Monty Hall Problem

Anything you want to talk about.

Is it to your advantage to switch your choice?

Your odds of winning are increased by switching.
38
76%
Your odds of winning are increased by keeping your original choice.
3
6%
Your odds of winning are the same (50/50) whether you switch or not.
9
18%
 
Total votes : 50

Re: Monty Hall Problem

Postby Jegpeg » 18 Nov 2017, 04:55

The flaw in your argument is here:

Tarsier wrote:The odds that a passenger will choose the first passenger's vacant seat are equal to the odds that he will choose the last passenger's vacant seat. Since each has equal odds, and those are the only two seats that will affect the outcome of the puzzle, the probability is 50/50.


While this is true for most passengers it is not true for the first passenger where the odds of the first passenger's seat (his own) is 0 (because the question says he chooses a different seat at random) but he probability he chooses the last passenger's seat is 1/199.
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Re: Monty Hall Problem

Postby Tarsier » 18 Nov 2017, 05:40

Jegpeg wrote:The flaw in your argument is here:

Tarsier wrote:The odds that a passenger will choose the first passenger's vacant seat are equal to the odds that he will choose the last passenger's vacant seat. Since each has equal odds, and those are the only two seats that will affect the outcome of the puzzle, the probability is 50/50.


While this is true for most passengers it is not true for the first passenger where the odds of the first passenger's seat (his own) is 0 (because the question says he chooses a different seat at random) but he probability he chooses the last passenger's seat is 1/199.


Actually, you're right, but it's because you are stating the problem different than I've seen it traditionally, in that it isn't specified that the first passenger "instead of sitting in their own seat sits in a different seat at random." In other words, the first passenger also has a 50/50 shot, if he doesn't choose a different seat, of randomly picking his own seat or the seat of the last passenger.

So, with the stipulation that he DOES pick a different person's seat, I agree with you the odds wouldn't be 50/50 anymore. I think that, having encountered (my stated version of) the problem before, I skipped over the the crucial detail of your presentation of it.

So, would there then be a 198/199 chance that the last passenger will have a 50/50 chance of getting his own seat?

EDIT: I see that you already caught that you misremembered the problem. Somehow I read past that when keeping up on my phone earlier. My apologies.
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Re: Monty Hall Problem

Postby NoPunIn10Did » 18 Nov 2017, 05:47

Yeah I was really confused when you kept saying the math was easy. The problem as stated screws things up quite a bit.
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Re: Monty Hall Problem

Postby mjparrett » 18 Nov 2017, 17:10

Is anyone still on this thread?

A disease affects 1 in 100,000 people. It is fatal.

There is a test to tell you if you have the disease or not. It is 99% accurate (as in it tells you the genuine answer, diseased or not, 99% of the time).

You take the test. A week later your results come back and it is bad news; you have the disease.

Should you be worried for your life?? Are you disease ridden and death awaits?!
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Re: Monty Hall Problem

Postby Jegpeg » 18 Nov 2017, 20:56

Tarsier wrote:So, would there then be a 198/199 chance that the last passenger will have a 50/50 chance of getting his own seat?


Correct, which is a probability of 99/199.
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Re: Monty Hall Problem

Postby Jegpeg » 18 Nov 2017, 21:12

mjparrett wrote:Is anyone still on this thread?

A disease affects 1 in 100,000 people. It is fatal.

There is a test to tell you if you have the disease or not. It is 99% accurate (as in it tells you the genuine answer, diseased or not, 99% of the time).

You take the test. A week later your results come back and it is bad news; you have the disease.

Should you be worried for your life?? Are you disease ridden and death awaits?!


