Choosing a Game Show Door

Off Topic
41

[quote]LIFTICVSMAXIMVS wrote:
Pookie, please tell me you do not believe the wiki site. There are only two choices. In fact, there are always only two choices because the host will always eliminate one bad option.

Your computer program is wrong because you start with incorrect assumptions – that there are three choices – when in fact there are always only two.[/quote]

Let’s try the 100 door version.

I have picked a winning door. The md5 hash for the solution is: 890fd093d2f3f4e1d00e35d1784d0a41 (so that you can verify at the end that I didn’t change the winning door on the fly.) (Web site: md5 Hash Generator )

Pick a door between 1 and 100.

I will then open all remaining doors except one and give you the option to switch.

Sticking with your initial choice should give you a 50/50 odd of winning, right?

Let’s see how many goats you win before you get it.

42

[quote]LIFTICVSMAXIMVS wrote:
johnnytang24 wrote:
No, you start with 3 choices. The program acts exactly as the game show would, step by step.

I understand what you think is happening but it isn’t.
what is happening:
First pick – 1/3 + 1/3 + 1/3 = 1
second pick – 1/2 + 1/2 = 1

What you think is happening: second pick – 1/3 + 2/3 = 1

You assume that the other choice besides the one you made becomes 2/3 because the host eliminates one. That is incorrect because there are ONLY TWO choices. The question you need to ask is how did the other choice become 2/3. It would have to apply to both selections since both must have an equal weight of being correct and that cannot be because their probabilities need to add up to 1.

You are putting way too much thought into this and forgetting simple probability. That wiki is wrong.

The only way the probability will be 2/3 being right is if you get to make 2 selections from the get. You, in fact, do not get that option.[/quote]

I can’t tell if you are just trolling now, or you really are that unwilling to understand, so I’ll stop now.

43

[quote]LIFTICVSMAXIMVS wrote:
johnnytang24 wrote:
No, you start with 3 choices. The program acts exactly as the game show would, step by step.

I understand what you think is happening but it isn’t.
what is happening:
First pick – 1/3 + 1/3 + 1/3 = 1
second pick – 1/2 + 1/2 = 1

What you think is happening: second pick – 1/3 + 2/3 = 1

You assume that the other choice besides the one you made becomes 2/3 because the host eliminates one. That is incorrect because there are ONLY TWO choices. The question you need to ask is how did the other choice become 2/3. It would have to apply to both selections since both must have an equal weight of being correct and that cannot be because their probabilities need to add up to 1.

You are putting way too much thought into this and forgetting simple probability. That wiki is wrong.

The only way the probability will be 2/3 being right is if you get to make 2 selections from the get. You, in fact, do not get that option.[/quote]

i see what youre saying. so since its really 50/50 all along you think its wiser to take another shot.

i follow you on that part but heres where i veer off, you only have the 2 choices to start, forget about the revealed door as you said. its going to be a 50/50 regardless because whether or not you pick one door or the other it doesnt matter, theres still another door.

44

wait nvm, i just re-read what you wrote and what you said makes perfect sense, i thought you were saying something else.

45

[quote]LIFTICVSMAXIMVS wrote:

I understand what you think is happening but it isn’t.
what is happening:
First pick – 1/3 + 1/3 + 1/3 = 1
second pick – 1/2 + 1/2 = 1

What you think is happening: second pick – 1/3 + 2/3 = 1

You assume that the other choice besides the one you made becomes 2/3 because the host eliminates one. That is incorrect because there are ONLY TWO choices. The question you need to ask is how did the other choice become 2/3. It would have to apply to both selections since both must have an equal weight of being correct and that cannot be because their probabilities need to add up to 1.

You are putting way too much thought into this and forgetting simple probability. That wiki is wrong.

The only way the probability will be 2/3 being right is if you get to make 2 selections from the get. You, in fact, do not get that option.[/quote]

We give up man… Just know that you are wrong.

46

[quote]johnnytang24 wrote:
I’ll stop now.
[/quote]

Thank you. You are just confusing people with stuff that isn’t really so complicated you need a computer program to figure it out.

Garbage in, garbage out.

If you start with incorrect theory your solutions will always be incorrect.

47

[quote]Affliction wrote:
We give up man… Just know that you are wrong.
[/quote]

Being a mathematician you should be able to offer a formal proof like the one I did a few posts back.

48

[quote]johnnytang24 wrote:
Try putting this in an html file, it runs the game 1,000,000 times and checks to see how many times you would win if you switched:

Results:

[/quote]

Thanks for the coolest post of the thread.

