oh no you didn't....

[CPA]

http://en.wikipedia.org/wiki/Monty_Hall_problem

It that's a Smart car, I'd rather have the goat.

 

[Rakehellion]

As before, you are totally ignoring initial conditions. What you just stated is not the same as the Monty Haul problem at all.

It's the exactly same logical problem. You bought a lotto ticket, you're shown another one, and there are 175 million losers who have been wiped from the equation.

 

[3chordcharlie]

It's the exactly same logical problem.

You have to be trolling this.
No one coordinated enough to type words is this thick.

 

[KeithTalent]

It's the exactly same logical problem.

Except it's not. :hmm:

KT

 

[DrPizza]

I feel sad when I can't tell if someone's just trolling & I should ignore them, or if they truly are stupid & I should help them.

 

[Rakehellion]

^Idiots who think they're right.

 

[kranky]

^Idiots who think they're right.

Of course, the whole thing is interesting because it's completely counter-intuitive. So naturally it seems "wrong" at first. You need to think this problem through, you'll get it.

But it's not smart to call people idiots when they ARE right. Really, no kidding, no trolling, in all seriousness - it's better to switch. It is.

 

[Rakehellion]

But it's not smart to call people idiots when they ARE right. Really, no kidding, no trolling, in all seriousness - it's better to switch. It is.

That hasn't been sufficiently explained. The counterintuitive nature of the problem disregards the simple solution.

And no matter how right you think you are, you're still wrong if you can't articulate your position.

 

[HamburgerBoy]

That hasn't been sufficiently explained. The counterintuitive nature of the problem disregards the simple solution.

And no matter how right you think you are, you're still wrong if you can't articulate your position.

lol when you're disregarding everything that happens prior to the final pick. You're right: it's a 50/50 chance if you drop the ticket you started with (after the other 174,999,998 are eliminated) onto the winning ticket, suddenly experience amnesia, and have to pick up one of the two. But that's not the situation here.

 

[JujuFish]

I love these threads because it brings out people who are not only mathematically incompetent, but vehemently stand by their position. Very entertaining.

 

[Cuda1447]

http://www.businessinsider.com/the-most-controversial-math-problems-2013-3#

#2

most controversial math problems, I don't have to click it to know. 1=.99999 right?

 

[Pheran]

That hasn't been sufficiently explained. The counterintuitive nature of the problem disregards the simple solution.

And no matter how right you think you are, you're still wrong if you can't articulate your position.

When several people have correctly explained the solution until they are blue in the face but you still don't get it, the problem is you.

But hey, since you are able to purchase Powerball tickets that have a 50% chance of winning the jackpot, why aren't you out rolling in cash?

 

[Baked]

The birthday one's stupid. I have 40+ employees and none of them have the same birthday. I've interviewed people with my same birthday though.

 

[MrColin]

Funny how this time the Monty Hall problem spawned the war instead of .999 = 1. It's also funny how adamantly people insist on being wrong. OK guys, put it to the test then:

Monty Hall Simulator

I wish I could find a simulator that supports more than 3 doors, because that would make it incredibly obvious that switching is better.

With regard to #2,

0.999... = 1 > 0.999

 

[Rakehellion]

lol when you're disregarding everything that happens prior to the final pick. You're right: it's a 50/50 chance if you drop the ticket you started with (after the other 174,999,998 are eliminated) onto the winning ticket, suddenly experience amnesia, and have to pick up one of the two. But that's not the situation here.

It's irrelevant what happens before the final pick because we're standing with two countable numbers.

There's a 100% chance that either my ticket or the other is the winner, so explain how mine is one in a million and the other is a sure thing. Sounds like you don't understand probability.

 

[3chordcharlie]

It's irrelevant what happens before the final pick because we're standing with two countable numbers.

There's a 100% chance that either my ticket or the other is the winner, so explain how mine is one in a million and the other is a sure thing. Sounds like you don't understand probability.

No, the 'game' was *always* going to end up with your ticket and another ticket, and one of them the winner.

The 'one ticket' represents the chance that the entire group of 'tickets you did not pick' contained the winning ticket. If any of those (N-1) tickets was the winner, then switch and you win.

