Where's the cheese?

We seek an equation in terms of $p$ . Note that, if the mouse goes to either side of $0$ , he must go through $0$ to get to the other side. So we act like the mouse stays on one side, as if it moves to the other it has a new movement path.

Begin with the movement where the mouse eventually ends on $0$ . We can go $0\rightarrow1\rightarrow0$ , $0\rightarrow1\rightarrow2\rightarrow1\rightarrow0$ , $0\rightarrow1\rightarrow2\rightarrow1\rightarrow2\rightarrow1\rightarrow0$ , and so on. The probabilities are $(\frac{1}{2})^2,(\frac{1}{2})^4,(\frac{1}{2})^6$ , and so on. Summing these gives us an infinite geometric series; using the formula $\frac{a}{1-r}$ for this series, we obtain $\frac{\frac{1}{4}}{1-\frac{1}{4}=\frac{1}{3}}$ .

However, we were only considering one side of $0$ ; we multiply by $2$ to get both sides, since they are symmetric. Note that when the mouse gets to $0$ , we are back at probability $p$ . So we have that $p=(2)(\frac{1}{3})(p)+\text{ other terms}$ .

The only other place where the mouse can end up is $3$ (or $-3$ , but again we only consider one side). Our sequence of movement paths is $0\rightarrow1\rightarrow2\rightarrow3,0\rightarrow1\rightarrow2\rightarrow1\rightarrow2\rightarrow3,$ and so on. Again we get a geometric sequence, this time $(\frac{1}{2})^3,(\frac{1}{2})^5,(\frac{1}{2})^7,\cdots$ . Using the same formula for the summation, we get $\frac{\frac{1}{8}}{1-\frac{1}{4}}=\frac{1}{6}$ . We again multiply this by $2$ due to symmetry.

So we have the equation $p=\frac{2}{3}p+\frac{1}{3}$ , which implies $\frac{1}{3}p=\frac{1}{3}\Rightarrow p=1$ , so our answer is $100p=\boxed{100}$ .

At first I was confused, because it wasn't stated how many times Bernoulli would flip a coin... I then decided that it must be infinitely many times.

If the coin is truly random, then in an infinite sequence of H or T, there must exist every subsequence of H's and T's. Thus, the mouse token must eventually reach a cheese token, making the probability $p = 1$ .

$\lfloor 100p \rfloor = \lfloor 100\cdot1 \rfloor = \boxed{100}$

When Bernoulli is on point 0 the probability he will end on -1 or 1 is 1 so we let A be the probability of Bernoulli getting to 3 or -3 from 1 or -1 respectively. Let B be the probability of Bernoulli getting to 3 or -3 from 2 or -2 respectively. Then we get the two equations:

$A=\frac{1}{2}A+\frac{1}{2}B$

$B=\frac{1}{2}A+\frac{1}{2}$

Solving these equations will give $A=1$ . It is guaranteed that Bernoulli will land on 1 or -1 in his first coin flipping therefore $p=1 \times A = 1$ . Hence the answer is $⌊100p⌋=100$

If the mouse never reaches the cheese, this requires an infinite number of moves, so the probability is 0. If the mouse reaches the cheese, there is a finite number of steps, so the probability is not 0. Clearly only the nonzero terms will add to 1, so the mouse will always reach the cheese. $\lfloor100\cdot1\rfloor = 100$

This is a one dimensional random walk. In a one dimensional random walk, each point on the $\mathbb{Z}$ line is visited an infinite number of times. The result is called the gambler's ruin. Thus the probability that the mice will eventually reach is $1$ because it will cross the values $3$ and $-3$ an infinite number of times. The probability that the mouse will reach these points in $k$ steps is not $1$ .

Before we start blindly approaching the problem, we have to think for a minute.

Bernoulli essentially flips the coin forever. So we can easily see that at one point there will be $3$ more heads than tails or $3$ more tails than heads. Thus we get that the probability that the mouse gets the cheese is just $1$ . So ${1}\cdot{100}$ rounded down is just $\boxed {100}$

The probability that the mouse will get cheese is 1(1/2+1/2).Hence, 100*p=100.

The probability of the mouse not getting the cheese is $(1/2)^{n}$ , where n tends to infinity. Hence the probability of the mouse not getting the cheese tends to zero and hence, p =1. Thus [100p] = 100.

Since flipping a coin may result in 3 or more consecutive heads or tails, the mouse WILL ALWAYS able to get it either of the 2 cheese. p = 1 because the event will surely happen. Therefore [100p] = [100(1)] = [100] = 100 [x] is the floor function of x