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Author Topic: To All You BFM Geniuses  (Read 23547 times)

Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #15 on: July 24, 2009, 06:03:50 PM »
On number 5, if you have two die, and one is a 2, there's only one possible scenario where you could get a total value of 6, which is with a 4 on the other die. The die are mutually exclusive so the first die has no impact on the second, making the chance 1/6, or the chance that it was a 4.

How confident are you in your answer, percentage-wise?  :winkgrin:

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #16 on: July 24, 2009, 06:21:42 PM »
Do you want me to make a program to stimulate it?
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #17 on: July 24, 2009, 06:32:47 PM »
It sounds like you want to make a bet.  If you are, I'll take it.

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #18 on: July 24, 2009, 09:00:09 PM »
Simulated it just now 10 million times. And let's see how many times it added up to 6 given that the first one was 2... what do you know, 1666242, roughly 1/6 with a very low degree of uncertainty. :P
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #19 on: July 24, 2009, 09:17:39 PM »
Simulated it just now 10 million times. And let's see how many times it added up to 6 given that the first one was 2... what do you know, 1666242, roughly 1/6 with a very low degree of uncertainty. :P


You might want to check your program for any logical error.  Post your source code and I'll point it out for you.

Come on now Edison, all those advanced math classes you took in high school... you of all people should know that mathematics can be very counterintuitive and is full of paradoxes, so you can't always rely on "common sense" or intuition.. I'm sure you'll figure out why it can't be 1/6.  If you give up, let me know and  I'll explain why. :)


EDIT:
Here's a screen of my simulation I wrote in C++. As you can see, the probability gets closer and closer to my answer of 18.18% as the number of trials increase, in accordance with the laws of probability.

« Last Edit: July 24, 2009, 09:30:52 PM by »X« »

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #20 on: July 24, 2009, 10:29:42 PM »
In my opinion, that's a different question than what I was answering, where it's the same dice every time that you see. They were probably aiming for your interpretation, but I still hate how they can never be specific as to what they want, like "What's the probability that one of the dice is a 2 and the sum is 6." Just ask exactly what you're looking for.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #21 on: July 24, 2009, 11:49:17 PM »
I believe the question is clearly asking... if someone rolls two standard dice, and you see one of the dice lands on a 2, then what is the probability the other dice landed on a 4.

Most people would say something like, "hmmmmm... the two dice are independent, and there's 6 sides on the other dice, and so there obviously must be a 1/6 chance of getting a 4 on the other dice." It would make perfect sense, except that it's not true.  The question is do you know why? ;)

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #22 on: July 25, 2009, 01:57:53 AM »
Except I read it as you are seeing the same dice every time in all these different situations, and not whichever has landed on two. So if you label each die as one and two, then it only counts as seeing die one as being a 2, but not seeing die two as a 2. I viewed it as in any scenario, it was a specific die you were seeing, not whichever one was a 2. That's just what the question meant to me. Thought I made it clear that I knew why you get 2/11 when I said what a clearer question would be with the whole fact that either die could be a 2.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #23 on: July 25, 2009, 11:21:54 PM »
You are complicating things again. If two dice are rolled together, if you see a 2, then the probability is very slighly greater the other dice is not a 2.

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #24 on: July 26, 2009, 12:54:37 AM »
No it's not. You get 2/11 when you're looking at the chance that they add up to 6 with a 4 and a 2.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #25 on: July 26, 2009, 01:25:13 AM »
No it's not. You get 2/11 when you're looking at the chance that they add up to 6 with a 4 and a 2.

If you put 2/11 in your calculator, it will equal to 18.18%, my original answer from the beginning :)

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #26 on: July 26, 2009, 02:23:04 AM »
I know that 2/11 is .181818181818. What I am saying is that one die rolling a 2 does not make the other die less likely to roll a 2. What I'm saying is that you're looking at the situation where either die can be a 2, while I'm looking at it as only a specific die can be a 2. So in your interpretation of either die possibly being a 2, it is 2/11, while in my interpretation of only a specific die, it is 1/6.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #27 on: July 26, 2009, 02:48:15 AM »
I know that 2/11 is .181818181818. What I am saying is that one die rolling a 2 does not make the other die less likely to roll a 2. What I'm saying is that you're looking at the situation where either die can be a 2, while I'm looking at it as only a specific die can be a 2. So in your interpretation of either die possibly being a 2, it is 2/11, while in my interpretation of only a specific die, it is 1/6.

Let's be realistic. If you go to a casino, and a dealer rolls two dice, you can't really tell if it was dice A or dice B that landed on a 2.  But the fact that you saw a dice landed on a 2, you can predict the other dice is less likely to be a 2 also.  The question isn't saying that the dealer rolls one dice at a time.  It's rolled together, so they aren't independent of each other. There's no ambiguity here.

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #28 on: July 26, 2009, 02:53:43 AM »
They are completely independent events. These dice aren't entangled. Your wording is basically saying that the probability for each number is 2/11 for every number except for 2, which is 1/11, which is not correct.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #29 on: July 26, 2009, 03:20:42 AM »
They are completely independent events. These dice aren't entangled. Your wording is basically saying that the probability for each number is 2/11 for every number except for 2, which is 1/11, which is not correct.

Actually, it is correct. Go to the store and buy 2 dice and conduct the experiment and you'll be amazed at the result.  Since 16.67% and 18.18% are so close together, make sure your number of trials is sufficiently large enough so that statistically, you'll have a high level of confidence in your answer. Test your alternative hypothesis. :)

Or..... I guess taking some time off of Physics and spending more on Advanced Probability Theory wouldn't hurt.

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