Monday, February 04, 2008

Are there any Statisticians who read this blog and can friggin explain this to me?!

I've been sitting in on a Stats 221 class. Most of it is just new terminology for stuff I already knew. But today's lecture blew my mind a bit.

It has to do with probability. We played the game that you can find on this website.

I think I sort of understand. But when I tried to explain the reasons that I thought I understood to the professor, he said I was "prettymuch *mostly* right." But not *totally* right?! And he never really bothered to explain the real reason.

So if you can explain this to me, I will send you a virtual kiss.

Ow. My brain hurts.


bff said...

i hate that applet. I think there's something stupid going on because only a total moron could lose. If I pick #1 and it's NEVER the money, then I know #2 or #3 is the one: EVERY time I saw a donkey I picked the cash on the second click. Why did this happen? shouldn't the chances of getting it right when the choices are down to 2 be 50/50? WHy was I always getting the cash on the second try? Where's the "chance" in that?

k said...

It has something to do with the fact that in the initial pick, there is a 2/3 chance of hitting a donkey for each individual card. But when they reveal one of the donkeys, your initial pick hasn't changed it's odds of getting a donkey (2/3) but because you now know two boxes with a 2/3 chance of getting the donkey, the other box has to have a 2/3 chance of getting the cash. So the odds of winning are *always* greater if you switch boxes on the second click. (by 2/3; and I did check the applet to make sure that over the long haul the odds of winning were 2/3)

that's what I think, at least. Obviously I wouldn't have posted this thing if it made total sense to me.

Anonymous said...


When you select the first door, you create two groups. Your group has just one door, and the second group has 2 doors. Your group has 1/3 probability of being correct. The second group has 2/3 probability of being correct. Well that 2/3 probability doesn’t change just because the wrong door is exposed, that second group still has a 2/3 probability of being the correct group.

Also, keep in mind that the door opened by the host is not random. The host always exposes a wrong door. That is why the probability of the second group doesn’t change after the host exposes a wrong door.

I hope this is comprehensible enough. I wouldn’t want your brain to explode.

Love Dad

k said...

Ah, thanks, dad. I see where my explanation went wrong. I told the professor that when you revealed the card, you created two groups. One with the original 1/3 chance and one with a 1/2 chance. So I was right except for the number.

k said...

now I gotta figure out how to send that virtual kiss...