I am assuming there is no reason (before the test) to have heightened my probability above 1 in 100,000 for example if it is a contagious disease and I have been in contact with a known sufferer, (in reality such a test is only likely to be taken if there is)

My odds of getting the having the disease is up to almost 1 in 1000 a cause for concern that might worry some people but still fairly long odds
The chances of me having the disease and testing positive = 0.00001 * 0.99 = 0.0000099
The chances of me not having the disease and testing positive is 0.99999 * .01 = 0.0099999

Probability of having the disease given I have tested positive is 0.0000099 / 0.0099999 = 0.00099 (to 5dp)

Actually this is one of the reasons people are not routinely tested for a disease when the is a low probability that you actually have it.
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Re: Monty Hall Problem

Postby Keirador » 31 Dec 2017, 23:15

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Re: Monty Hall Problem

Postby Tarsier » 02 Mar 2018, 16:28

So I saw a post on Facebook about tricky math problems, and the illustration was for the Monty Hall problem. On my way to drop the kids off at school, I thought about it some. I posted that problem in this forum and we had some debate about it. Everyone's pretty logical here, though, and even those who struggled with it a bit saw their way into the logic eventually. However, I also posted on a different website/forum where I've seen a less logical basis for a lot of posts, and the arguments were more fierce. Some absolutely refused to accept the answer, even after the extreme scenarios of 1,000,000 doors were presented. Even with those odds against picking the ONE door with a car on the first try, some posters were insisting that, if Monty Hall eliminated 999,998 of the doors, the odds were still 50/50 and no door had any advantage with probability.

It occurred to me this morning that those same posters would (hopefully) have trouble with that conclusion, and realize the benefit of a switch, if we tried going in the opposite extreme:

Suppose you are on Let's Make a Deal, and Monty shows you ONE door. He tells you there is a car behind ONE of those doors. Which door would you pick?

Um ... I'd pick the only door possible - the ONE door.

That's right. And what would the probability be that you picked the door with the car.

100%, dude. It's the only door.

Okay. Now, as a fun twist, Monty Hall adds ANOTHER door. He reminds you that the car is only behind one of the doors. Any other doors hide a goat. He invites you to keep your original pick or switch. Are your odds better one way or the other? The car remains where it was at the start of the game.

I would stay the same - with the door I picked when there was only one door.

But there are two doors, now. The odds are 50/50, right?

No. I picked my door when I had 100% odds of getting it right. I'll stick with that one.


Just as, from this extreme, the odds don't change that you should keep the door you picked, from the other extreme (1,000,000 doors), the odds don't change that you probably picked wrong. Nor does it change with only three. In other words, your original pick was biased by a different probability that influences how correct you probably were with that selection (assuming that the rules are fair and the host knows what is behind the doors).

Thoughts? Is this a better explanation for those who struggle wrapping their mind around it?
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Re: Monty Hall Problem

Postby Keirador » 03 Mar 2018, 00:59

The beauty of the Monty Hall problem is that you don't need to find a different way to explain. If somebody is earnestly trying to understand the problem, they can play the game. Just get three index cards as your "door" stand-ins, some marks on one side as your goats and car, and keep picking, switching or staying as you wish and keeping track of the results. Eventually, they'll get it. That is, if they're earnestly trying to understand, and my guess is that they're not.
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Re: Monty Hall Problem

Postby Tarsier » 03 Mar 2018, 01:11

Keirador wrote:The beauty of the Monty Hall problem is that you don't need to find a different way to explain. If somebody is earnestly trying to understand the problem, they can play the game. Just get three index cards as your "door" stand-ins, some marks on one side as your goats and car, and keep picking, switching or staying as you wish and keeping track of the results. Eventually, they'll get it. That is, if they're earnestly trying to understand, and my guess is that they're not.


I'm not sure I totally agree. I've seen people who struggled with it adamantly for a while, but eventually, something clicked and they were delighted to have figured it out. I've been that way with some puzzles I couldn't wrap my mind around at first. It's counter-intuitive, and intuition is a tough obstacle to work past.

I agree that "playing the game" proves the probability, but for a lot of people (myself included), it is only intellectually satisfying once the reason, and not just the outcome, is understood. Otherwise, I sit up at night trying to work it out in my head.
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