4 doors, 1 car

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 a b \frac ab , where a a and b b are coprime positive integers, what is a + b ? a+b?


More probability questions:

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The answer is 11.

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1 solution

Geoff Pilling
May 6, 2016

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 1/4 . So, by symmetry, the remaining 3 / 4 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 (3/4)/2 = 3/8 of having the car behind them.. And 3 + 8 = 11 3+8=\boxed{11}

What do you mean by "this doesn't effect the probability of the car being behind the door you picked"? Assuming we ve known that there s no car in door four.. Why it`s not 1/3?? Thanksss

Muhammad Edgar Sena - 5 years, 1 month ago

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Yeah, its definitely counter intuitive. This is an extension of the Monty Hall problem which has confused people for a long time!

So, let me try to explain a bit better... So you start off and pick one of four. So your probability of being right so far is 1/4. Then, the host reveals one without the car. Now you switch.

So, lets look at all the possibilities:

1) You already had picked the car and you switch. Clearly you lose.

2) You didn't pick the car. So, now you know its behind one of the two doors that neither you or the host opened. So you have a 50/50 shot.

Since there were originally 1 way to do the first one, and 3 ways to do the 2nd one (i.e. 3 ways you could have chosen a door with no car behind it), your chances of winning the car are

P(win car) = (1/4) 0 + 3/4 (1/2) = 3/8.

Make sense?

Lemme know if I should clarify anything...

And thanks for trying out the problem!

Geoff Pilling - 5 years, 1 month ago

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Yeah, clearly I guess it makes sense.. I've read about monty hall and it's tricky if you think shortly.. Sankyu thenn ^^

Muhammad Edgar Sena - 5 years, 1 month ago

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