Quant interview problems

The lower bound is $0.02$ and the upper bound is $1$.

Consider a Markov chain with two states. Either we are with the umbrella $S_1$ or we are without the umbrella $S_0$. We set $P_i$ the prob of being is state $S_i$ and $p_{ij}$ the prob of transitioning from state $S_i$ to state $S_j$. We note $p_{00} = 0, p_{01} = 1, p_{11} = p, p_{10} = 1-p$. Then we get the set of equations $P_0 = P_1 \cdot ( 1- p ); P_1 = P_0 + P_1 \cdot p$, together with $P_0 + P_1 = 1$. Then note $\mathbb{P}[\text{Getting Wet}] = P_0 \cdot p$. This results in the probability of getting wet being $\frac{1-p}{2-p}p$.

Build a directed graph from the integers found in the subsets. The neighbours of an integer are the integers on its right in any of the given subsets. Then use topological sorting.

We define the constants: total amount of steps on the escalator $M$, and the speed of the escalator $v_e$. Then we have $$M = t \cdot (v_e + v_b)$$ with $v_b$ the speed of Bob and $t$ the amount of time to reach the top. The first time, the total time passed was $20 / 1 = 20$ seconds and the second time $30 / 2 = 15$ seconds. We then get the system of equations $M = 20(v_e + 1), M=15(v_e + 2)$, from which follows $M=60$ and $v_e=2$. We then fill in for $v_b = 3$ and find $t = 12$, meaning Bob walked $36$ steps. If $v_b = \infty$, then clearly the escalator did not contribute, so Bob walked all $M=60$ steps.

This answer is incomplete:
You clearly never play paper if the opponent never plays rock. So you have to choose the probability $p$ by which to play rock (and play scissors with prob $1-p$). The opponent chooses simultaneously paper with probability $q$ and scissors with probability $1-q$. The equilibrium comes with $p = q = \frac{1}{3}$. The expected win rate is then $\frac{4}{9}$ and your lose rate is $\frac{1}{9}$. You are willing to pay up to $\frac{1}{3}$ to play.

Note that $\mathbb{P}[\text{Survival}] = \mathbb{P}[ \text{3 out of first 4 are caught}] + \mathbb{P}[ \text{Not 3 our of first 4 are caught & you are not caught}]$. The first probability on the right hand side equals $\mathbb{P}[\text{3 out of first 4 are caught}] = (1 / 2)^3 + 3 \cdot (1 / 2)^4 = 5 / 16$. The second probability on the right hand side then equals to $\mathbb{P}[\text{Not 3 our of first 4 are caught & you are not caught}] = (1 - \frac{5}{16}) (1 / 2) = 11 / 32$ Which implies $\mathbb{P}[\text{Survival}] = 11 / 32 + 5 / 16 = 21 / 32 = 0.65625$. A little over $0.5$ is to be expected from the problem statement.

Yes. The expectation of both players is the same, namely $3 \cdot 3.5 = 10.5$, and they are both distributed symmetrically around their expectation.

The game is not fair anymore. Say $X_i$ is the distribution of the $i$-th player. We now compete not against $X_2$, but against $max(X_2, X_3)$. It becomes clear when considering not one extra, but 100 extra players. The probability of one in $101$ players rolling d20s with a result of 19 or 20 is $1 - (0.9)^{101} \approx 1$. We cannot win whenever at least one player rolls a 19 or 20, and the probability of this happening is already more than $0.99$, so the game is clearly unfair for extra players. The more extra players, the least fair it becomes.

First one has a higher probability of winning. This can be seen using tree structures. Alternatively one can see this as a markov chain, consisting of states $000, 001, 010, 011, 100, 101, 110, 111$. We start in a random state. The state $011$ can only be reached from $001$ with a probability of $0.5$, but $001$ is itself a terminating state, so in fact $011$ can only be reached upon initiation. The probability of this happening is $\frac{1}{8}$. In all other cases, we will reach $001 before we reach $011$, meaning we win with probability $\frac{7}{8}$

There are two legit positions on the table, these are 3cm and 13cm. In both positions, they fall off with probability $0.5$. This implies a geometric distribution (with parameter $0.5$) of the amount of steps. And thus an expectation of $1 / 0.5 = 2$ steps before falling off.