[Guest]

This is very similar to a recent puzzle but I have made it slightly more complex:

There are 3 rooms and a mouse. Inside the first room there is some cheese. The mouse has poor memory and does not remember what rooms he has already been into. If he goes into the first room he searches for 3 minutes and gets the cheese. If he goes into the second he spends 4 minutes and then leaves. If he goes into the third room he spends 5 minutes and then leaves. The mouse continues to enter rooms until it has the cheese.

HOWEVER, THE MOUSE NEVER RE-ENTERS THE ROOM IT JUST LEFT.

So what is the average time it takes for the mouse to get the cheese now?

[Guest]

Question: is it not possible for the mouse to take an infinite amount of time to find the cheese. Thus making an average inaccurate? I just figured a continues loop, since possible, should throw off the entire equation.

[Guest]

There are 5 possible ways the mouse could go if he never re-enters a room. Room A - 3 minutes. Room B then A - 7 minutes. Room C then A - 8 minutes. Room B then C then A - 12 minutes. And Room C then B then A - 12 minutes. The average of 3, 7, 8, 12, and 12 is 8.4. So the average time he takes to find the cheese would be 8.4 minutes.

[Guest]

Question: is it not possible for the mouse to take an infinite amount of time to find the cheese. Thus making an average inaccurate? I just figured a continues loop, since possible, should throw off the entire equation.

The longer it takes the less likely this becomes. So the chance that the mouse takes for ever would be 0 which would cancel out the infinite amount of time... if that makes sense.

So I guess the answer is no

[Guest]

There are 5 possible ways the mouse could go if he never re-enters a room. Room A - 3 minutes. Room B then A - 7 minutes. Room C then A - 8 minutes. Room B then C then A - 12 minutes. And Room C then B then A - 12 minutes. The average of 3, 7, 8, 12, and 12 is 8.4. So the average time he takes to find the cheese would be 8.4 minutes.

Ah, sorry I had a feeling my OP was misleading.

The mouse CAN re-enter a room but not when it has just left. eg. the mouse enters room 3, then room 2 and then can go into room 3 again. But it can't go to room 3 and re-enter immediately.

[Guest]

Well technically the mouse could go into the second room, come out, go into the third, come out go into the second, come out go into the third, come out go into the second..............................(riddle inside a riddle, how many times did i miss use comas?)

And as for this being "unlikely":

You never said that the mouse is less likely to go into a room she/he previously did, so the fact that it is "less likely" is only logic from our world being placed onto the mouses world. If she/he forgets she/he forgets, meaning it is just as likely for him to enter any of the two rooms. I know its far fetched but completely possible (similar to flipping a coin and always getting heads)

Edited April 18, 2009 by amfavs

[Guest]

Assuming the room the mouse enters each time is random:

4.25

[Guest]

Well technically the mouse could go into the second room, come out, go into the third, come out go into the second, come out go into the third, come out go into the second..............................(riddle inside a riddle, how many times did i miss use comas?)

And as for this being "unlikely":

You never said that the mouse is less likely to go into a room she/he previously did, so the fact that it is "less likely" is only logic from our world being placed onto the mouses world. If she/he forgets she/he forgets, meaning it is just as likely for him to enter any of the two rooms. I know its far fetched but completely possible (similar to flipping a coin and always getting heads)

Assuming the mouse's choice is always random (between all three rooms for the first choice, and between the two remaining rooms after he has completed a fuitless search), it is possible the the mouse will keep alternating between the two rooms without the cheese. However, the longer this goes on, the lower the probability. As the number of time the mouse atlernates between these two rooms approaches infinity, the probability approaches zero. Do the math and you will get the answer in my previous post.

[Guest]

Assuming the room the mouse enters each time is random:

4.25

This isn't right. What method did you use?

[unreality]

isn't this exactly the same as this?

[Guest]

isn't this exactly the same as this?

Nope. There is a subtle difference which makes this one more complicated - the fact that the mouse cannot immediatly re-enter the same room.

[Guest]

This isn't right. What method did you use?

Ah, I think I see the error in my ways -- after the first choice (with a 1/3 probability) each additional choice is a 1/2 probablilty, since the mouse cannot go back into the room he just left. (My original answer used 1/3 probability for each choice).

9 minutes:

it is an inifinite series -- the average time is the time for each possibility times its probability:

3*(1/3) + (7+8)*(1/3)*(1/2) + (12+12)*(1/3)*(1/2)^2 + (16+17)*(1/3)*(1/2)^3 . . .

This can be reduced to -1 + the infinite series (9*(x-1)+6)*(1/3)*(1/2)^(x-1) as x goes from 1 to infinity.

(There may be a simpler way to express the formula.)

[Guest]

Ah, I think I see the error in my ways -- after the first choice (with a 1/3 probability) each additional choice is a 1/2 probablilty, since the mouse cannot go back into the room he just left. (My original answer used 1/3 probability for each choice).

9 minutes:

it is an inifinite series -- the average time is the time for each possibility times its probability:

3*(1/3) + (7+8)*(1/3)*(1/2) + (12+12)*(1/3)*(1/2)^2 + (16+17)*(1/3)*(1/2)^3 . . .

This can be reduced to -1 + the infinite series (9*(x-1)+6)*(1/3)*(1/2)^(x-1) as x goes from 1 to infinity.

(There may be a simpler way to express the formula.)

