Monty Hall variation The question is a variation of the classic Monty Hall problem. Should one switch their first choice of door one, if there are four doors, with two cars and two goats, the second door containing a goat. How would one solve this using fractions?
 A: So the setup: You know there is some arrangement of two cars and two goats behind four doors.  You pick a door, call it $\rm A$ and then another door opens to reveal a goat.   You wish to know if you should stay with $\rm A$ or switch to one of the other two doors, call it $\rm B$.
edit: To be clearer, let $A$ be the event of a car being behind the first door you choose, and $B$ be the event of a car behind behind whichever other door you choose should you switch after one of the goats is revealed.
There is equal chance that a car is behind the door you first picked.
$$P(A)=\tfrac 12$$
If you were right first time, then there's one goat and one car behind the other two doors. $$\mathsf P(B\mid A)=\tfrac 12$$
If you were wrong, then both cars hide behind the other doors. $$\mathsf P(B\mid A^\complement)=1$$

 $$\mathsf P(B)~=~P(A)\mathsf P(B\mid A)+\mathsf P(A^\complement)\mathsf P(B\mid A^\complement) \\= \tfrac 24\tfrac 12+\tfrac 24\tfrac 11 \\ = \tfrac 34$$

A: Same answer as Graham Kemp, but explained in another way. We assume that, as in classic Monty Hall, the host will always reveal a goat from the remaining doors regardless of what you caught.
$1)$ In $1/2$ of the time you get a car in the first selection. Then one of the two goats is revealed and the other two closed doors will have "goat" and "car". You can get each with probability $1/2$ by switching, so:
__$1.1)$ In $1/2 * 1/2 = 1/4$ of the time you get car by switching
__$1.2)$ In $1/2 * 1/2 = 1/4$ of the time you get goat by switching.
$2)$ In $1/2$ of the time you get a goat in the first selection. Then the other goat is revealed and the other two closed doors will have "car" and "car". So, in this entire $1/2$, you get a car by switching.
In total, you win a car by switching $1/4 + 1/2 = 3/4$ of the time, so it is better to switch.
A: Suppose we switch. The only way to get a goat is to choose a car originally, that is $\frac12$ chance, and then after seeing one of the goats, choose the door with the other goat, that is $\frac12$ chance. That would be $\frac14$ chance of getting a goat. That leaves $\frac34$ chance of getting a car.
Suppose we don't switch. Then there is a $\frac12$ chance of getting a car.