You only win by sticking with your original ticket if you beat the original odds and picked the winner. This is not 50%.

To get back to 50% you have to choose randomly between your two remaining options.

 

[Pheran]

With regard to #2,

0.999... = 1 > 0.999

Yes, of course I meant to write .999..., not just .999. I edited the post.

 

[Pheran]

No, the 'game' was *always* going to end up with your ticket and another ticket, and one of them the winner.

The 'one ticket' represents the chance that the entire group of 'tickets you did not pick' contained the winning ticket. If any of those (N-1) tickets was the winner, then switch and you win.

You only win by sticking with your original ticket if you beat the original odds and picked the winner. This is not 50%.

To get back to 50% you have to choose randomly between your two remaining options.

At this point, he is either intentionally trolling or just incredibly dense. In either case, further argument is a waste of time.

 

[3chordcharlie]

At this point, he is either intentionally trolling or just incredibly dense. In either case, further argument is a waste of time.

I'm pretty sure it has to be trolling. I have seen grown adults unable to understand this problem though.

To me the issue with the original version of the problem has always been one of psychology. There's actually a fairly large chance that you picked the right door (1/3) and you're going to feel 'dumb' if you switch and lose, even though it was the correct way to play.

Most of us don't get repeated chances to play dominant strategy games like this for high stakes. You don't get two-thirds of a new car for playing properly. You either get the car, or you don't.

 

[purbeast0]

anyone who doesn't understand the monty hall problem is probably one of those people who is just not a "math" person. the slide that shows the grid of possible outcomes, as well as the 1 in 50 slide, explain it very well in laymans terms.

 

[Pheran]

To me the issue with the original version of the problem has always been one of psychology. There's actually a fairly large chance that you picked the right door (1/3) and you're going to feel 'dumb' if you switch and lose, even though it was the correct way to play.

Right, that's why I recast it into the Powerball problem, since everyone knows your chance of buying a winning Powerball ticket is pretty damn low.

 

[Rakehellion]

The 'one ticket' represents the chance that the entire group of 'tickets you did not pick' contained the winning ticket. If any of those (N-1) tickets was the winner, then switch and you win.

No, it's only one ticket. The other tickets don't exist.

To get back to 50% you have to choose randomly between your two remaining options.

That's pretty much what probability is. There's no certainty either way.

If you're playing a shell game with two shells, you take your pick, and the guys says "Oh, but there used to be a million other shells in this game" that still doesn't affect your odds.

 

[TheVrolok]

The problem is much simpler. You aren't 'trying to pick the right door' in the first step.

You are dividing the total into two groups:

One is a group of 1 and the other is a group of (N-1).

Now decide whether you want 1/N or (N-1)/N chance of winning, and act accordingly.

Here's another explanation.

How does a "switcher" win. He needs to pick a wrong ticket, and then the ticket remaining in the bowl will be the winning ticket. So the probability of a "switcher" winning is 174,999,999/175,000,000 (the probability of your first pick being the wrong ticket).

How does a "non-switcher" win. He needs to pick the right ticket, because then the ticket in the bowl will be a losing ticket. So the probability of a "non-switcher" winning is 1/175,000,000 (the probability of your first pick being the right ticket)

To get back to 50% you have to choose randomly between your two remaining options.

I remember when one of my math teachers first presented this to us back in the day, and at first, I was confused, too, because it is somewhat counter-intuitive. However, it should click if one thinks about it long enough. I think the three quotes above explain it quite nicely.

 

[blackdogdeek]

I think the three quotes above explain it quite nicely.

I think the picture that shows the three possible outcomes should be sufficient.

 

[Rakehellion]

Right, that's why I recast it into the Powerball problem, since everyone knows your chance of buying a winning Powerball ticket is pretty damn low.

Same concept applies. Even though your odds were one in a million, you know that was a farce.

You know for a fact that:
John Smith on Cherry street didn't win. That makes your odds go up.
Your lotto pool at the office didn't win. That makes your odds go up even more.
Jenny who prayed to Jesus for the winning numbers didn't win. That makes your odds go up.

And so on until you're left with two tickets. You aren't at one in a million if all other other tickets have been voided.