Would one of you who still is on the WOV forum please tell "Indignant" (really a good handle he picked) that he will have to choose which set of results he wants to use when he rolls two dice: either the horizontal results or the vertical results from his chart. You cannot use both to come up with an answer for this problem.
As I mentioned before, the chart can be used to answer a different question.
"Indignant" has responded to my post on the WOV forum. (I am sure he is boring the hell out of everyone there.) He wrote: "Isn't the whole G.D. friggin' chart - all 36 outcomes - in play when rolling two dice?"
Indignant, good buddy, the whole friggin' chart with all 36 outcomes does not apply to the question. One die has already been identified as a 2, which leaves the other die with its 6 faces to be considered.
Granted, this is not a great example of probability theory, but it is common sense.
Indignant99 from the WOV forum has joined our forum but it appears he has not posted yet. He did respond to my post above on the WOV forum and here is what he posted (in bold):
Isn't the whole G.D. friggin' chart - all 36 outcomes - in play when rolling two dice?
Yes, the result itself lies on either the horizontal, or the vertical, or lo-and-behold the intersection. But none of the ELEVEN rolls are ruled out.
But before the cup is removed, and the dice revealed, is the result gonna be one of the horizontal ones? Or one of the vertical ones? Huh, genius?
(The answer is: Yes, it's gonna be either one of the horizontal, or vertical, ones.)
Before cup removal, but after "deuce" announcement, exactly which restricted zone - horizontal vs. vertical - have we been incarcerated in?
The peeker/announcer did not inform "it's vertical," nor "it's horizontal."
What he actually informed, was "it's in the criss-cross."
After the cup is removed to reveal the dice, we do not enter into some restricted fantasy zone (horizontal versus vertical). The outcome is done. It is what it is.
Unfortunately, in the real world you cannot use both the vertical and the horizontal charts. That's good for theory, but in the real world with two physical dice choose either the vertical or the horizontal.
And in the real world, dice do not appear on charts with multiple options for what appears on their faces.
I'm afraid Alan is effectively trolling his own forum at this point. He has to have realized by now he is wrong and his "interpretation" is wrong. He hasn't answered what his interpretation of the question has to do with answering the more simpler question:
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"
and by now I just assume he isn't admitting he's wrong, because it will make him feel ashamed or something.
kewl are you really asking this question again?
Have you answered it yet?
I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select the guy with the red shirt from group 1, what are the odds of blindly picking the matching shirt from group 2.
Hint--it ain't 1 of 11.
Really? You guys pride yourselves with good comprehension/writing skills, yet you post an analogue which is completely not what the question asks?Originally Posted by regnis
Hint--it ain't asking what are the odds of blindly picking the matching shirt from group 2.
It asks if you blindly pick a person from each group and at least one of them is wearing a red shirt, what are the odds they are both wearing red shirts. Don't you see the difference?
Gee..
regnis has it correct. That is the question. kewl, you are answering a different question with your 1/11 response.
How come? Where did you read this? Where does it say you get to choose a die? You assume that the statement "at least one die is a 2" gives you the right to set one of the dice and ask a question about the other. Your assumption is flawed. And you are looking at this at very wrong angle. An elementary one, a child will do better than this.Originally Posted by Alan Mendelson
That is why the proposed bet will lose player's money. That's a fact(given sufficient enough trials, for trolls like RS).
And any amount of denying can't change this fact. The intrinsic probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2 is 1/11. It is intrinsic probability. Not a game of words, not a trick.
And you continue to fail to understand that once a die shows a 2, it can't change to accommodate the other 5 faces. A 2 is a 2 is a 2. Now ----what are the odds the other die is a 2.Originally Posted by kewl
My shirts weren't bleeding madras (that shows some age-huh)---so once it was red it was always red. What are the odds that I pick a red shirt from the other group.
You are blindly picking that first red shirt. You could have picked blue or yellow. Once you did, what are the odds of blindly picking the matching shirt.
Back to the dice---the friend peeked and said one is a 3. What are the odds the other is a 5? Still 1 of 6.
. The answer to that one is 1/6Originally Posted by regnis
That's a different question - one that assumes the peeker will announce the value of a random die even if there is no 2 under the cup. The answer to that one is 1/6.
This is not the question asked here though.
It's what they do folks. When they can't understand or accept that the rest of the world sees something in the question that they don't see, some of them turn to frustration and vulgarity, some shout out insults because it makes them feel better, and some just get plain hurt by it period. Why else do you think they frequent the most liberal....most theoretical gaming forum ever?
How is it different if the peeker said one is a 3 instead of a 2. The question is still what is on the other die. That is always going to be 1 of 6.Originally Posted by kewl
kewl you are simply making the same mistake that the Wizard and all the others made: you ignored the condition presented in the question. This is why I asked the Wizard and everyone else to do a video so we could see using real, physical dice how you interpret the question and the condition -- and you all made same mistake: you wouldn't accept that you can't change the value of a die.
Go back to the Wizard's video and what does he do? First he shows a die as a 2, and then he goes through the math mumbo-jumbo of changing the value of that die.
Of course you guys make the claim that we don't know which of the two dice is the 2. But it doesn't matter when you have a problem involving only two dice. Of course, you can't see that.
You ask: Where does it say you get to choose a die? I am going to ask you a related question: where does it say you can first have one die showing a 2 and and get the results, and then switch the 2 to another die and get different results and then add the two sets of results?
I am going to ask you to use your common sense for a moment: take two dice and if I tell you one of those two dice is showing a 2 pick one of the dice and set it as a 2. With one die known to be a 2, how many options are there on the other die in this two dice puzzle?
Again, what you are doing is finding the odds with one die being a 2, and then repeating the procedure for finding the odds when the other die is a 2, and then you are adding them together to get 1/11.
Let me rephrase.Originally Posted by regnis
If the question was:
You roll a pair of dice and no matter what the outcome is the peeker will announce the value of one of the dice (at random). What are the odds that there will be a: pair, a (announced value) and 1, a (announced value) and 2, a (announced value) and 3...whatever),
then the answer will be 1/6.
And if the question was (in your case with 5 and 3):
You roll a pair of dice. The peeker makes announcement when at least one 3 or one 5 is under the cup. What are the odds that there are exactly 5 and 3 under the cup?
then the answer is 1/11.
Now, compare the second question with the one asked here. Keep in mind that there was discussion about whether the peeker announces a value every time or only if there is a 2 under the cup and it was agreed upon, and Alan concurred, that the case in discussion is the one where the peeker only announces if there is a 2. Otherwise he says nothing.
Last edited by kewl; 05-17-2015 at 10:25 PM. Reason: one 3 or one 5
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"Originally Posted by Alan Mendelson
Your answer?
If you asked me that at a craps table I would say 1/36 because we know that in craps the chances of throwing 2-2 are 1/36.Originally Posted by kewl
If you asked me this: two dice have been rolled and one die came up showing a 2. What are the chances that the other die is showing a 2? I would answer 1/6.
Now to be specific about your question: "What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?" As soon as you told me one die is a 2, there is a 1/6 chance that the other die is a 2.
I say we put an end to this exercise in futility and wait for a video. From Alan.
He does exactly like in the proposed bet in this thread's post#1. Throws dice, see if there is a two and counts the number of times where there is 2-x and 2-2.
Does it lots of times. Then see what happens with the ratio.
The ratio will tell us all, truthfully, what are the true odds when we throw a pair of dice and at least one of them shows 2, that they are both 2.
That's it. No more mumbo-jumbo.
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