View Poll Results: Please choose what you believe are the proper answers.

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  • It is okay to double-count a die in this two-dice problem.

    2 20.00%
  • It is NOT okay to double-count a die in this two-dice problem.

    3 30.00%
  • The original question is the same as having a spinner on a table.

    4 40.00%
  • The original question is NOT the same as having a spinner on a table.

    4 40.00%
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Thread: How do you interpret the "dice problem"?

  1. #1
    Let me just sum this up and return to the basic question. Just in case you missed the other threads this is the original "dice problem" as posted on the Wizard of Vegas message board:

    You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

    What is the probability that both dice are showing a 2?


    There are two basic "sides" giving different answers: one side says the answer is 1/11 and one side says the answer is 1/6.

    Video explanations for both sides have been presented.

    In case you missed them, here they are again. First the video from Michael Shackleford who says the answer 1/11:



    And here is my video and I say the answer is 1/6:



    At first blush you can see a problem: Michael willingly "double counts" a die, but I don't. Both of us show at least one die rolled with a two and while I remove the die with a 2 to show the answer is 1/6, Michael keeps the die showing a 2 and "double counts" it to increase the number of options to be 11. Clearly he needs two dice to show 11.

    I've asked this before and I am yet to get a straight answer: why are you allowed to "double count" a die in a two dice problem? I don't double count my die showing a 2 but Michael "double counts" his die showing a 2.

    And so I need someone from the 1/11 "side" to clearly explain how they justify using one die "twice" to solve their problem? And explain why when the value of a die is known that they are allowed to change it?

    The second conflict between the 1/11ers and 1/6ers is the interpretation of the original question (above, in bold).

    I consider that question to be the same as this event and question at a craps table -- and this is an event that we've all seen a few times:

    The shooter throws two dice and one die immediately comes to rest showing a "2" but the second die spins like a top. What are the odds that the second die will also come to rest showing a 2?

    If you think this craps table event is the NOT THE SAME as the original question please explain why?

    I added a poll reflecting these questions. Please participate.

  2. #2
    Alan, the reason you double count is the question refers to "probability" aka "odds". What that means is across an infinite number of independent events what is the ratio of two 2s appearing vs. one 2 and another (1,3,4,5,6) value. In no way does this prescribe any particular die to be a 2.

  3. #3
    So it's okay to count the same die twice? And to assign different values to that same die?

    I guess probability is different from elections where you can only vote once.

  4. #4
    Actually, Arci, you're only partly correct. The most common usage of "probability," the one that appears first in any dictionary, is not math related. The math-related definition is usually the second or third definition.

    So, technically (and I know some math dudes are -- if nothing else -- technical), the most common interpretation of this problem should "probably" be the non math interpretation. What would this mean? It means the tricky question, which does not say "an infinite number of throws or peeks or slams of the cup," but instead refers in present tense to a single throw, could easily be interpreted as referencing a single throw.
    Last edited by redietz; 05-21-2015 at 07:59 PM.

  5. #5
    Look, the "set-up" part of the question is not in mathematical terms. It's in common language terms referring in present tense to one roll of the dice and one peek. Obviously, if the author was interested in spelling out that this is a mathematical probability question from the beginning, he would have written something referring to plural throws, peeks, and so on. Nothing prevented him from doing this. He wasn't paying by the buck for words. So clearly the author wanted the illusion of a single throw in the readers' minds, then switched gears at the end and asked a (math) probability question, which refers to more than one event.

    If you use the non math definition of "probability," which is logical enough here, the cute thing is that the answer to the question would be the same for both the 1/11 and 1/6 camps. What is the probability of the other die being a 2? Not very.

    Kumbaya.
    Last edited by redietz; 05-21-2015 at 08:39 PM.

  6. #6
    The shooter throws two dice and one die immediately comes to rest showing a "2" but the second die spins like a top. What are the odds that the second die will also come to rest showing a 2?

    If you think this craps table event is the NOT THE SAME as the original question please explain why?
    This is not the same question because it does not include every situation that the original question does. The original question considers rolls where at least one die is a 2. The spinner example misses the rolls where the first die immediately comes to rest on a number other than 2 and the spinner eventually settles on 2.

  7. #7
    Originally Posted by 1in11 View Post
    The original question considers rolls where at least one die is a 2. The spinner example misses the rolls where the first die immediately comes to rest on a number other than 2 and the spinner eventually settles on 2.
    Why did you change the conditions in my "spinner" example? I specifically wrote above:

    Originally Posted by Alan Mendelson View Post

    The shooter throws two dice and one die immediately comes to rest showing a "2" but the second die spins like a top. What are the odds that the second die will also come to rest showing a 2?
    Do you care to change your position?

    The "spinner" situation is exactly what the original problem question asks us to consider. It has always been about one die, and never about "double counting" any die to come up with your 1/11 Rube Goldberg answer.

  8. #8
    Originally Posted by arcimede$ View Post
    Alan, the reason you double count is the question refers to "probability" aka "odds". What that means is across an infinite number of independent events what is the ratio of two 2s appearing vs. one 2 and another (1,3,4,5,6) value. In no way does this prescribe any particular die to be a 2.
    Does your answer change considering this is just ONE ROLL and not an "infinite number"? Are you allowed to "double count" on ONE ROLL ??

  9. #9
    I did not change the conditions on your example. I showed a condition that your example does not consider, but is covered by the conditions of the original question.

  10. #10
    Originally Posted by Alan Mendelson View Post
    So it's okay to count the same die twice? And to assign different values to that same die?
    Yes, you are assigning different values on different throws. Think VP. You aren't limited in the number or types of aces you will get on future deals because you got one on the current deal.

