Cheese, mouse and a cat

Question

I have a problem and I don't know how to solve it because I don't know where to start. If we have the following situation:

Room 1-Room 2-Room 3-Room 4-Room 5

There is a little mouse in room 4 and he always forgets in which room he has been when going to the next room. In room 5 there is a big hungry cat waiting for him and in room 1 there is cheese. What is the chance he will get the cheese and not being eaten by the cat? My error solution: If I go like $\displaystyle \left(\frac{1}{2}\right)^{3} + \left(\frac{1}{2}\right)^{4} +$ (endless possibilities). I know that this is not the way I should calculate it. This is I think a geometric distribution because it is memoryless. On the other hand I can use binomial distribution to calculate it but there are endless possibilities..

I just need a push in the right direction. Thanks in advance.

In your formulation, can the rat always go from Room X to Room Y for every $X\neq Y$, that is, for every two different rooms there is a door that connects them? — Marra, Feb 19 '13 at 18:57
If you are in room 3, then you can go in room 2 and 4 etc. But if you are in room 2, you can't go in room 4 for example. — Asa1789, Feb 19 '13 at 19:00
For a more general setting with more rooms, see http://math.stackexchange.com/questions/288298/symmetric-random-walk-with-bounds?rq=1 — Ross Millikan, Feb 19 '13 at 19:00
You need to assume that the rat always moves and that the mouse chooses directions equally at random. — Thomas Andrews, Feb 19 '13 at 19:04
@Ross: That question treats the expected number of steps before eating or being eaten, not the probability for one of them occurring. — joriki, Feb 20 '13 at 09:57

score 4 · Answer 1 · answered Feb 19 '13 at 18:59

4

HINT: If he started in Room $3$, symmetry shows that his probability of getting the cheese would be $\frac12$. Starting in Room $4$ he either reaches Room $3$ on his first move or falls prey to the cat, each with probability $\frac12$.

Added: I’m assuming that the rooms are arranged in a line, as in the diagram in the question, and that the mouse is equally likely to go to the left and to the right.

answered Feb 19 '13 at 18:59

Brian M. Scott

616,228

I don't think so. He might just hang about in Rooms 2,3 and 4 forever. – Tara B Feb 19 '13 at 19:00
1

@Tara That occurs with probability 0. – MJD Feb 19 '13 at 19:00
Did you take in account that the rat could enter in a infinite loop like Room 2-Room 3-Room 2-Room 3... – Marra Feb 19 '13 at 19:00
1

@TaraB: His probability of doing so is $0$. – Brian M. Scott Feb 19 '13 at 19:00
@Brian Did OP specify that the probability of moving left is the same as of moving right? – MJD Feb 19 '13 at 19:01
@MJD: Yeah, I just realised that! – Tara B Feb 19 '13 at 19:01
1

@MJD: I’m assuming that that’s part of the import of memoryless. – Brian M. Scott Feb 19 '13 at 19:01
At least I wasn't the only one. =] – Tara B Feb 19 '13 at 19:02
1

@BrianM.Scott I don't agree. A memoryless actor could be equipped with, say, a die that has "left" on two faces and "right" on four. – MJD Feb 19 '13 at 19:02
@MJD: Of course that’s possible. But in the absence of a statement to that effect, the default assumption seems to me to be a symmetric random walk. – Brian M. Scott Feb 19 '13 at 19:03
Couldn't the little mouse just use its scent to figure this out? :P – Marra Feb 19 '13 at 19:03
@ Brian M. Scott: By the symmetry the chance going to the next or the previous room is equal. But the thing that bothers me is that th amount of possibilities like room 2,3,2,3,2,3,2,3,2 and then 1 etc etc. Should I take in count the amount of possibilities or is that not relevant here? I think it is but I am not sure.. – Asa1789 Feb 19 '13 at 19:09
@Asa1789: The probability that the mouse stays in Rooms $2$-$4$ forever is $0$. – Brian M. Scott Feb 19 '13 at 19:12
Calculate the chance that the mouse gets to Room 1 in exactly 3, 4, 5, etc, steps, then define a serie that depends on $n$. – Marra Feb 19 '13 at 19:15
@ Brian M. Scott: I didn't ment that he stayed between two rooms forever but that there are many many possibilities and I don't know how to handle that. – Asa1789 Feb 19 '13 at 19:15
Use factorials. – Marra Feb 19 '13 at 19:16
@Asa1789: The symmetry argument lets you avoid dealing with the various cases: you don’t have to handle them. – Brian M. Scott Feb 19 '13 at 19:16
@ Gustavo Marra: getting in room 1 could be done even in 10^10 ways, right? – Asa1789 Feb 19 '13 at 19:18
@Asa1789: There are infinitely many different ways to reach Room $1$. – Brian M. Scott Feb 19 '13 at 19:19
Exactly @Asa1789. Though you probably could find an expression for the mouse to reach Room 1 in $n$ movements, or at least find a upper bound for it and show that it converges. – Marra Feb 19 '13 at 19:20
Thanks all of you! I will now try to use symmetry to solve the problem. – Asa1789 Feb 19 '13 at 19:22
@Asa1789: You’re welcome. It really is the simplest solution, though there are certainly others that consider the process in more detail. – Brian M. Scott Feb 19 '13 at 19:24

Christian Blatter · Answer 2 · 2013-02-20T09:39:46.937

1

Denote by $p_k$ the probability that the mouse will finally get the cheese when she sits in room $k$. Then $$p_1=1,\quad p_2={1\over2}p_1 +{1\over 2}p_3,\quad p_3={1\over2}p_2+{1\over2} p_4,\quad p_4={1\over2} p_3+{1\over2}p_5, \quad p_5=0\ .$$ Now find $p_4$.

edited Feb 20 '13 at 09:39

answered Feb 19 '13 at 19:24

Christian Blatter

226,825

Cheese, mouse and a cat

2 Answers2