You find yourself on a game show, and the host presents you with four doors. Behind three doors are an assortment of gummy bears, and the remaining door has a pure gold car behind it!

You pick a door, and then the host reveals a door behind which he knows is only a couple of red gummy bears. Then you are given the choice to stick with your first choice or switch to one of the other two unopened doors.

If the probability that you will win the car if you switch is $\frac ab$ , where $a$ and $b$ are coprime positive integers, what is $a+b?$

More probability questions:

The answer is 11.

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The car initially could be behind doors 1, 2, 3, or 4. Without loss of generality, lets say you pick door 1. The probability that you the car is behind the door you picked is 1/4.

Now, again, without loss of generality, the host can reveal that there is no car behind door 4. This doesn't effect the probability of the car being behind the door you picked,. That is still $1/4$ . So, by symmetry, the remaining $3/4$ probability (the probability that it is not behind the door you picked initially) is divided between the remaining two doors. So, they will each have a probability of $(3/4)/2 = 3/8$ of having the car behind them.. And $3+8=\boxed{11}$