Correct

[Guest]

I understand that mathematically it gets closer to zero, but you asked for the average. Logically, there could be an infinite number of times that the mouse almost reaches infinity. Thus the average could correctly be 0.0002. I think you are using Math while I am using Logic. To ask an average on an unlimited test seems logically impossible.

[Guest]

I understand that mathematically it gets closer to zero, but you asked for the average. Logically, there could be an infinite number of times that the mouse almost reaches infinity. Thus the average could correctly be 0.0002. I think you are using Math while I am using Logic. To ask an average on an unlimited test seems logically impossible.

Not at all. The "average" would be what you get if you were to do this in experiment lots of times or the value it would be tending towards. I don't know how esle to explain it. The chance that it takes a very long time is very unlikely and so has little weighting.

[Guest]

I understand what you are saying, but there is still too much assumed. Average is the middle of a set of values. Now, since there was never a set limit or max. amount created for the set, technically the set could consist of three attempts. Hypothetically speaking, the mouse could find the cheese the first time for all three example attempts, meaning that the correct average (for this example) is 3 minutes. On the other hand, it should also be noted that any other value could replace each one of these three. I understand that mathematically (with correct limitations) this problem is completely possible. However since there are not proper limitations my answer stands correct, 3 minutes.

Edited April 18, 2009 by amfavs

[Guest]

I understand what you are saying, but there is still too much assumed. Average is the middle of a set of values. Now, since there was never a set limit or max. amount created for the set, technically the set could consist of three attempts. Hypothetically speaking, the mouse could find the cheese the first time for all three example attempts, meaning that the correct average (for this example) is 3 minutes. On the other hand, it should also be noted that any other value could replace each one of these three. I understand that mathematically (with correct limitations) this problem is completely possible. However since there are not proper limitations my answer stands correct, 3 minutes.

Hmm... When you say "Average is the middle of a set of values" it soudns like your getting confused between mean and median.

Well anyway, I wrote a quick program to simulate this scenario. It iterates it 100,000 times and accumulates the times taken and then divides this by 100,000 (ie average). And it comes at 9 minutes (to at least 2 sf) everytime. There is no more I can say on this. This was the value I was intending people to work out.

[Guest]

Ouch!

I am not getting confused, I never said that the middle had to be in that set of numbers.

Well if you said that our mouse had to attempt to get the cheese 100,000 times then your answer would be 100% correct. Unfortunately, there were no limits meaning that my 3 minutes as well as your 9 minutes are both correct.

Great!

We may never agree. You think mathematically, I think logically. One is not greater than the other.

Edited April 18, 2009 by amfavs

[Guest]

Well i'm sorry. I really cant see where your coming from here. I still dont understand how you got an average time of 3 minutes. Maybe someone else can explain?

[Guest]

I understand what you are saying, but there is still too much assumed. Average is the middle of a set of values. Now, since there was never a set limit or max. amount created for the set, technically the set could consist of three attempts. Hypothetically speaking, the mouse could find the cheese the first time for all three example attempts, meaning that the correct average (for this example) is 3 minutes. On the other hand, it should also be noted that any other value could replace each one of these three. I understand that mathematically (with correct limitations) this problem is completely possible. However since there are not proper limitations my answer stands correct, 3 minutes.

I think it is safe to assume when the question asked for the "average" time, it meant the population average (mu in statistical terms) rather than a sample average. The population average would represent the average of all possible combinations -- this can be calulated by finding the limit of the infinite series (see my earlier post). The only other assumption you have to make is that the mouse's choice is completely random.

Logic in it's purest form can be expressed mathematiclly.

[Guest]

Well i'm sorry. I really cant see where your coming from here. I still dont understand how you got an average time of 3 minutes. Maybe someone else can explain?

Okay, average is specific to a set a data. Since your set of data is limitless, so is the average. Your average is correct when there are 100,000 attempts. And although it is thought that averages generally stay the same I am sure that if you were to run the same test with 3, 50, 100, 150, 200, 1,000 .... attempts each average would be different. All I am saying is, if you want there to be ONE correct answer then the question needs more limitations. Previously I attempted to express this thought by using a very unlikely (although possible) extreme.

Thank you,

Amfavs

Edited April 18, 2009 by amfavs

[Guest]

I think it is safe to assume when the question asked for the "average" time, it meant the population average (mu in statistical terms) rather than a sample average. The population average would represent the average of all possible combinations -- this can be calulated by finding the limit of the infinite series (see my earlier post). The only other assumption you have to make is that the mouse's choice is completely random.

Logic in it's purest form can be expressed mathematiclly.

Understood... However i have never felt it "safe to assume"

[Guest]

This is very similar to a recent puzzle but I have made it slightly more complex:

There are 3 rooms and a mouse. Inside the first room there is some cheese. The mouse has poor memory and does not remember what rooms he has already been into. If he goes into the first room he searches for 3 minutes and gets the cheese. If he goes into the second he spends 4 minutes and then leaves. If he goes into the third room he spends 5 minutes and then leaves. The mouse continues to enter rooms until it has the cheese.

HOWEVER, THE MOUSE NEVER RE-ENTERS THE ROOM IT JUST LEFT.

So what is the average time it takes for the mouse to get the cheese now?

So the mouse enters the first room, and finds the cheese after three minutes. And it does not need to enter the second and third rooms, so the average time for the mouse to find the cheese stays three minutes.

Edited April 18, 2009 by Eternal child