There's a prob­lem of­ten used to show the un­in­tu­itive na­ture of prob­a­bil­i­ty, which has be­come very well known.

In that prob­lem a con­tes­tant in a gameshow has to choose be­tween three doors (A,B,C), on one there is a car, on the oth­er two are goat­s.

Af­ter the con­tes­tant choos­es, the host opens an­oth­er door and shows a goat.

Then, the host of­fers the con­tes­tant the chance to switch his closed door for the oth­er closed door.

Should he switch?

The in­tu­itive an­swer is "it does­n't mat­ter", be­cause there's two doors and one car, so it's a 50-50 chance.

But the re­al an­swer is that it does mat­ter, be­cause it's a 33-67 chance!

While it's sim­ple to show this to be the case to a sta­tis­ti­cal­ly-e­d­u­cat­ed dude, it's some­what hard­er for a lay­man.

In fac­t, I think most ex­pla­na­tions suck.

Here's my shot at it:

If you were of­fered the chance to switch be­tween your closed door and the oth­er two closed doors, would you take it?

The in­tu­itive an­swer to that is of course, yes, be­cause it's 67-33 for the car to be on the oth­er two doors.

Now, re­gard­less of where the car is, can the host open one of those two doors and show a goat? Of course, yes.

So, would you feel your odds went down be­cause the host showed one of your two closed doors had a goat be­hind it? No, be­cause he could al­ways do that, and you know there was (at least) one goat there!

So, what dif­fer­ence does it make if one door is open or not?

I don't ex­pect this to con­vince any­one, re­al­ly, but just in case, I have a python im­ple­men­ta­tion of this prob­lem (goat­car.py :-) if any­one wants it, if em­piri­cism can con­vince you ;-)