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itachi-san

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Posts posted by itachi-san

  1. Just a little reinforcement on the 'eye' answer. For the second line, you would have to assume the person had both of their eyes. What if they got into one of those literal 'eye for an eye' ordeals? :)

  2. Nice try with 'an eye'

    The saying 'an eye for an eye' doesn't usually include the word 'take', and I'd say that it usually isn't kept. Usually isn't never though. i.e. sometimes what's taken in revenge can be kept, particularly if the people were actually taking each other's eyes out :o

    Eye works well with the second part though

  3. I did the opposite. I took a square and looked at the possibility of how many positions could land on it. I'm still thinking I'm right. Black plus white doesn't have to add up to 49 because the ruler is not the same as just picking squares.

  4. what does:

    multiple squares can connect to multiple black spaces

    mean?

    Oh i get it! Makes more sense if you say 1 square can connect to multiple black spaces

  5. #1 is 23/49 by the same reasoning:

    There are 12 white squares that have 2 possibilities (at the corners)

    There are 16 white squares that have 3 possibilities (by the centers of each boundary line)

    There are 5 white squares that have 4 possibilties (in the center)

    =24 + 48 + 20 = 92 possibilities

    divided by 196 (49 possibilities for the beginning of the ruler * 4 possibilities for the end)

    =92/196

    =23/49

  6. For #2, the probability of landing on a black square is 12/49.

    Reasoning: take each 4 square black section. 1 square (the outer-most) has 2 possibilities, 2 squares have 3 and 1 square (the inner-most) has 4.

    Added together, this makes 12 * 4 groups of black squares = 48.

    And the probability for the ruler is 1/49 for the first square * 1/4 for where it lands (only 4 possible places) = 196

    48/196 = 12/49

  7. I strongly disagree with 50% and the coin flip analogy. With this problem the tank needs to get across multiple bridges (at least 3) to get across.Therefore, you have to take all previously crossed bridges' probabilities into account. The shortest number of bridges the tank can cross is 3, but that doesn't mean it will cross 3. I'm not sure what the math would be to determine the number of bridges crossed, but the answer is at the most 1/2 * 1/2 * 1/2 = .125, though I'm certain the answer is lower because the amount of bridges to be crossed is higher than 3 since the original question doesn't ask for the least amount. Can anyone help on the math for that? (is it really 7-3=4? and if so, why?)

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