49

[quote]LIFTICVSMAXIMVS wrote:
If you start with incorrect theory your solutions will always be incorrect.[/quote]

Nice of you to notice, since you’re the one with incorrect theory.

The question is not “pick one of two doors,” it’s “would you like to switch?”

How about you pick that door from the hundred…

50

[quote]LIFTICVSMAXIMVS wrote:
Affliction wrote:
We give up man… Just know that you are wrong.

Being a mathematician you should be able to offer a formal proof like the one I did a few posts back.[/quote]

Without doing a formal proof, you are wrong because there is a change of state, but not where you think it is. The door you originally picked will ALWAYS have a 1/3 chance of being correct, no matter at what point in the game you are, unless the winning option has been revealed.

The “change of stage” as you call it, happens when the host eliminates a wrong answer from the sample space. Therefore, the remaining sample space (originally consisting of two doors, now consisting of one) retains its probability of 2/3. So the host has conveniently left you the choice of:

A) Sticking to your guns and keeping your original door. You will win 1/3 of the time, as obviously you had a 1/3 chance to begin with.

B) Changing your choice. You will win 2/3 of the time. Precisely BECAUSE the odds must add up to one, as you so eloquently pointed out.

The most telling explanation is the 1000 doors variation of this.

You would have a 1/1,000 chance of picking the correct door initially. If the host opens 998 doors, all of them with goats behind them, the door that you chose first will still have a 1/1,000 chance of being the one that conceals the car, but the other remaining door will have a 999/1,000 probability of being the door that is concealing the car. Here switching sounds like a pretty good idea… no?

I’m sorry, I don’t know your qualifications, and I’m not going to bore you with mine; just know that there are people far smarter than EITHER of us who categorically agree with me.

51

[quote]pookie wrote:
LIFTICVSMAXIMVS wrote:
pookie wrote:
Switching takes your chances from 1/3 to 2/3, so yes, switching is good.

After door #3 was open by the host his chances became 1/2.

What does that change if he picks door #2?

Argh.

Have you read the rest of the thread?

Let’s try one more way:

You pick one door out of three. Your door has a 1/3 chance of being right. The remaining group of two non-picked doors has 2/3 chance of being right, right?

You have 1 group of 1 door with 1/3 the chances and a second group of two doors with 2/3 chances.

The host asks you if you’d prefer the second group with 2/3 the chances instead of your current group with 1/3 the chances. To do that, he eliminates one of the bad doors from the second group and offers you to switch to the remaining one.

As a group, the remaining door and the opened one still have 2/3 of the chances of having the prize. That’s why you have two closed doors with one of them (your initial pick) having 1/3 chance of winning and the remaining door having 2/3 chance of winning.

If you still don’t get it, well program yourself a simulation and find out by running a few million trials. You’ll get: “not switching” wins 33.3% or the time, “switching” wins 66.6%. It’s a mathematical fact.
[/quote]

Pookie, you are summing probabilities that are not related to each other.

Try this mental exercise. Assume there are two people separated by a wall so that they do not know what the other is doing. The first guy gets to pick 1 out of three.

The probability is: 1/3 right + 1/3 wrong + 1/3 wrong = 1

The host then eliminates one incorrect choice and tells the other guy separated by the wall, with no knowledge of what choice the other guy made, to pick one out of the two remaining.

The odds are: 1/2 right + 1/2 wrong = 1

The odds do not change just because there is only one guy picking nor are they different because the other guy has no knowledge of what is picked first – the situations are analogous to each other. The fact remains that after each possibility of picking an incorrect choice are removed the odds need to be recomputed. Your script does not account for that.

52

This last post will prove to be your downfall, my friend. You created a significantly different scenario. WE DO know what the first choice was, and what the sample space was, unlike the person separated by the wall.

In a vacuum, if the person behind the wall was offered a 1 out of 2 choice after 998 doors had been eliminated, of course the odds reduce to 50/50. He has no idea the sample space was originally that large. He just knows that behind one door is a car. We have the advantage of watching the doors (and the odds) unfold.

53

Listen, I’ll play the 1000 door variation of this game with you all day. You stick to your initial choice, I’ll switch my choice after 998 of the doors have been eliminated. I WILL WIN 999/1000, you WILL LOSE 999/1000. You can’t argue with this. You are wrong.

54

[quote]Affliction wrote:
Listen, I’ll play the 1000 door variation of this game with you all day. You stick to your initial choice, I’ll switch my choice after 998 of the doors have been eliminated. I WILL WIN 999/1000, you WILL LOSE 999/1000. You can’t argue with this. You are wrong.[/quote]

Let’s play this game for cash, lifty. Every time you win, I’ll give you $100. Every time I win, you only owe me $1. Deal?

