Catch that mouse!

We label the holes $A,B,C$ in some order and show that the best strategy is choosing one hole and sticking to it.

Clearly, at $t = 1$ , the probability of finding the mouse is $\frac{1}{3}$ . Let's assume we pointed our flashlight at hole $B$ . If we use the alleged best strategy (that is, if we continue to point the flashlight at $B$ ), the probability of finding the mouse at $t = 2$ is $\frac{2}{3}\cdot\frac{1}{2} = \frac{1}{3}$ because the mouse was not in $B$ at $t=1$ (which has a probability of $\frac{2}{3}$ of happening) and so it has a probability of $\frac{1}{3}$ of being in $B$ at $t=2$ .

If we use a different strategy (that is, if we switch which hole we are pointing our flashlight at), the probability changes to only $\frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}$ . To illustrate this, assume we switch to $A$ at $t = 2$ . If the mouse was in $A$ at $t=1$ (which has a probability of $\frac{1}{2}$ of being the case, since the mouse was not in $B$ and must therefore have been in either $A$ or $C$ ), then it can't possibly be in $A$ at $t=2$ because the mouse switches holes every time $t$ increases. If the mouse was in $C$ at $t=1$ , then it has a probability of $\frac{1}{2}$ of being in $A$ at $t=2$ . This gives a total probability of $\frac{2}{3}(\frac{1}{2}\cdot 0 + \frac{1}{2} \cdot \frac{1}{2}) = \frac{1}{6}$ for the mouse being in $A$ at $t=2$ .

This argument can of course be extended to other times. If we select a hole at time $t = i$ and do not find the mouse, the probability that the mouse will be in that hole at $t = i + 1$ is twice the probability that it is in each of the other two holes, and so picking a single hole and sticking to it is the best possible strategy. QED

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We proceed to find the expected value of the amount of time it takes to find the mouse using this strategy. The probability that we find the mouse at $t = 1$ is $\frac{1}{3}$ . With probability $\frac{2}{3}$ , we will have to check the hole again, and so the probability of finding the mouse at $t=2$ is $\frac{2}{3} \cdot \frac{1}{2}$ . In general, the probability of finding the mouse at time $t = i, i>1$ is $\frac{2}{3} \cdot \frac{1}{2^{i-1}}$ . Thus, our desired expected value is $\frac{1}{3} \cdot 1 + \frac{2}{3} (\sum_{i=2}^{\infty} (\frac{1}{2^{i-1}} \cdot i))$ . We evaluate the sum using simple arithmetic/geometric sequence techniques and find that it is equal to $3$ .

Thus, $\frac{a}{b} = \frac{1}{3} + \frac{2}{3} \cdot 3 = \frac{7}{3} \to a + b = \fbox{10}$ .

At $t_1$ , we can not do better than probability $\frac{1}{3}$ . If we don't find the mouse, we keep shining our flashlight into the same hole until we find it.

Given that we didn't find the mouse at time $t_i$ , it will appear in our hole at time $t_{i+1}$ with probability $\frac{1}{2}$ . We can not do better than that - not even if we know exactly where the mouse was as at time $t_i$ .

So we wait $1$ second at first, after which with probability $\frac{2}{3}$ we're not done yet. From that point onwards there is an expected waiting time $x$ of $2$ seconds ( $x=1+\frac{x}{2}$ , so $x=2$ ).

$E = 1 + \frac{2}{3} \times 2 = \frac{7}{3}$

It is easy to see that the probability of getting the mouse on the first second is $\frac{1}{3}$ , on the second second it is $\frac{2}{3} (\frac{1}{2})$ , on the third second it is $\frac{2}{3} (\frac{1}{2})^2$ and so on if you play optimally (That is, if at $t = 2$ the mouse is at hole $3$ , you don't then point the flashlight on hole $3$ at $t = 3$ ). So we need to compute

$\frac{1}{3} + 2(\frac{2}{3})(\frac{1}{2}) + 3(\frac{2}{3})(\frac{1}{2})^2 + ...$

Pull out $\frac{1}{3}$ and $\frac{2}{3}(\frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + ...)$ . The latter expression, as it is geometric with $|r| < 1$ , equals $\frac{2}{3}( \frac{\frac{1}{2}}{1 - \frac{1}{2}}) = \frac{2}{3}$ . Now, the leftover sum is the following: $\frac{2}{3}(\frac{1}{2} + 2(\frac{1}{2})^2 + 3(\frac{1}{2})^3 + ...)$

This sum can be written as an infinite sequence of infinite geometric sequences, namely $S_1 = \frac{2}{3}(\frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + ...), S_2 = \frac{2}{3}((\frac {1}{2})^2 + (\frac{1}{2})^3 + ... ...)$ .

Using the formula for infinite geometric sequences this equals

$\frac {2}{3} (\frac {\frac {1}{2}}{\frac {1}{2}} + \frac{(\frac{1}{2})^2}{\frac{1}{2}} + ...)$

Modifying it further, this sequence equals $\frac{2}{3}(1 + \frac{1}{2} + (\frac{1}{2})^2 + ...)$ . Now it is clear that this sequence is ALSO geometric, giving us $\frac{2}{3}(\frac{1}{1 - \frac{1}{2}}) = \frac {4}{3}$ .

Now, adding this back to the two components of the equation we pulled out, we get $E = \frac {4}{3} + \frac{1}{3} + \frac {2}{3} = \frac{7}{3} \rightarrow a + b = 10$ .

Simple problem if you understand it,

it's *1+ $\frac{2}{3}$ 2= $\frac{7}{3}$ **

7+3=10

Optimal strategy is to keep observing at the same hole itself.

The probability of observing the mouse at a particular hole at $t=1$ is 1/3.

If not observed in that hole, with probability 2/3, from one of the other two holes, the mouse can appear with equal probability of 1/2 in the hole being looked at or the other of the two other holes. Thus the probability of observing the mouse at $t=2$ is $\frac{2}{3} \times \frac{1}{2}.$

Probability of not finding the mouse at the given hole at $t=2$ is the same as above. This process can be repeated once again, and the probability of finding the mouse at $t=3$ at the hole being observed at is $\frac{2}{3} \times \frac{1}{2} \times \frac{1}{2} = \frac{2}{3} \times \frac{1}{2^2}.$ This process goes ad infinitum.

Thus expected observing time is given by the series $\langle t \rangle = \frac{1}{3} \times 1 + \frac{2}{3} \times \frac{1}{2} \times 2 + \frac{2}{3} \times \frac{1}{2^2} \times 3 + \cdots.$ This sum can be expressed in the form $\langle t \rangle = \frac{1}{3} + \frac{2}{3} \left( \sum_{n=1}^\infty \frac{n+1}{2^n} \right)$

The sum in the parenthesis can be evaluated using standard tricks with geometric ratio variable $x = 1/2$ $S(x) = \sum_{n=1}^\infty (n+1) x^n = \frac{\partial}{\partial x} \sum_{n=1}^\infty x^{n+1} = \frac{\partial}{\partial x} \frac{x^2}{1-x}.$ Thus, with $S(x) = \frac{2x}{1-x} + \frac{x^2}{(1-x)^2}, \qquad S(1/2) = 3.$

Therefore the expected time of first observation of the mouse is $\langle t \rangle = \frac{1}{3} + \frac{2}{3} \times 3 = 2 \frac{1}{3} = \frac{7}{3}.$