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Author Topic: To All You BFM Geniuses  (Read 23599 times)

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #30 on: July 26, 2009, 03:46:21 AM »
I still don't think you see what I'm saying. In my interpretation, were I to test it using dice, I would have one red and one blue. I'd only look at the red one to see if it's a two. If it was, I would then see if the other is a four. In this way, I would get 1/6. In your interpretation, I would examine both dice, and if either was a two, then I'd look at the other and see if the other was a four. In this way, I would get 2/11. I ran simulations for both of these methods of looking for a two, and the first gave 1/6 while the second gave 2/11.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #31 on: July 27, 2009, 09:31:07 PM »
I do get exactly what you are saying. Your interpretation is wrong. The question says nothing about a red die and a blue die, looking a red die first, and then the blue die. The answer would be trivial if that was the case.

Quote
In your interpretation, I would examine both dice, and if either was a two, then I'd look at the other and see if the other was a four.

What would be the point of predicting the "other" die if someone already knows the value of both dice already? It makes no sense at all. Your wrong assumption of my interpretation will still produce the correct answer by coincidence though. However, the question is asking to predict the value of the "other" die by knowing the result of just one die that you see.

Oh, and if you ever gamble before, almost all of the time the dice will have the same color and are almost indistinguishable from one another.

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #32 on: July 28, 2009, 12:19:18 AM »
My point is that in my mind, a theoretical simulation of this would have the dice rolling the same way exactly every time with different results, and only one specific die would be seen. It's as if you were to repeat the same exact motions of the dice every time, but they result in different numbers. In no way is it an incorrect interpretation of the problem, even if it may not be the interpretation that the askers meant. And the part you quoted was me simplifying what the difference between our two interpretations even more, since it didn't seem like you understood what I was saying, and I don't know that you do now for sure. I have a completely correct interpretation of your interpretation, which is that either dice could have been seen, and that you can't restrict yourself to the possibility of only ever seeing one of the dice. It's not by coincidence that they have the same value, as the meaning behind them is the same; they have the exact same method of simulation in a computer program. And so what if that would be trivial? Some of the other questions presented were, in my mind, trivial as well.

My presentation of different colored dice was to illustrate my interpretation more clearly, but you're taking it somewhat too literally. To simulate it by hand, since you can't have a way to make sure that they person happens to see a specific die every time, they'd have to be different colors for my interpretation, as trivial as it may be. The triviality of a question has never stopped people from asking it.
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #33 on: July 29, 2009, 09:54:38 PM »
I wasn't taking what you said literally; I was merely trying to point out a flaw in your logic. That's all   :winkgrin:

Anyways...

Original Question:

5) Suppose a dealer for a casino rolls two standard dice. You saw one of the dice; it was a 2. What is the chance that the sum of the face value of the two dice adds up to 6?


I'd prefer to calculate things in my head, but as they always say, "When in doubt, list them all out." Of course, this problem has a small sample space so listing them is reasonable.

So the question says that two dice are rolled together and "you saw one of the dice; it was a 2."  Since there 2 dice, and there 6 possible values on each die, then there are 36 possible outcomes. I don't think anyone will disagree on that, for it's just basic math. So here are sample space:

{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}


Since you saw a die with a "2", then one of these 11 outcomes must have occured:


{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}


And from those 11 possible outcomes, only 2 of them add up to 6.

{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}


That is why your answer of 1/6 is incorrect. The problem is very counterintuitive, just like the Monty Hall problem. But I have to give it to you, you have good debate skills.

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #34 on: July 29, 2009, 11:31:43 PM »
Except you still don't see what I'm trying to say. In my interpretation, where only a specific die can be seen as having a two, this is what you get:

{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}

There are 6 scenarios where the first die is a 2, and only one of them where the second die is a 4, giving 1/6. The scenarios where the second die is a 2 do not count in my interpretation as they are not on the die that you saw.

I already knew how you got your answer. The point that I've been trying to get across is that we're answering different questions.
« Last Edit: August 01, 2009, 08:55:33 PM by BFM_Hydra »
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Offline jim360

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Re: To All You BFM Geniuses
« Reply #35 on: July 30, 2009, 04:21:46 AM »
I figured that to resolve this once and for all I'll just go back to basics.

