Three Rooms, Safety, and Death

Note that if you are on $R_2$ , The probabilities of ending up in $S$ or $D$ are equal, since they are both two rooms away.

So the probability of surviving once you get to $R_2$ is $\frac{1}{2}$ .

When you are on $R_1$ , you have a probability of $\frac{1}{2}$ that you escape, and a probability of $\frac{1}{2}$ that you escape with a probability of $\frac{1}{2}$ .

So the probability that you escape is $\frac{1}{2}+\frac{1}{2}*\frac{1}{2}=\frac{3}{4}$ .

Since $3$ and $4$ are coprime, the required answer is $3+4=\boxed{7}$ .

Let $P_1$ , $P_2$ and $P_3$ be the probabilities of escape if you are in room 1, 2 and 3, respectively.

If you are in room 1, you have a $\frac{1}{2}$ chance of choosing door $S$ and a $\frac{1}{2}$ chance of choosing room 2, so we have $P_1 = \frac{1}{2}+\frac{1}{2} P_2$

If you are in room 2, you have a $\frac{1}{2}$ chance of choosing room 1 and a $\frac{1}{2}$ chance of choosing room 3, so we have $P_2 = \frac{1}{2}P_1+\frac{1}{2} P_3$

If you are in room 3, you have a $\frac{1}{2}$ chance of surely dying and a $\frac{1}{2}$ chance of choosing room 2, so we have $P_3 = \frac{1}{2} (0)+\frac{1}{2} P_2$

Solving the three equations simultaneously, we arrive at $P_1=\frac{3}{4},P_2=\frac{1}{2},P_3=\frac{1}{4}$ . Thus the answer is $3+4=\fbox{7}$

I'll touch on two solutions aside from the ones already posted. The first one is a bit technical, as it requires knowledge of martingale theory, but the second is pretty intuitive as it follows from Newtonian mechanics.

Solved using Stochastic Processes:

This is equivalent to a symmetric random walk with barriers at -1 and 3 starting from 0. Since the process is a martingale, we can use the optional stopping theorem (Doob's optional sampling theorem) to solve for the probability we survive:

$P(survive) \times -1 + (1-P(Survive)) \times 3 = 0$ .

Solving for P(Survive), we get $\frac{3}{4}$ , so $3 + 4 = 7.$

Solved using Basic Mechanics from Physics:

Alternatively, consider we had a see-saw of length 1, with a 1 pound kid sitting on one end and a 3 pound kid sitting on another end. To balance, you'll notice we have to place the fulcrum closer to the fatter kid since he's heavier. The distance from the 1 pound kid to the fulcrum is the probability that we survive (fulcrum is placed at distance $\frac{3}{4}$ from the 1 pound kid) .

There's a $\frac{1}{2}$ chance of immediate survival, and a $\frac{1}{2}$ chance of moving to $R_2$ . If you go to $R_2$ , you are situated exactly halfway between life and death, so there is a $\frac{1}{2}$ chance of survival.

Summing up the cases gives $\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4}$ , so the answer is $3+4 = \boxed{7}$ .

Because at first time, probability 0.5 go to S, 0.5 go to R1. R1 is in the middle of S and D => Probability 0.5 will be distributed equally between D and S. So probability to be survive is: $0.5+0.25=0.75=\frac{3}{4}=> p+q=\boxed{7}.$

THERE ARE 5 ROOMS..LET ME START WITH R1 THEN I GOT 4 DOORS TO ENTER IF I CHOOSE D THEN I SHALL NOT SURVIVE. THEREFORE PROBABILITY NOT TO SURVIVE IS 1/4 PROBABILITY FOR SURVIVE 1-1/4=3/4=p/q p+q=3+4=7