[Guest]

I searched and didn't find this as asked before.

A mouse has 3 rooms to go into.

There is a mouse and 3 rooms. It wants to find cheese. Only room 1 has cheese in it.

If it enters this room, after searching 3 minutes, it finds the cheese and eat it.

If it enters room 2, after spending 4 minutes for search, it goes out.

If it enters room 3, after spending 5 minutes for search, it goes out.

The mouse does not remember that it was in rooms 2 and 3 after it

goes out of them, and it continues going in and out until it

finds the cheese. (It can go into the same room again and again.)

What is the average time for the mouse to find cheese?

Edited April 17, 2009 by nobody

[Guest]

assuming each room has equal likelihood of being selected before selecting the room with cheese

room 1's cost = 3 mins

room 2's cost = 4 + 3 mins

room 3's cost = 5 + 3 mins

giving a total of 18, breaking it into an average = 6 mins

anywhere near to the solution? or is the answer infinite?

Edited April 17, 2009 by tarunark

[Guest]

assuming each room has equal likelihood of being selected before selecting the room with cheese

room 1's cost = 3 mins

room 2's cost = 4 + 3 mins

room 3's cost = 5 + 3 mins

giving a total of 15, breaking it into an average = 5 mins

anywhere near to the solution?

Your this statement is true:

assuming each room has equal likelihood of being selected before selecting the room with cheese

sorry, rest is false.

You say room2's cost is 4+3 but after going out from room 2,it can enter to room 2 again or to room 3 ???

[Guest]

Your this statement is true:

assuming each room has equal likelihood of being selected before selecting the room with cheese

sorry, rest is false.

You say room2's cost is 4+3 but after going out from room 2,it can enter to room 2 again or to room 3 ???

i see your point, then there could be an infinite loop situation as well, and then and infinite time spent before entering room 1.

besides that i accept my answer may not be correct, and for sure incorrect in the first post, 3 + 7+ 8 = 15 ??? what was i thinking,

will think about it a little more, any pointers?

[Guest]

i see your point, then there could be an infinite loop situation as well, and then and infinite time spent before entering room 1.

besides that i accept my answer may not be correct, and for sure incorrect in the first post, 3 + 7+ 8 = 15 ??? what was i thinking,

will think about it a little more, any pointers?

Yes, sure it may be an infinite loop; but this doesn't means that an average can't be calculated.

While flipping a coin it may show allways heads infinitely, this is not impossible, but on average its odds is 1/2.

Really I can give no clues, I wish but I can't find anything to point.

It has a quite simple solution. Indeed I didn't solve it myself, and I suppose who will post that simple solution, eather faced with this before, or is a quite clever guy.

[Guest]

It would take 12 minutes on average.

if T is the Time to find the cheese, you'd get the formula:

T = 1/3 * (3 + (4+T) + (5+T))

T = 12

[Guest]

12. Boiled down, its:

3 + 1.5 * 2/3 * 9 = 12

Is it coincidence that if you just add up the 3 amounts of time for the 3 rooms, you get 12 as well?

[Guest]

It would take 12 minutes on average.

if T is the Time to find the cheese, you'd get the formula:

T = 1/3 * (3 + (4+T) + (5+T))

T = 12

Yup, this is the simplest solution. Cong...

Now I saw Judix's post, result is correct but needs some explanation, maybe.

[Guest]

It would take 12 minutes on average.

if T is the Time to find the cheese, you'd get the formula:

T = 1/3 * (3 + (4+T) + (5+T))

T = 12

I did not even think of this procedure before, but it most definitely make sense.

If we call the average time to find the cheese T,

then we can say that the amount of time to find the cheese for

Room 1 - 3min, because that is where the cheese is

Room 2 - 4 + T, because after he exists the rooom he will still have to basically start over

Room 3 - 5 + T, same reasoning as Room 2

So if T is equal to the average of the times need to find the cheese, starting whith each room then.

T = (3 + (4+T) + (5+T))/3

so we get

T = 4 + T*2/3

carry over the T's, and get

T/3 = 4

so

T=12

I hope that provides an explanation

[Guest]

Average time = SUM (2/3)^i + 1.5*SUM i*(2/3)^i - where i starts from 0 (zero) - ( ^i means at the power of i).

Fairly simple to get to. A bit more complicated to solve it, but it can be done.

The answer is: 16.5 minutes.

I do not believe that there is a simple solution. The proof is kind of lengthy.

[Guest]

Average time = SUM (2/3)^i + 1.5*SUM i*(2/3)^i - where i starts from 0 (zero) - ( ^i means at the power of i).

Fairly simple to get to. A bit more complicated to solve it, but it can be done.

The answer is: 16.5 minutes.

I do not believe that there is a simple solution. The proof is kind of lengthy.

Hmm... I dont know. I think the above answers of 12 seconds makes sense to me and the solution that Diloul gave was quite simple.

[Guest]

Average time = SUM (2/3)^i + 1.5*SUM i*(2/3)^i - where i starts from 0 (zero) - ( ^i means at the power of i).

Fairly simple to get to. A bit more complicated to solve it, but it can be done.

The answer is: 16.5 minutes.

I do not believe that there is a simple solution. The proof is kind of lengthy.

The simple solution above, by diloul is accepted as correct in a serious math forum.

Really I also thougth that this problem would have a complicated solution, but the simple one as 12 seems to be correct, and no objection has come in that forum.