Ok---my final post on this subject. I roll a 2. What are the odds it becomes a 6 so that we can satisfy the 1-11ers?
I set my die on 2 one thousand times and it hasn't changed even once. Is my sample size too small??
Ok---my final post on this subject. I roll a 2. What are the odds it becomes a 6 so that we can satisfy the 1-11ers?
I set my die on 2 one thousand times and it hasn't changed even once. Is my sample size too small??
Why would you (and Alan) expect a die to change it's value? Don't get ridiculous now.Originally Posted by regnis
Instead, you need to only comprehend that you are dealing with two dice and either one of them can become a two. That doesn't give the ability to a die to change it's spots, don't get silly. Instead it makes it twice as possible to get a 2 (and twice as possible to miss a 2 because its "on the other die" lol), than with one die only (the error you all make when concentrate on one die and forget that there were two dice to begin with).
Last edited by kewl; 06-03-2015 at 02:04 PM.
Kewl there were two dice at the start of the problem. At least one of them is a 2. We can subtract one of two dice from consideration. The remaining die has six faces. Consider this:
2(6 faces) - 1(6 faces) = 1(6 faces) or 1/6
Why can we remove one die? Because we are told at least one die shows a 2. Does it matter which of two dice we remove? No because either die would not effect the answer or the question.
That is what you do when you fail to freeze at least one die as a 2.Originally Posted by kewl
And you also can't rotate the non-2 in the real physical world. But if I remember correctly, this is what you did in your video and is your "proof".Originally Posted by Alan Mendelson
Failure in comprehension. What we do is give you a reminder there are two dice to consider and not one. You can't come after the throw and eliminate anything. That's the outcome now. You don't mess with the outcome. You calculate probability. And it's done before the roll.Originally Posted by Alan Mendelson
Yes I did. To show that it had six faces giving us 1/6 for the second die. The die showing a 2 remained a 2.Originally Posted by RS__
Obviously you must alter reality. Thanks for admitting that's what you do.Originally Posted by kewl
Originally Posted by regnisAww…I had answered your questions after you cried about people refusing to do so. Yet, you're refusing to answer my questions.Originally Posted by regnis
Last edited by Zedd; 06-03-2015 at 02:29 PM.
Huh? What?Originally Posted by Alan Mendelson
Give 'em heck, Alan.Originally Posted by Alan Mendelson
Kewl you wrote "it's done before the roll." That alters reality.
The original problem specifically states that two dice were rolled and at least one is showing a 2. Whatever you thought prior to that no longer matters. Now you have to deal with at least one die showing a two. The question now centers on the second die in the problem which may or may not be a two and the chances for that second die to be a 2 is 1/6.
By figuring probability before the roll alters the reality of the question. You are changing the facts and the conditions of the problem.
As I have said a hundred times, your 1/11 is "good math" but for a different question -- perhaps for a question that asks: "how many different combinations of two dice have at least one 2 and of those combinations how many would show 2-2?" That would be your 1/11 answer.
The only way to resolve this problem is to actually have Alan bet some of his money on the 1/6 odds.
That won't resolve anything because anything can happen when two dice are rolled. I know that and you know that. In fact, since the bet with the Wizard that we decided on has a very limited number of rolls -- and only a $50 lunch was at stake -- I've already conceded defeat.Originally Posted by arcimede$
More importantly, can you show the proof of 1/11 using two dice, with the "2" being fixed?
There is no "fixed" 2 in the original question. Are you trying to change the question to match your answer?Originally Posted by Alan Mendelson
It's very simple. We could do it over skype.
- I roll two dice in a cup and peek under and tell you whether at least one 2 is present.
- If so, the bet is on.
- When uncovered if 2-2 shows, I owe you $80.
- When uncovered if 2-x shows (where x is a non 2), you own me $10.
- We run the test for 100 bets. That is 16 cycles of your 1/6 odds. The result should be very close to statistical reality. If you want to do more rolls that is fine with me. the more the merrier.
- When all is done the loser writes a check to the winner.
Arc I have something that won't cost you any money or much time. Why don't you make a video using two dice -- with at least one die coming to rest as a 2 and without changing that die showing a 2 show me how your answer is 1/11?
You can do it yourself. Set up your camera over a table.Originally Posted by Alan Mendelson
- Throw a pair of dice
- Every time at least one of them is a 2 put a mark on a piece of paper (that meets your first criteria ... "with at least one die coming to rest as a 2)"
- Every time both of them are a 2 put a mark in a another column. Never is any die changed after coming to rest.
