Nihl's riddle: You're in small labyrinth w/ 4 ways around you. At end of each is...

Andu

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Riddle:
You are in a small labyrinth.
There are 4 ways around you.
At the end of each way there is a room.
Three of four rooms are empty, while 4th have exit from the labirynth.
One empty room what is opposide to the room with exit is always red color.
During one average labyrynth run, how many rooms you will met before finding exit?

nihl riddle.jpg


I'm not good in probability, I have tried to solve this and I'm stuck at this part.

nihl riddle tries.jpg

And I think it's
(100% + 50% + 50% + 33%) / 4 = 58,33% of rooms

But also:
In 4 possible solutions we entered a total of 8 rooms, and in 4 solutions there are 16 rooms in total. So 8 / 16 = 50% of rooms

I need's your help.
 
Last edited:
And what are you allowed to do? Pick one way once? Then the probability is 1/4.
 
Riddle:
You are in a small labyrinth.
There are 4 ways around you.
At the end of each way there is a room.
Three of four rooms are empty, while 4th have exit from the labirynth.
One empty room what is opposide to the room with exit is always red color.
How much chances do you have to find a correct way?

View attachment 9679


I'm not good in probability, I have tried to solve this and I'm stuck at this part.

View attachment 9680

And I think it's
100% - [ (100% + 50% + 50% + 33%) / 4 ] = 41,66%

But also:
In 4 possible solutions we entered a total of 8 rooms, and in 4 solutions there are 16 rooms in total. So 8 / 16 = 50%

I need's your help.

The question is ambiguous. I suppose "how much chances" is supposed to mean "what is the probability"; but it is unclear whether "to find a correct way" means on the first try (1/4), or eventually (1), or something else. And your work suggests that you are taking to mean something related to "expected number of tries".

Where did this come from, and what is the exact original wording?
 
And what are you allowed to do? Pick one way once? Then the probability is 1/4.

The question is ambiguous. I suppose "how much chances" is supposed to mean "what is the probability"; but it is unclear whether "to find a correct way" means on the first try (1/4), or eventually (1), or something else. And your work suggests that you are taking to mean something related to "expected number of tries".

Where did this come from, and what is the exact original wording?

Sorry, was unable to edit topic. Had to wait for moderator to aprove it.

The question is:
How many rooms in percent (25% equals to 1 room) you will met in average labirynth run, from start to exit.
 
The question is:
How many rooms in percent (25% equals to 1 room) you will met in average labirynth run, from start to exit.

Given that this is the question, what you did almost makes sense. But the numbers you averaged are not 25% for one room, and so on! You took visiting one room as 100%.

There's another issue, though: your 4 cases are not the correct cases. I would label the two empty rooms as, say, A and B, and then if you choose either of those, you still have to choose the next room to try randomly. Or perhaps you are assuming a specific strategy, in which when you have no new information you always take the next room clockwise, so that if you choose A (the northeast) your next choice is X, and if you choose B(southwest), your next choice is R. In that case, your chart is valid (except the last column).

Clearly you are allowed to use a strategy (using the information from the red room); it will be interesting to see whether this clockwise strategy changes the probabilities!

See what you get.
 
There is probability 1/4 that you will choose the path directly to the exit. That requires going down one tunnel.

There is probability 1/4 that you will choose the path to the red room. Then you turn around and go directly to the exit. That requires going down 3 tunnels (to the red room, back to the center, to the exit).

There is a probability 1/2 that you will choose the path to one of the other rooms. Having done that, turn around and go back to the center. At this time you know the exit is either to the right or left. There is now a probability 1/2 you choose the path to the exit. So there is a 1/4 probability you will go down 3 tunnels. But there is also a probability 1/2 you will choose the path to the red room. You will then turn around and go back to the center then to the exist. So there is a probability 1/4 you will go down 5 tunnels.

The expected number of tunnels you will have to go down is
(1/4)(1)+ (1/4)(3)+ (1/4)(3)+ 1/4(5)= 1/4+ 3/4+ 3/4+ 5/4= 12/4= 3. On average you will go down 3 tunnels before you find the exit.
 
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