Math probability dilemma

No, the other door has a 99% chance of being the correct door. The door you have initially chosen has a 1% chance of being the correct door. This makes it very obvious that you should always switch.

But the context has changed. There is now only 2 doors, and it must be one of them. In the context of all the doors then yes, there is only a 1% chance. But in the context of the reduced doors it only has a limited chance of being one of the 2 doors.

It seems to me a bit like flipping a coin. Just because I have fliped 5 heads in a row, it does not increase the chance of me fliping a tainls next flip. I still have just a 50% chance.

If one of the doors #1 to #99 is the correct door the following is true:
You switch -> you win
You do not switch -> you lose

I don't understand. If I switch, might I not switch to one of the 98 remaining doors that does not contain a prize, and therefore lose?

~Josh

I'm sorry I wasn't clear. The host opens all doors, except the one you have chosen and the door that contains the prize, or, in case you choose the door with the prize, all doors except the one you have chosen and a random other door. Then you get the option to switch doors or not.

Zaigas/My example on the doors problem I feel puts it in perspective. You should switch because the probability that your first choice is the right one is less then the probability that another one is the right one.

The host that knows which door is right opening all of the doors left except for one, just groups all of the rest of the doors into one door. Of course you should switch.

On the subject on free will/determinism, heres a thought exercise.
Suppose free will doesn't exist, everything that happens is already determined by the past. Suppose that this makes it possible to build a computer which scans your brain and predicts what you are going to do in a certain situation with 100% accuracy.

You are then placed in a room with two boxes, and you are allowed to pick one of the boxes, or both of the boxes. One of the boxes is guaranteed to have $100, the other box may have $1,000,000, or nothing.
The computer(which has already scanned your brain) puts $1,000,000 in the second box, only if it predicts that you are going to pick only that box.

So what would you do? If the computer wasn't accurate 100% of the time, but 90% of the time, or some other percentage, how would this effect your decision?

One way of looking at this problem is that since the computer has already either put the money in the box, or not. You wil always be better off picking both boxes. $100 > $0 if there is no money in the one box, $1,000,100 > $1,000,000 if there is.

On the other hand, if the 100% accurate in knowing what you are going to do, then you should pick just the one box, because if you pick both you will only get $100 ($1,000,000 > $100).

My initial thought was: You have a 1 in 100 chance of picking the right door to begin with, fine. Now all the other doors except the one you picked and one other door. Sticking with the one you picked originally is the same as "choosing" that door, there is still a 50/ 50 chance of getting it.

My second thought was: So, out of all those other doors Monty just left that one mystery door. Unless you got extremely lucky the first time, he is naturally not going to open the other 98 doors and expose the prize. The odds are 99 in 100 that he has just left the prize door.

Very deceptive at first, very interesting. Has it been used in a boardgame?

Sometimes it's difficult to separate what we make up and whats real, I'll give you an example:

There was three board game designers making some dice game prototypes giving the publisher 10 dices each for him to play with (total 30 dices), it turns out the publisher only needed 25 dices but since he couldnt distribute the 5 dices he wanted to give back evenly, the publisher dicided to keep 2 for himself and only give back 3 so the board game designers could get one dice each back.

Since the board game designers had first given the publisher 10 dices each but got one back they had effectivly only given 9 dices each (10-1=9) to the publisher, or a total of 27dices if you will. (9*3=27)

Making the total sum of dices the publisher had; 9*3=27 plus the 2 dices the publisher kept = 29 dices...(27+2=29)

Hmmm 27 dices given and 2 kept adds up to 29 not 30!!
Where da heck did the last dice go? *looks around on floor*