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Ecoris
Promising
Supreme Hero
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posted January 21, 2011 07:11 PM |
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Edited by Ecoris at 23:52, 21 Jan 2011.
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Actually, there will be 21 combinations with 7 buildings: 6+5+4+3+2+1+0 = 21.
In general, you wish to make combinations of N buildings when you have B different buildings to choose from. The number of such combinations is given by
B! / ( N! * (B-N)! )
When B = 7 and N = 3, you get 35 combinations. (7*6*5 / 3*2*1).
Edit: See also Binomial Coefficients
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted January 21, 2011 07:36 PM |
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Also known as "X choose Y".
I wrote more about them on page 30 of this thread (post 10). Although, my figures seem to have vanished, so I'll have to look into that.
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ihor
Supreme Hero
Accidental Hero
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posted April 03, 2011 10:16 AM |
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Who if not me can revive this thread ?
I don't want to return to the thing I posted half a year ago just because I even don't remember all that. And I don't know whether dimis and Ecoris still here, but at least I saw Corribus latest posts.
So here's a puzzle .
You are playing a lottery, so you can choose one of two boxes. You also know for sure, that the amount of money in one of the boxes is exactly twice more than in another. After you choose one of the boxes, you are offered to change your choice and take second box or to not to do that and select first one.
Question: should you change your mind or not and why?
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 03, 2011 05:53 PM |
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Hi ihor - nice to see you again. I know that dimis still lurks but he doesn't post. Revival of the math thread might lure him out of retirement. Haven't seen ecoris. And as you pointed out, I'm still here. For now.
Your question reminds me of the famous Monty Hall problem. Let me think about it for a bit.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted April 03, 2011 09:39 PM |
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familiar faces
Hey ihor,
Nice seeing you posting. As of the problem, I know this one, so I am not going to spoil the fun (at least for now).
Regarding the previous problems, I spent a day during Xmas, but I didn't solve all of them. I remember that I also had a couple of questions, but I don't remember the questions any more. Anyway, may be I 'll come back to them around June and actually post something.
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The empty set
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Ecoris
Promising
Supreme Hero
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posted April 04, 2011 10:45 AM |
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I still visit these boards every once in a while, mostly via email notifications.
I'm not sure I understand the problem correctly: There are two boxes, one contains twice as much money as the other. I pick one of them. Then what? Am I just given the choice of picking the other one instead without any further knowledge?
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted April 04, 2011 11:16 AM |
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Yeah, this question only makes sense if there's a third (empty) box, which is opened after you choose (aka. the Monty Hall problem).
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What will happen now?
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted April 04, 2011 06:30 PM |
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Here's the way I read the problem. Let's say you're the one disbursing the cash. If you only offer to let them switch boxes when they pick the double cash one, in the long run it means you keep more money for yourself because some people might switch to the lower cash one. So I'd so no, don't switch.
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 04, 2011 06:47 PM |
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If I'm reading this right, it doesn't matter whether you switch or do not.
Let's say one box has X dollars in it and the other one has 2X.
I have a 50% chance of picking the one with X and 50% chance of picking the one with 2X.
If I always DO NOT switch, I will, on average, get 1.5X dollars. ([X + 2*X]/2)
If I always DO switch, I will also, on average, get 1.5X dollars.
([2*X + X]/2)
So it doesn't matter if I switch or not.
That seems way too simple so I imagine I'm reading the problem incorrectly.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted April 04, 2011 07:02 PM |
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I hope I am not spoiling this for ihor. It is exactly what he wrote. Two boxes, and once we choose we are (always) allowed to change our mind. The question is, should we ? Yes, no, why ?
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The empty set
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ihor
Supreme Hero
Accidental Hero
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posted April 05, 2011 12:40 PM |
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Hi again,
this problem is a bit paradoxal. The calculations presented by Corribus are correct, but look at the question from another side. Imagine, that host shows you that there are 100$ in box you have chosen. Should you switch then?
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smithey
Promising
Supreme Hero
Yes im red, choke on it !!!
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posted April 05, 2011 01:34 PM |
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It's basic statistics, the answer is YES always change !!!!
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 05, 2011 03:31 PM |
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Quote: Hi again,
this problem is a bit paradoxal. The calculations presented by Corribus are correct, but look at the question from another side. Imagine, that host shows you that there are 100$ in box you have chosen. Should you switch then?
Well I guess the problem facing the decider is that he doesn't know whether $100 is the higher award or the lower award.
Often, revealing additional information like this changes the problem. I'll have to think of it again in light of this new information.
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted April 05, 2011 03:54 PM |
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Quote: It's basic statistics, the answer is YES always change !!!!
That is only true, as Corribus states, as long as additional information is provided. If there are only two boxes, no additional information is provided.
Even if you get to see that there is $100 in your box, that doesn't provide any information unless you know what the maximum win is or total amount of money (in which case you then have the obvious answer). Unless of course you take into account that money only comes in discrete amounts - i.e. you can only have (I don't know - what's the smallest coin in US currency?) $0.01, $0.02, $0.03, etc. which means there is a limited amount of values available < $100, but an infinite amount of values > $100, which would then imply that statistically you should change.
