A brain teasing probability puzzle
- Thread starter Neki
- Start date
-
Guest, it's time once again for the massively important and exciting FoH Asshat Tournament!
Go here and give us your nominations!Who's been the biggest Asshat in the last year? Give us your worst ones!
But that is the premise of the actual Monty Hall problem. The OP did not correctly state all the premises, and indeed strictly speaking he posted a common variant of the problem (there are dozens), but everyone is working under the assumption that he was posting the actual Monty Hall problem, but did so incorrectly. I think he even confirmed this somewhere in the thread. The classic Monty Hall problem shouldn't be confused with the game show from which the name is derived, Monty Hall himself acknowledged that the game show didn't match up to the problem because the choice wasn't always offered but that by the problem as stated in the mathematical community switching was always the correct choice.
Szlia
Member
- 6,634
- 1,376
Oups! My bad for typing it in a hurry. I edited the original post to put the correct 1.25 X.
You are unto something here: the idea that things go wrong because X does not represent the same thing in all its instances. Yet it does not fully elucidates the paradox. Anyway, for more:Two envelopes problem - Wikipedia, the free encyclopediaSecond, please stop abusing that poor defenseless X.
Given:
2X = $10000
and:
.5X = $5000
solve for X...
Whoops again.
Melvin
Blackwing Lair Raider
- 1,399
- 1,168
A more correct way of fixing the two problems I pointed out would be to change the .5X term to something sane, namely X. So now that you have an internally consistent definition of wtf X actually is, you have a choice between X and 2X dollars, with an average of 1.5X, and just as much motivation to switch as to say "meh, fuck it."
Tuco
I got Tuco'd!
- 47,937
- 82,629
I don't get the trick here but I'll respond humbly with my response at first glance.
If A=100
and B=200
If P(A)=0.5
and P(B)=0.5
The expected result from envelopes 1 and 2 would be: P(A) * A + P(B) * B = 150.
Because the expected results from each envelope are the same, switching or not switching would give the same result. So mathematically it doesn't matter, thus my recommendation would be to either pick the closest envelope or to play this game:
iannis
Musty Nester
- 31,351
- 17,657
That just seems like a situation where you don't have enough information to predict an outcome. That doesn't seem like a paradox. It seems like a simple choice.
What went wrong is asserting that a switch results in 1.25x. It doesn't. It results in either 2x or 1/2x.
It really shines through if you put something in the envelopes that isn't numerical. One has a yellow ribbon, the other has a blue ribbon, so logically you've got 1/2 a chance of picking a green ribbon.
What went wrong is asserting that a switch results in 1.25x. It doesn't. It results in either 2x or 1/2x.
It really shines through if you put something in the envelopes that isn't numerical. One has a yellow ribbon, the other has a blue ribbon, so logically you've got 1/2 a chance of picking a green ribbon.
Szlia
Member
- 6,634
- 1,376
The wikipedia page mentions this, but then it takes it a step further: you open the envelope you picked and it contains... say... $100. So now you know the other envelopes contains $50 or $200 so switching is risking losing $50 for a chance at winning $100, a deal that seems difficult to rationally refuse. We are back to the previous logic, but without the shady use of X. And the paradox is back, because there is no scenario where opening the envelope makes you discover a sum that does not make you want to switch... so why open the envelope in the first place?
I find Iannis' "it's not the right tool set for the problem at hand" rebuttal interesting, but also somewhat problematic: what are the criteria to determine when statistical tools are no longer effective? We have here a choice, partial information, numbers... so it's less clear than with colored ribbons.
Tuco
I got Tuco'd!
- 47,937
- 82,629
Wait are there three envelopes with 50, 100 and 200? Or are there two envelopes with 100 and 200?
iannis
Musty Nester
- 31,351
- 17,657
It seems sorta like that "halve the distance the rock never hits the tree" thing.
You -can- model it that way and retain integrity and it is utterly proper. But it's not useful or practical because it is not the best possible indicator of the reality of the situation being modeled.
What the best model is... you'd have to ask a mathematician. Or maybe just somebody smarter than me. I don't pretend to know. What I do know is that that 1.25x doesn't make the right sort of sense and I suspect it's just because there isn't enough information available to make a model from.
You -can- model it that way and retain integrity and it is utterly proper. But it's not useful or practical because it is not the best possible indicator of the reality of the situation being modeled.
What the best model is... you'd have to ask a mathematician. Or maybe just somebody smarter than me. I don't pretend to know. What I do know is that that 1.25x doesn't make the right sort of sense and I suspect it's just because there isn't enough information available to make a model from.
Tuco
I got Tuco'd!
- 47,937
- 82,629
Ok, I still contend that my above answer is correct. The value of each envelope is the same (3/2X, so if X=$100 each envelope contains a green ribbon with $150), so it makes no sense to switch.
Did I get the correct answer?
Eomer
Trakanon Raider
- 5,472
- 272
I've skimmed it, and you've repeatedly claimed that the second choice is completely independent of the first. That's false. Repeatedly claiming otherwise doesn't change that. But it's cool bro, we all know you're trolling. Carry on.
You are adding a really specific addition to the problem that is in no way mentioned or implied.
Yeah, I don't quite get it. Just because there is a wikipedia page about it doesn't mean it doesn't arise from improper starting assumptions. You can't just set up a problem where x equals two different things and claim paradox when the formula doesn't resolve.
Oh hang on, now you've confused me again. Maybe there is more to this than I thought.
Okay, here's the problem I think. You don't have a 50% to gain from switching because you originally selected an envelope. You have already introduced a 50% chance that you've selected the higher envelope amount to begin with.
I found an explanation that's better than any I could make:
chthonic-anemos
bitchute.com/video/EvyOjOORbg5l/
- 8,606
- 27,290
All of these cars, goats, and envelops filled with cash. At this point, mathematically speaking, I think it's reasonable to rob the guy.
Reasonably.
Reasonably.
Haast
Lord Nagafen Raider
- 3,281
- 1,636
Hoss
Make America's Team Great Again
- 27,621
- 16,059
This thread does have a proper paradox that I don't think anyone answered.
You have 100 doors, 1 prize, a host and 2 contestants. Each contestant picks a door, and the host opens up 98 doors to reveal the prize is not behind any of them. Will both contestant increase their odds of winning the prize by switching?
Tuco
I got Tuco'd!
- 47,937
- 82,629
Hehe that's a good question. I didn't see it before. I'll have to think about it.
At first glance I'd say no. The probability that contestant A or B picked the correct door is 1%. So you don't improve your odds by choosing a different door.
At second glance a huge difference between the previous question is that the host picks all the doors not chosen, which means that one of the contestants picked correctly. This would happen only 2% of the time making the entire game show very silly. But given that this only happens 2% of the time I'd still be confident in not changing.