The first time I heard of the goat problem, a deceptively simple exercise, I flat out refused to believe the solution. A friend and I had been just about to go to a bar, but that plan had to be cancelled in favour of spending 3 hours to prove it to me. See how well you do:
Imagine a TV game show where the winner chooses between 3 doors. Between one door is a car, between each of the other two doors there’s a goat.
The contestant chooses one door, and the host then opens another door behind which there is a goat. This is always possible since there are two goats and one car.
The host will now give the contestant the option of sticking with the door she has already chosen or switching to the one door still unopened. What should the contestant do? There are of course three possible answers:
1: The contestant should stick to the first choice
2: The contestant should switch
3: It doesn’t matter
What do you think? The answer can be found here and it WILL surprise you. I LOVE it when things get counter-intuitive.
A warning though: Bringing this riddle up may cause aggravation. I have seen people flat out refuse to acknowledge the solution and get very frustrated in the process.
UPDATE: Tveskov pointed me to this online version of let’s make a deal, which let’s you try out the game for yourself and keeps track of the stats for you. From the site:
Despite a very clear explanation of this paradox, most students have a difficulty understanding the problem. It is very difficult to conquer the strong intuition which most students have in this case. As a challenge to students who don’t believe the explanation, an instructor may ask the students to actually play the game a number of times by switching and by not switching and to keep track of the relative frequency of wins with each strategy.
The goats have been replaced by donkeys, but don’t let that confuse you.
This should not be so hard to explain….
If the game show host had only just put a mark on one of the doors left, and given you the choice of selecting the other door left or sticking with your first choice – then the odds would be the same: one third for each door, and it would not matter whether you switched or not.
But in this case the game show host adds information into the equation. He actually tells you something. He reveals one of the "bad doors". This means you have one door that you chose – which remains odds "one in three" that the prize is behind.
And you have the other door – which is now ….
arghhh!!!!
I don’t get it!
Is the door left odds "two in three" or is it "one in two" that the good prize is behind it?
Actually I think it is two in three. Because the game show host can only say something about the two doors not selected by you (which are – combined – odds "two in three"). And the question he answers is: which of the two doors is odds "two in three"?
Man you’re right. It is mindboggling.
Love to read your blog Alexander.
best
Gunnar
You’ve got it, Gunnar.
Here’s the way I think about it:
1: When does it NOT pay, to change your initial selection?
– When you found the car on your first attempt.
2: What are the odds that you found the car on the first try?
– 1 in 3
3: So there’s a 2/3 chance, that switching will get you the car.
Which is just a different way of sying what you said.
And thanks for the kind words!
Heh, it’s a classic. And you’re right – it can take many beers in many bars to get everyone in a group of people to agree about the answer…
What about the mention of Marilyn vos Savant
What if we consider playing the game a second time. I now know that the host will open a door I have not chosen. So initailly I guess in my head that I should choose Door 1 so I tell the host that I choose Door 2. 1 of 2 things could happen, each as likely, either the host opens Door 1 showing a goat (it turns out I was wrong about door 1) or the host opens door 3. Have I loaded the chance of Door 1 being the car door by out witting the host as I know their behaviour? If so I can increase the chance of an event without actually knowing before hand which door has a car behind it, this sounds like magic to me?
in my experience, a huge problem with folks understanding this thought puzzle is that people don’t quite understand the conditions for the host opening a door. Suppose, for example, the car is behind door A. If you pick door B, the host will open door C, since that is one of the doors you didn’t pick AND it doesn’t have the car. If you had instead picked door C, the host would open door B. If you pick door A, the one with the car, the host may open either door B or C, and it doesn’t matter which, since they are both substantially car-free.
Without knowing where the car is though, which of the 3 doors we choose first doesn’t matter a bit since they all have the same probably of having a car. So instead of worrying about whether A B or C has the car, we only consider our two options for strategy: switch or stay.
If we stay, we will end up with the car IF and ONLY IF our first choice was the door with the car.
if we switch, we will end up with the car IF and ONLY IF we first choose a door withOUT the car.
Since there is a 1/3 chance of choosing the car door first, there is a 2/3 chance of choosing a non-car door first. The switch strategy, therefore, results in a 2/3 chance of winning the car.
I didn’t understand the answer to this problem initially either though. Even after reading vos Savant’s explanation.
How I got my head around this was thus.
Say there were 100 doors, 99 have goats and 1 car.
The host shows you 98 goats so you are left with 1 goat and 1 car behind the remaining 2 closed doors.
What were the chances of you picking the car originally? 1 in 100.
Therefore, statistically you would have picked a goat 99 times in 100.
That is why you must switch – because you had a 99-1 chance of picking the car. Now you’ve switched that to 1-99!
The priciple remains the same with 3 doors, as it does with 100.