What Day Is It?

I had quite a weekend, demonstrated by the fact that it's Tuesday and I am just getting around to posting.

Went up to Brooklyn on Friday night and met up with Prof. Bitch, who was in town for the Blog Sheroes thing.  Just a few years ago, I went to Brooklyn all the time, so it was nice to go back.  We went out for good Mexican food in Park Slope with Bitch's sister-in-law.  It's only a couple hour drive, so I got back here around 3-4 a.m.

Saturday, a friend of mine and I decided it had been too long since we had just gone out drinking, and it would be a while before we really got the opportunity again, so we did that.  Unfortunately, it started pelting down rain just as we were about to go out, so we had to wait for it to slow down, and drank a bottle of wine while we waited.  Finishing that, we noticed the rain had stopped, so we proceeded to hit four different bars in about 4 hours, during which time we lost our judgment completely.  I won't go into the details, but we were ... silly ... in public and just out of control by the time we got home.  I had forgotten just how comfy a floor was when you're really drunk.  Very comfy.  Let's just draw a veil of silence over the rest.

Speaking of stupid things, I recently was re-introduced to the limits of my mathematical abilities by the Monte Hall Problem.  Here's the basic statement of the problem:

In the show, a contestant was given a choice of three doors of which one contained a prize.  The other two doors contained gag gifts like a chicken or a donkey.  After the contestant chose an initial door, the host of the show then revealed an empty door among the two unchosen doors, and asks the contestant if he or she would like to switch to the other unchosen door.  The question is should the contestant switch.

So, I was adamant that it made no difference whether the contestant switched doors or not.  I reasoned that after one of the "wrong" doors was opened, then there was one prize door and one gag gift door left.  I thought there was no reason to believe that the door the contestant chose was more or less likely to be the prize door, so the odds were even and therefore there was no reason to switch. 

Of course, the correct answer is that the contestant should always switch doors.  It drove me insane until I had read several explanations.  The one I found most understandable is below the fold.

When the contestant initially picks a door, the odds that she's picked the right door are 1/3.  When the host of the show reveals one of the "wrong" doors, that doesn't change that probability - the odds that the door the contestant has chosen is the prize door are still 1 in 3.  The odds that the "wrong" door that the host reveals is the prize door does change - from 1/3 to 0/3.  Since the probablities must sum to 1, then the odds the other remaining door is the prize door become 2 in 3.  Therefore, the contestant should switch and we would expect that switching would get the prize in 2 of 3 cases. 

Comments

[bitchphd]

Ok, I think that is an example of a math problem that *seems* to work but in fact doesn't b/c something is being left out--the factor of time. The odds that the original choice was correct are 1/3 and that doesn't change; but given that a door has been opened, the *problem* changes. The real question isn't to "hold on to" the 1/3rd door; it's to choose between two doors, either one of which has a 1/2 chance. In practice, that means you either stay with your "original" choice or switch, but in point of fact it isn't your "original" choice any more. It's a new one, and whether you switch doors or not, your odds are now 1/2.

[Mithras]

A more extensive explanation is here.

[Ilkka Kokkarinen]

Here's an even shorter explanation.

By not switching, you win iff you picked the right door to start with. Probability of this is 1/3.

By switching, you win iff you picked a wrong door to start with. Probability of this is 2/3.

[Mithras]

That's perfect, Ilkka. Thanks.