A math question

[Brightguy]

Both problems make sense. The point is, you never said anything about Monty in your description of the problem, so your solution is incorrect; the correct answer is "it doesn't make a difference".

[CodeMonkey2000]

Well, when you first start you have a 2 in 3 chances that of picking a cup with no coin. When you realize that one of them doesn't have a coin your chances of having no coin is still the same, since you chose it when there were still three cups (2/3 with no coin). So you can't have a 50/50 chance because of the initial scenario.

[A.J]

this (as most of you already know) is a problem whose solution is argued by the best of the mathematicians in the world!

both of the point of views make sense, but I say that the other cups has better odds (with similar reasoning to what codemonkey had)

[Insectoid]

Geez! Forget my story already and just go with the monty hall one! I don't care if mine was flawed, the question is the same! So my reasoning IS correct from my point of view and I don't care if you think otherwise, just vote in the friggin poll!

(prepare for sarcasm)(sarcasm initiated)
For a Brightguy, you aren't a very bright guy.
(Sarcasm terminated. Normal state restored)

[Brightguy]

The question is NOT the same.

Question 1: After choosing a cup, one of the other cups is removed (it happens to be one which did not contain the token).
Question 2: After choosing a cup, one of the other cups is removed (specifically chosen not to contain a token).

What's the probability that the chosen cup contains the token? In Question 1 the probability is 1/2 and in Question 2 the probability is 1/3.

The scenario is simple enough that you can draw a probability tree showing every possible outcome: branch on token location and removal choice (you can also branch on original selection choice, but that's irrelevant as following branches will be symmetric). Once you have the probabilities of every possible outcome, take the sum of the ones where the token was chosen and divide by the sum of those consistent with the given information.

[r691175002]

Brightguy @ Mon Jun 02, 2008 7:55 pm wrote:

The question is NOT the same.

Question 1: After choosing a cup, one of the other cups is removed (it happens to be one which did not contain the token).
Question 2: After choosing a cup, one of the other cups is removed (specifically chosen not to contain a token).

What's the probability that the chosen cup contains the token? In Question 1 the probability is 1/2 and in Question 2 the probability is 1/3.

The scenario is simple enough that you can draw a probability tree showing every possible outcome: branch on token location and removal choice (you can also branch on original selection choice, but that's irrelevant as following branches will be symmetric). Once you have the probabilities of every possible outcome, take the sum of the ones where the token was chosen and divide by the sum of those consistent with the given information.

Intent does not affect probability. Question 1 and Question 2 are the same as soon as it happens to be one which does not contain the token.

[rizzix]

r691175002 @ Mon Jun 02, 2008 8:20 pm wrote:

Intent does not affect probability.

It most certainly does.

[richcash]

[quote="r691175002 @ Mon Jun 02, 2008 8:20 pm"]

Brightguy @ Mon Jun 02, 2008 7:55 pm wrote:

Intent does not affect probability. Question 1 and Question 2 are the same as soon as it happens to be one which does not contain the token.

No, they're not the same at all. If you imagine a similar scenario with many, many cups it will become blatantly obvious.

[r691175002]

[quote="richcash @ Mon Jun 02, 2008 8:40 pm"]

r691175002 @ Mon Jun 02, 2008 8:20 pm wrote:

Brightguy @ Mon Jun 02, 2008 7:55 pm wrote:

Intent does not affect probability. Question 1 and Question 2 are the same as soon as it happens to be one which does not contain the token.

No, they're not the same at all. If you imagine a similar scenario with many, many cups it will become blatantly obvious.

I will say it again, intent does not affect probability (At least in the classical sense). As soon as the incorrect alternate choices have been revealed (Whether by random chance or by purposeful selection) the problem becomes the same.

We have three cups. You choose a cup (Probability of 1/3). Now the guy randomly picks out a door and opens it, and it turns out that he opened an incorrect door (Aka does not have the prize). Explain to me how this is in any way different.
How does the game host stumbling upon opening the door without the prize change the probability of your selection being correct from 1/3?

If the host opens door C, the host has opened door C. Why he did it has no bearing on the question whatsoever.

It does not matter if the game host knows ahead of time which choice he is supposed to reveal. If he reveals the correct one in terms of the question (Aka the one that has not been chosen, and that does not contain the prize) the problem is a monty hall problem.


Lets look at your questions:
Question 1: After choosing a cup, one of the other cups is removed (it happens to be one which did not contain the token).
Question 2: After choosing a cup, one of the other cups is removed (specifically chosen not to contain a token).
Explain to me how the end result of each of these options is different. In both cases the same cup will have been removed. Probability isn't going to go "Hey! He guessed! That means it doesn't count!".

[tenniscrazy]

if you are just picking a random door then you have a chance to pick the one with the prize behind it.

when you are picking one that you know doesn't have a prize behind the door than if you switch doors it's like you are choosing both of the doors you hadn't chosen in the first place...get it?

so if you randomly choose a door then you could open up the door that has the prize behind it..screwing up the whole question. But if you did happen to randomly choose a door that did not have a prize behind it your chances left would be 50/50. But if it was random than its kinda like you choose 2 doors...

[richcash]

It does matter. I'm generally not good at explaining but I'll try with this example.

Imagine 100 people getting a number from 1 to 100. They all have a 1% chance each. Now 98 numbers are eliminated and there's 2 left. You're still in it. You and the other guy each have a 1% chance? No, you each have a 50% chance.

Now let's imagine it's your birthday and everyone wants to see you in the final 2. A host knows which number won and eliminates 98 people whose number wasn't right. He guarantees to not eliminate your number. You don't have a 50% chance. You still have a 1% chance obviously.

That's the difference. I don't know if I explained it well enough.

[tenniscrazy]

don't you have 100% chance jk

[richcash]

tenniscrazy wrote:

don't you have 100% chance P jk

Well, yes, if the host cheats and gives you the winning number in the beginning.

But what I meant was the host knows the winning number from a draw beforehand and then people pick their numbers (or are assigned randomly). The host can just eliminate 98 right off the bat without eliminating you (for suspense purposes ). But that doesn't make you 50/50, you're still 1% and the guy who survived all those eliminations is 99%.


EDIT-The difference with randomly revealing a cup versus knowingly removing wrong ones is that your cup does have a chance of being eliminated, if that randomly revealed cup had the token. In the other one where the host is knowingly removing a wrong cup, yours had no chance of being eliminated.

If you pick a card from a standard playing deck it's 1/52 chance of being Ace of Hearts. If someone then randomly reveals 50 cards and none of them are Ace of Hearts then it's a 50% chance you'll have Ace of Hearts, right? But what if someone was looking at the cards and threw away 50 cards from the deck that were all NOT Ace of Hearts? Then it would still be 1/52 as it was initially, obviously.

[Insectoid]

All right, it was specifically chosen so as not to be the prize. So shut up and use the monty hall one, and foget it.

Just don't mention my story anymore.

[richcash]

insectoid wrote:

All right, it was specifically chosen so as not to be the prize. So shut up and use the monty hall one, and foget it.

Just don't mention my story anymore.

I never said your story was wrong once. I was having a friendly debate with r691175002 about the difference between the 2 questions posed by Brightguy. I know you meant Monty Hall, you clarified that pages ago. The discussion has now branched into the difference between these two problems.

Why can't we explore the difference between this problem and closely related problems?