What is the probability that two dice add up to 6 given that one of them is a two?

Then using that lovely Bayes' formula for these circumstances gives:

P(6 given a 2) = P(6)*P(a 2 given 6)/P(a 2) = (5/36)*(2/5)/(11/36) = 2/11.

QED.



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Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #36 on: July 30, 2009, 05:23:19 AM »
I've never denied that. I explicitly stated earlier that the probability that two dice add up to 6 given that one of them is a 2 is 2/11. I can find the quote if you want. What I've said multiple times and will continue saying is that I read (present tense) the problem as asking if a specific die is 2, and in any simulation, it is that die that is 2. In my interpretation, it's not "one of them is a two," but, "this specific one is a two, and the other can be a two or not, but I don't really care." I'll say it one more time, we are answering two separate questions here.
« Last Edit: July 30, 2009, 05:28:13 AM by BFM_Edison »
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #37 on: July 31, 2009, 05:20:37 PM »
Edison, you have taken so many positions throughout this thread that you seem to confuse people. You seem to be all over the place. You frist claimed the answer was 1/6, then when I said it was 2/11, you asked me if I wanted to see you "make a program to simulate it" to show my answer of 2/11 is incorrect, then you claimed to have the "completely correct interpretation of [my] interpretation" that I would have to "examine both die" to calculate the answer, which I showed to not be true.



EDIT: I missed you post, three360. Thank you for giving a new perspective. <3

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #38 on: July 31, 2009, 07:06:36 PM »
That's not what I said at all. I was saying that checking both dice for a two is the same problem in essence as the probability that you roll a 2 and a 4. I never said you had to, but that it is the exact same question in essence. The only time I've changed a position is when I realized that you interpreted the question itself differently, resulting in a different, non-trivial answer. I'm done, explaining this bores me >.>

Btw, the way 360 did it was the way I solved it for your interpretation as well.
« Last Edit: July 31, 2009, 07:47:19 PM by BFM_Edison »
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #39 on: July 31, 2009, 09:54:56 PM »
It's okay, Edison. One day you'll be able to correct me when I'm wrong.

Oh and the PM you sent me.... LOL I'm up for it anytime  :P

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #40 on: August 01, 2009, 03:41:46 AM »
No one's wrong here >.>
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Offline jim360

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Re: To All You BFM Geniuses
« Reply #41 on: August 01, 2009, 01:31:53 PM »
One way of explaining why I think the 2/11 interpretation is right is that, in the context of the other questions on the paper, it is clear that a non-trivial answer is required. I initially just looked at the question and said "1/6" but that requires no thought at all. This is a paper with some hard questions on and I can't honestly think that they would expect you to be able to answer a question by just looking at it - or, at least, that there would be a question on dice asking essentially, "what is the probability of getting a 4 on a fair die?", a question I'd hope that most 11-year-olds could answer. By introducing the 2nd die then it is clearly supposed to complicate the maths involved or lengthen the working.
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Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #42 on: August 01, 2009, 02:27:39 PM »
2, 3, and 4 seem pretty trivial to me, as does 6 if you know how to fit polynomials to specific points.
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Offline BFM_Hydra

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Re: To All You BFM Geniuses
« Reply #43 on: August 01, 2009, 08:57:50 PM »
Guys, seriously.

Both of you are right. You just looked at the question in different ways. If you rolled two random dice, the probability that you rolled a 4 after rolling a 2 is 1/6. The probability that they equal 6  GIVEN THAT EITHER is a 2 is 2/11. You're both right in different ways. Move on, seriously. LOL! :LOL:

I think 2/11 is the corret answer on the paper though.  :interesting:


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Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #44 on: August 02, 2009, 06:08:49 AM »
Guys, seriously.

Both of you are right. You just looked at the question in different ways. If you rolled two random dice, the probability that you rolled a 4 after rolling a 2 is 1/6. The probability that they equal 6  GIVEN THAT EITHER is a 2 is 2/11. You're both right in different ways. Move on, seriously. LOL! :LOL:

I think 2/11 is the corret answer on the paper though.  :interesting:

That's what I've been saying in almost every single post D:
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