    Originally Posted by Alan Mendelson View Post
    I guess probability is different from elections where you can only vote once.
    That's right.

  11. #11
    Originally Posted by 1in11 View Post
    I did not change the conditions on your example. I showed a condition that your example does not consider, but is covered by the conditions of the original question.
    WRONG. At least one die must show a 2. So in the case of the spinner one die is showing a 2.

  12. #12
    Originally Posted by redietz View Post
    Look, the "set-up" part of the question is not in mathematical terms. It's in common language terms referring in present tense to one roll of the dice and one peek. Obviously, if the author was interested in spelling out that this is a mathematical probability question from the beginning, he would have written something referring to plural throws, peeks, and so on. Nothing prevented him from doing this. He wasn't paying by the buck for words. So clearly the author wanted the illusion of a single throw in the readers' minds, then switched gears at the end and asked a (math) probability question, which refers to more than one event.
    When he asked about probability he put the question in context. Probability only relates to multiple events. As has been mentioned repeatedly this is a trick question that is meant to fool a person who doesn't think it through.

  13. #13
    Originally Posted by Alan Mendelson View Post
    Does your answer change considering this is just ONE ROLL and not an "infinite number"? Are you allowed to "double count" on ONE ROLL ??
    When you specifically ask for the "probability, that requires infinite rolls to determine.

  14. #14
    No, arci, "probability" does not, in general common usage, necessarily refer to multiple events. I'm sure you have checked the definition in the dictionary and realize this. The math-oriented definition of probability, which is the second or third definition, does. So the author of the dice question leaving the entirety of the rest of the question without a mathematical reference, and then sliding in the word "probability" at the end, does not necessarily put it in context, and does not necessarily refer to multiple events.

    Math guys are going to orient to the math-definition of "probability," but that is not the definition listed first in dictionaries. If you want to argue that dictionaries are incorrect, so be it.

    Everybody is on Alan's case for not admitting he is "wrong." Now let's see, after checking the dictionary, how many math guys are willing to say, "You know, I was wrong about the first and primary definition of probability."

    I have a feeling the math guys are going to say "I was wrong" in a fashion very similar to The Fonz (yeah, I'm dating myself).
    Last edited by redietz; 05-22-2015 at 02:37 PM.

  15. #15
    Originally Posted by Alan Mendelson View Post
    WRONG. At least one die must show a 2. So in the case of the spinner one die is showing a 2.
    How does the case where the die that immediately comes to rest on a number other than 2 and the spinner eventually settles on 2 not satisfy the condition that at least one of the dice is a 2?

  16. #16
    Originally Posted by 1in11 View Post
    I did not change the conditions on your example. I showed a condition that your example does not consider, but is covered by the conditions of the original question.
    The original question says at lease one die is a 2. The spinner situation is covered. If the first die was not a two in the spinner situation the original question would not be applied.

    Now what's your point?

  17. #17
    Originally Posted by arcimede$ View Post
    When he asked about probability he put the question in context. Probability only relates to multiple events. As has been mentioned repeatedly this is a trick question that is meant to fool a person who doesn't think it through.
    And what about the mathematicians who over-thought the question? As soon as I saw the Wizard video with double counting I knew they over-thought the original question or just ignored it. They were conditioned to look at probability formulas but failed to see the simple basic question.

    They could not see the forest for the trees.

  18. #18
    Originally Posted by Alan Mendelson View Post
    The original question says at lease one die is a 2. The spinner situation is covered. If the first die was not a two in the spinner situation the original question would not be applied.

    Now what's your point?
    If the first die is not a 2, then there is still the possibility of the original question applying. That is my point.

  19. #19
    Originally Posted by redietz View Post
    No, arci, "probability" does not, in general common usage, necessarily refer to multiple events.
    Really? Could you provide an example. In every case I know it refers to the likeliness of a particular result given multiple possible results. In math the number is generally computed assuming an infinite number of possible results. The only difference I see is the math definition is more rigorous.

    Originally Posted by redietz View Post
    I'm sure you have checked the definition in the dictionary and realize this. The math-oriented definition of probability, which is the second or third definition, does. So the author of the dice question leaving the entirety of the rest of the question without a mathematical reference, and then sliding in the word "probability" at the end, does not necessarily put it in context, and does not necessarily refer to multiple events.

    Math guys are going to orient to the math-definition of "probability," but that is not the definition listed first in dictionaries. If you want to argue that dictionaries are incorrect, so be it.

    Everybody is on Alan's case for not admitting he is "wrong." Now let's see, after checking the dictionary, how many math guys are willing to say, "You know, I was wrong about the first and primary definition of probability."

    I have a feeling the math guys are going to say "I was wrong" in a fashion very similar to The Fonz (yeah, I'm dating myself).
    Every definition I saw talks about "chance". Given this is a problem which is clearly related to math and gambling there's little else it could mean other than the mathematical definition. Your approach is like complaining "what is 2+2?" should be considered a non math question because "plus" could have multiple meanings. Absurd.

  20. #20
    Originally Posted by Alan Mendelson View Post
    And what about the mathematicians who over-thought the question? As soon as I saw the Wizard video with double counting I knew they over-thought the original question or just ignored it. They were conditioned to look at probability formulas but failed to see the simple basic question.

    They could not see the forest for the trees.
    Alan, you are the only one "over thinking" this problem. While it is easy to miss the meaning when first read, it should be completely obvious after someone points out either die could be the 2.

    Alan, what is the probability of being dealt 4oak in VP? Does it have anything to do with a single deal?

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