purbeast0
No Lifer
- Sep 13, 2001
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Originally posted by: Argo
Originally posted by: purbeast0
Originally posted by: Argo
Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.
As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:
Close your eyes and click any random door (without looking) and just to get it to reveal a bum door.
Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.
It does in this case. When you picked the first door you had 33% chance of winning. Now MOnte elimated the bad door, but he can't elimanate the door you already chose, so that door still has 33% chance of winning. However, the other door now has 63% chance of winning because it represents both doors.
Think about it this way: Mary, Joe and Frank each chose an equal pile containing 33 apples. But then Frank left and gave all his apples to Mary. It'd make sense for Joe to switch piles with her.
no i agree with 911TTZ.
think about this ... say you do not pick any door at all, and he opens 1 door. you are now left with a choice between 2 doors to choose.
that is the same exact thing as you picking one, and then he reveals one of the dud doors. either way he's going to remove one, leaving you with a choice of one or the other doors.
i do not follow the explanation either
Your example is flawed. First of all you picking a door and you not picking one is totally different scenarios. In the one case Monte has a choice of 2 doors to open, while in the other case he can either have a chocie of 2 or a choice of one. I'm tired of explaining it, but if you guys don't see this perhaps probability isn't your strong suit.
i do see it ... theoretically ... however in the very end, regardless, you are ALWAYS picking 1 of 2 doors, which would be a 50/50 chance.
also, with your little example, you left out one factor ... monte KNOWS which doors have nothing behind it, so he will ALWAYS choose a door that has nothing in it.
this is where the theory seems messed up:
say you have 100 doors, and 1 has a prize behind it. you pick 1 door, and monte removes 98 doors that all have nothing behind it (of course, he knows which one DOES have something behind it). so by this theory, you should switch doors because your odds will then be 99/100 of being right (since the other 98 wrongs were ruled out).
do you really thing switch doors EVERY time you will get the prize 99 times? i doubt it ...
(again, i might not fully understand the original theory so keep that in mind)