However, if that's the point of the question, it's incredibly vague.
EDIT > But still, if amounts are always 1:2, it's still only possible to have $ 50 or $ 200 in the other box, which still makes it completely random if you don't know anything about total gain. So yeah, doesn't make any change.
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What will happen now?
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Jabanoss
Promising
Legendary Hero
Property of Nightterror™
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posted April 05, 2011 04:08 PM |
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Quote: EDIT > But still, if amounts are always 1:2, it's still only possible to have $ 50 or $ 200 in the other box, which still makes it completely random if you don't know anything about total gain. So yeah, doesn't make any change.
Wait a second , couldn't that be key?
I mean either you gain $ 50 or $ 200, so if you take a new box it's mean value would be $ 125 which is above $ 100.
So it would be wise to change box as the probability to gain more is bigger then losing...
Just a thought.
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"You turn me on Jaba"
- Meroe
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted April 05, 2011 04:13 PM |
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Right, that's a good point, if you pick the other box there's a 50 % chance of winning $50 and a 50 % chance of winning $200, which on average is $125, which is more than what's in your current box.
However, doesn't that provide some sort of paradox? Because no matter what box you pick, your average gain of picking the other one would be 125 % of what's in the box you pick.
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What will happen now?
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Smithey
Promising
Supreme Hero
Yes im red, choke on it !!!
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posted April 05, 2011 04:17 PM |
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Edited by Smithey at 16:19, 05 Apr 2011.
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Quote: It's basic statistics, the answer is YES always change !!!!
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That is only true, as Corribus states, as long as additional information is provided. If there are only two boxes, no additional information is provided.
Even if you get to see that there is $100 in your box, that doesn't provide any information unless you know what the maximum win is or total amount of money (in which case you then have the obvious answer). Unless of course you take into account that money only comes in discrete amounts - i.e. you can only have (I don't know - what's the smallest coin in US currency?) $0.01, $0.02, $0.03, etc. which means there is a limited amount of values available < $100, but an infinite amount of values > $100, which would then imply that statistically you should change.
However, if that's the point of the question, it's incredibly vague.
EDIT > But still, if amounts are always 1:2, it's still only possible to have $ 50 or $ 200 in the other box, which still makes it completely random if you don't know anything about total gain. So yeah, doesn't make any change.
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I don't agree, Unless whoever asked the question brings in some hybrid/mutant catch/twist, it looks like statistics to me
The answer must be YES, once you've made the first choice, probability goes up hence you should always switch the boxes... well that's at least how I see things, considering I get paid for solving problems If I'm wrong... I'll quit my job and go on a travelling frenzy
P.S. please do tell me I'm wrong
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 05, 2011 04:18 PM |
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The problem with that Jabanoss is that at this point it's not random.
If the problem where: you open a box, see it's a $100 bucks and then you have the option of opening another box which randomly has either $50 or $200 in it, then yes, it'd be statistically correct to switch. But there's only one other box, and its value has already been determined before your opened even the first box. That is, the value is not randomly generated at the point of making a decision. So you don't really have enough information to decide whether to switch or not. Statistically speaking, there's no correct answer.
Unless these are quantum boxes and the values inside the boxes are in states of quantum superposition, much like Schroedinger's Cat.
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted April 05, 2011 04:24 PM |
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Quote: The problem with that Jabanoss is that at this point it's not random.
Isn't it still random? I mean there would be a 50 % chance that you picked the high box and a 50 % chance you picked the low box, right? And adding information that the box you have holds $100, does that skew chances of you having picked either the low or the high? I would say no?
I'm confused, I don't know what I think myself.
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What will happen now?
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 05, 2011 04:31 PM |
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@Alci
No, once you've already picked, it's no longer random.
Think of it this way.
Suppose you flip a coin. BEFORE you flip the coin, you have a 50% chance of getting heads and 50% chance of getting tails. AFTER you flip the coin, you've either got heads or got tails. There is no probability at this point because the probabilistic event occurred in the past. This is the case even if you haven't observed the result yet (flip the coin and cover it up, say). Now, if you flip the coin and THEN guess, and you devise a way to make your guess random (by a computer, say), then the probabilistic event is the guess and not the flip, so in this case you can still have a chance to be right after the flip. It all depends on what the probabilistic event is.
In the case of the box, someone puts a $100 dollar bill in one box and some other quantity in another box ($50 or $200, only this person has that information). Before you select, you have a 50% chance of picking the higher value or the lower value. Once you pick a box and open it, the determination is already made what is in the box you picked and what is in the other box. There's no more probability at this point, even if you haven't observed the state of the system. So if you choose the $100 box (randomly), the contents of the other box are not random. There is no probability. The only probability involved at this point would be if your decision to take the contents of the other box were determined randomly. In which case you don't have a choice so the problem is moot.
You can't speak of whether it is better to switch or not at this point because the contents of the other box have already been absolutely determined, and you don't have the information required to make any sort of informed decision. At this point it's just a non-probabilistic guess.
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