Monty Hall...ow ow ow ow

Originally posted by Nyxie
...but we also know prior to our choice that one of the other doors will be eliminated. so I think we have a two door problem the one we pick or the one remaining.

So, Nyxie, our door STARTS OFF with a 50% chance of winning does it?
If not, EXACTLY when does it's chances change to 50%. 🙄

.

Originally posted by Cribs
First, let's agree that the guard always tells the truth; the problem statement
doesn't really allow for him to guess or to lie.

I also think we all agree that if the guard has complete, a priori knowledge
about everybody's fate, then ...[text shortened]... guard, but only some gets passed
along to the prisoner.

Cribs

Ok - I think the source of our disagreement is this:

I am assuming that the guard is capable of lying, but does not have the authority to sentence prisoners to death himself.

You are assuming that the guard can sentence prisoners to death himself but is not capable of lying.

Given our assumptions I think we are both correct, however I think my assumptions are more within the capabilities of an average prison guard. As well as your interpretation implying a fairly unlikely allocation of responsibility to a guard, the Monty Hall parallel of it breaks a fairly implicit assumption of the Monty Hall problem: That the location of the prize is fixed.

Originally posted by iamatiger
the Monty Hall parallel of it breaks a fairly implicit assumption of the Monty Hall problem: That the location of the prize is fixed.

In the 3 Prisoners problem the prize is also fixed. The prize is survival and it has already been decided who will survive.

Originally posted by THUDandBLUNDER
So, Nyxie, our door STARTS OFF with a 50% chance of winning does it?
If not, EXACTLY when does it's chances change to 50%. 🙄

.

Yes if one third is always eliminated, then you have 2. So I am saying that the odds start out and remain 1/2, or one of two.

Originally posted by Nyxie
Yes if one third is always eliminated, then you have 2. So I am saying that the odds start out and remain 1/2, or one of two.

I would love to invite you to my casino. Oh, would we have a time...

Cribs

It's just my take on a problem. I did'nt say it was right, it's how I see it.
prisoner, a,b, c : f you are "a" you are one third of the equation, two thirds will die. 1/3 of a populace where 2/3 will die leaves you at 1/2 probaility, but that's just my take.

anyway I realize I am a tad silly so I'll go back to the lyrics thread and play there.

And cribs, if your casino will give me 1/3 disclosed card count and I hold another third in my hand, I'm on my way 😛 😵

Originally posted by Nyxie
Yes if one third is always eliminated, then you have 2. So I am saying that the odds start out and remain 1/2, or one of two.

So
if B asks the question his chances of surviving = 1/2
and
if B asks the question his chances of surviving = 1/2

Therefore together their chances of surviving = 1.5 (not 1) 😲

.

I first heard of this problem in a newspaper column by Marilyn vos Savant. It must have been in the early 1980s. My first reaction was: she's wrong! Gotta be wrong! How can it possibly pay to switch?

But over the next day or so I convinced myself that it really does pay to switch. At some point I even wrote a brief BASIC program to simulate the game, such that I could run the program and tell it to either switch or not switch. I ran enough simulations to show that it is better to switch.

As someone has already posted in this thread, a good way to convince yourself that it pays to switch is to consider the case of a million doors, with a prize behind only one of them.

Originally posted by Paul Dirac
As someone has already posted in this thread, a good way to convince yourself that it pays to switch is to consider the case of a million doors, with a prize behind only one of them.

...and after 999,998 doors have been opened there remains only your door and the other door.
If you play this game with pen and paper using 10 doors, it soon becomes obvious that the only way you can lose by switching is if you chose the door with the prize in the first place (so that when you switch you have the door with no prize). But the chance of that happening is only 10%. Hence you have a 90% chance of winning by not switching.

.

Originally posted by THUDandBLUNDER
...and after 999,998 doors have been opened there remains only your door and the other door.
If you play this game with pen and paper using 10 doors, it soon becomes obvious that the only way you can lose by switching is if you chose the door with the prize in the first place (so that when you switch you have the door with no prize). But the chance of that happening is only 10%. Hence you have a 90% chance of winning by not switching.

.

No, it is either behind your original guess or it isn't. Two states. Either or.
50/50. 1/2. Obvious.

Cribs

you know that sounds familiar...

"Hence you have a 90% chance of winning by not switching."

Sorry, that should be "Hence you have a 90% chance of winning by switching."

.

Originally posted by Nyxie
you know that sounds familiar...

Don't lissen to that jive-talking pimp! He jus' real mad 'cos he paid mo' dan $2048 to his father fo' his corner. (But hey, how did he know whose his father, know what I'm sayin'??) An' now doze ho's jus' don't be respectin' him, him havin' no pimpstick an' all.

.