probability that there is a letter in the 8th drawer?

And with it everything becomes clear at once, or not?

)))))))

after a short conversion, so get 8/16 = 1/2, my answer :)

from where 1/8 or 1/16....

 
Manov:

)))))))

after a short conversion, so get 8/16 = 1/2, my answer :)

whence 1/8 or 1/16....

I think you are already 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%)

 
Let's talk about petrol already. It hasn't gone up yet.
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