Tally your columns after 100 throws when a 2 is present and show us the video..
Is your reading comprehension that bad Arc? It appears so. Go ahead shoot the video with one die fixed as a two and show me your proof that there are eleven other possible faces of which 2-2 appears 1/11 times. Go ahead and do it. Yeah, do it just like the Wizard did it.Originally Posted by arcimede$
This thread has been brought to the attention of users at http://math.stackexchange.com/ That being said, I am a user there and have come to give an argument in favor of the "1-11'ers." I have multiple degrees in mathematics and my field is in combinatorics. This is precisely the type of question we would ask students in an undergraduate class involving basic probability.
The scenario (as I understand): You have two dice. You roll both dice and someone checks to see if there are any twos. If there are *no* twos, then the person verifying reveals the results of the dice and has the dice cast again.
Eventually the verifier will truthfully state "There is at least one two" at which point you ask the question of "What is the probability that *both* dice are two?"
Let us reimagine this scenario somewhat. To keep track of what is going on a bit more easily, let us imagine that the two dice are *different colors*, one red, and one green. Clearly, adding or removing color to the mix will not in any way change the probability of the situation.
As the dice are *fair*, every result is equally likely. We have the following results:
1-1. 1-2, 1-3, ...
2-1. 2-2, 2-3, ...
...
For a total of 36 equally probable scenarios. As per usual, our definition of probability in an equiprobable setting (such as this one) is Pr(A) = |A|/|S|, where A is the event in question, |A| represents the size of the event in question, and |S| represents the size of the sample space.
This question however is one about "conditional probability," written mathematically as Pr(A|B) "the probability of event A occuring given that we know that B occured" or more simply as "probability of A given B."
Let A represent the event that BOTH dice show a 2. Let B represent the event that "AT LEAST ONE" die shows a 2. Let us go one step further and also let C represent the event that the red die shows a 2, to explain where your answer comes from.
The question being asked is: "What is the probability that both dice show 2 given that at least one die (be it the red one *or* the green one) is confirmed to be a two", i.e. Pr(A|B)=?
The *definition* of conditional probability states that Pr(A|B) = Pr(A and B)/Pr(B). which, since we are in an equiprobable setting can be simplified further to be = (|A and B|/|S|)/(|B|/|S|) = |A and B|/|B|.
We now try to calculate |A and B| and also calculate |B|. How many outcomes have it such that "both dice are 2 AND at least one die shows a 2", the only outcome is 22, so |A and B|=1, and the probability Pr(A and B) = 1/36. Calculating the size of |B|, we look at all outcomes where there is at least one two. Referring to our chart above, we see that occurs when either the red die is 2 or when the green die is 2 (or both), for a total of 11 outcomes. (specifically 2-1. 2-2, 2-3, 2-4. 2-5, 2-6, 1-2. 3-2, 4-2, 5-2, 6-2). Remember: Each of these events is equally likely to be the one that has been tossed.
Thus, we calculate Pr(A|B) = |A and B|/|B| = 1/11.
----
Why are you convinced that the probability is instead 1/6? That is explainable as well. You seem to be confused as to the nature of how the dice are acting. You think "The *FIRST* die is a 2, the second die could either be a 2 or not", however how do you define the "first die." With colors on the dice, it makes it easier to see whats going on.
You seem to have mistaken the situation as: You have two dice. You roll both dice and someone checks to see if THE RED DIE IS A TWO. If the red die is not a two, then the person verifying reveals the results of the dice and has the dice cast again.
Eventually the verifier will truthfully state "The red die is a two" at which point you ask the question of "What is the probability that *both* dice are two?" I.e. Pr(A|C)=?
We calculate this the same way as before, noting that |A and C| = 1, as before, and |C| = 6. You get then that Pr(A|C) = Pr(A and C)/Pr(C) = (1/36)/(1/6) = 6/36 = 1/6.
Yes, of course the results of the dice themselves are independent, however an event which uses information about both dice simultaneously ("at least one die is a two") is no longer independent of the results of the dice.
Jihkro thanks for joining and for posting. Unfortunately everything you wrote above has been discussed before.
Our difference is over the wording of the question.
Those of us who say the answer is 1/6 believe that the question simulates the following:
Two dice are thrown on a craps table and one comes to rest on the 2 face. The second die spins like a top. What are the chances that the "spinner" will land on a 2?
We -- the 1/6ers -- see nothing to indicate that given two dice with one frozen to show a 2, could possibly result in 11 faces to be considered.
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