55

[quote]LIFTICVSMAXIMVS wrote:
The host then eliminates one incorrect choice and tells the other guy separated by the wall, with no knowledge of what choice the other guy made, to pick one out of the two remaining.

The odds are: 1/2 right + 1/2 wrong = 1[/quote]

That’s where you’re wrong. The odds at that point are 1/3 + 2/3 = 1. The second guy simply has no way of knowing which is which.

If he picks the same door as the first guy picked, he’ll win 33% of the time and if he picks the remaining door, he’ll win 67% of the time. Just because he’s late to the game doesn’t mean the odds for each door have changed.

If you go see a boxing match between two boxers you’ve never heard of before, that doesn’t suddenly make their odds of winning 50/50. Your ignorance of the actual odds doesn’t affect them in any way.

56

[quote]Affliction wrote:
This last post will prove to be your downfall, my friend. You created a significantly different scenario. WE DO know what the first choice was, and what the sample space was, unlike the person separated by the wall.

In a vacuum, if the person behind the wall was offered a 1 out of 2 choice after 998 doors had been eliminated, of course the odds reduce to 50/50. He has no idea the sample space was originally that large. He just knows that behind one door is a car. We have the advantage of watching the doors (and the odds) unfold.[/quote]

Knowledge of “sample space” does not change “real” space.

The probability that MATTERS is whatever the last state is.

Since the first guy doesn’t know if what he picks is correct it is completely analogous to the example I give where the second guy picks without any prior knowledge.

You are being thrown for a look by even analyzing the first pick since it is ONLY the second pick that matters.

57

[quote]Affliction wrote:
In a vacuum, if the person behind the wall was offered a 1 out of 2 choice after 998 doors had been eliminated, of course the odds reduce to 50/50.[/quote]

Those are the odds of the “new guy” picking the right door; they aren’t the odds of the original door (1/1000) and the remaining door (999/1000).

Nothing will change the odds of the doors.

All you can do is get a stranger off the street who doesn’t have the information necessary to determine the (most probably) winning door.

If that ever happens to you, the correct answer is “I pick the door the first guy didn’t choose.” :slight_smile:

58

Here’s your mathematical proof. Let’s start with 3 doors, behind one is Goat 1, behind another is Goat 2, behind a third is a car.

There is obviously a 1/3 chance of picking Goat 1, a 1/3 chance of picking Goat 2, and a 1/3 chance of picking the car. In total, there is a 1/3 + 1/3 = 2/3 chance of picking the goat.

Let us assume outcome 1 in which you pick a goat. There was a 2/3 probability that this would happen. Whenever this does happen, there is a 100% chance that the host will open the door with the other goat. 2/3 * 1 = 2/3. So there is still a 2/3 chance that your door has a goat.

Let x be the probability of a goat in the remaining door. Since there are only two doors remaining, 2/3 + x = 1, so x = 1/3. Your door has a 2/3 chance of being a goat, the other has a 1/3.

Outcome 2 has you picking the car. There is a 1/3 chance of this. The host then shows a goat, guaranteed. 1/3 * 1 = 1/3. Your door has a 1/3 chance of being a car. Following the same logic in outcome 1 shows that the other door has a 2/3 chance of having the car.

59

It is hard for me to elucidate exactly what I mean when trying to disprove your wall example, and we never stipulated certain things that would have a big effect on the outcome of that particular exercise.

Basically, if the host comes to the second person behind the wall and says, “I’ve got two doors, one with a car behind it, one without, pick one,” then the odds are 50/50. But you are eliminating the variable change aspect to the exercise, thus rendering it useless. There has to be a first pick, and then an elimination of X amount of doors, and then another pick opportunity for this to work.

Still, I’m going to stick to my guns and ask you to disprove the 1,000 door example. I know it’s impossible.

60

[quote]pookie wrote:
Affliction wrote:
In a vacuum, if the person behind the wall was offered a 1 out of 2 choice after 998 doors had been eliminated, of course the odds reduce to 50/50.

Those are the odds of the “new guy” picking the right door; they aren’t the odds of the original door (1/1000) and the remaining door (999/1000).

Nothing will change the odds of the doors.

All you can do is get a stranger off the street who doesn’t have the information necessary to determine the (most probably) winning door.

If that ever happens to you, the correct answer is “I pick the door the first guy didn’t choose.” :slight_smile:
[/quote]

Thank you pookie. You made the point I was grasping for. I did a poor job of wording that, I think I clarified with my post before this one.