Oxford PAT 2011, Question 11

2011_page8

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12 thoughts on “Oxford PAT 2011, Question 11

    1. Hi Alex. Good question — the answer is that when A=3 and B=4 not all the numbers between 0 and 35 come up with equal probability. Look at Alice’s answer above and my comment in response, which deals with this in more detail.

  1. Hi, I don’t think I quite understand your reasoning for this sentence:

    “Because the dice are digits in base six the total number is 6 times the first digit plus 1 times the second digit”

    The way I keep trying to look at it, you appear to be treating each number as a digit in a base-6 multiple-digit number – but the question uses a formula which simply sums their displayed numbers?

    Using base 6 makes sense but I just can’t get my head around its connection to the question.

    I appreciate the help!
    Best, Stephanie

    1. Hi Stephanie. Yes your analysis of what I am saying is right. Some ways of thinking about things work for some people and not others — this way works for me but it’s not the only way of thinking about the problem. So why not consider Alice’s solution, in the comments above, which is absolutely 100% right and doesn’t use the ‘base 6’ parallel.

    1. Hi Esther. I don’t think the text says “base digit 6”; you mean “digits in base six”? What I mean is that you can consider a die to represent a digit between 0 and 5 (where 1 dot represents the digit ‘0’, 2 dots represent the digit ‘1’, and so on, until 6 dots represent the digit ‘5’). If you line several dice up next to each other, you can represent a number in base 6, where each die is one of the digits of the number. So for example, if you throw a five in the big digit, and a two in the small digit, the digits are (5-1) and (2-1), so the number is the base six number 41, which in base 10 is 25.

  2. Sure. Here’s a more detailed explanation.

    The key point here is that the two dice need to be mapped on to the numbers from 0 to 35. This works when we consider the two dice as being digits in base six, but where each digit is a number one less than the face value of each die — so 1 on the die represents the digit 0, and d on the die represents the digit d-1, all the way up to 6 representing the digit 6-1, which is 5. Because the dice are digits in base six the total number is 6 times the first digit plus 1 times the second digit: so double 1 on the dice represents the two digits 0, which together represent the number 6×0 + 1×0, which is 0; double 6 on the dice represents the two digits 5, which together represent the number 6×5 + 1×5, which is 35.

    From this reasoning we get the equation S = 6(d1-1) + d2 – 1, and this simplifies to be S = 6d1 + d2 – 7. We have been told that the formula is S = Ad1 + Bd2 + C, so now we can just read off the values A, B and C: A = 6, B = 1, and C = -7.

    1. Hi!

      I might be wrong, but isn’t there a much simpler way to do this question? Can’t you just say 6A + 6B + C = 35 and A + B + C = 0. Then you’d rearrange that; A + B = (35 – C)/6 , A + B = – C, so 35 – C = – 6C, so C = -7. Then you have A + B = 7, but for the equal probability of each score you’d know A or B = 1, and B or A = 6?

      (Thank you so much for this website by the way! It’s massively helpful 🙂 )

      1. Hi Alice, thanks for your comment.

        I think your answer is a good one. You also make a good point that A, B could be 1, 6 or 6, 1 — in the terms of my answer that would correspond to the higher-order digit being d1 and d2 respectively. So you have found two correct solutions.

        Is your answer simpler than my answer? Well, I think your answer as you’ve written it could be accused of glossing over the details of why “for the equal probability of each score you’d know A or B = 1”. If you went into details about this you could either say “because when A, B = 1, 6 or 6, 1 you have a two digit counter in base 6” (which is basically what I have said), or you could show that there are numbers in the range that you can’t get with a combination other than 1, 6 or 6, 1 (e.g. you can’t get Ad1 + Bd2 = 8 with any other combination, and so you would never be able to get Ad1 + Bd2 – 7 = 1). If you did this I think the complexity of the answers would be about the same.

        I’m glad the website has been useful, and best of luck in the exam.

    2. Some background info which I found incredibly helpful:
      http://mathforum.org/library/drmath/view/55739.html

      To add on, I think how base 6 helps is this: we want the sum of dice to span from 0,1,2… to 35, but alas we only have 6 numbers to work with: 1-6. We also want each combo to give us a different number, for an equal probability of each result.
      (e.g. d1 = 2 and d2 = 3 can’t give us the same S value as d1 = 3 and d2 = 2)

      Base 6 helps us “map” each result to a unique number from 0-35. It expresses 0-35 using only 6 digits instead of the full 10.

      Base 6 calculations look like this: S = 6 x d1 + 1 x d2
      Where S is the unique number we want to know, and d1 and d2 are the dice.
      The question is simply asking “What is the sum of the dice if we expressed it in S = Ad1 + Bd2 + C form?”

  3. Please could you explain your working? Specifically how you get the values of A and B. I managed to find C, but struggling to understand how you can be sure about the other two numbers. Thanks

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