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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    11 Oct '11 11:12
    In a game the goal is to throw a six-sided die until you meet or exceed a given figure. If you meet it exactly, you win; if you go over, you lose. You can pick any number you want as long as it's reachable by two or more rolls, and not reachable with one, say, 50, and you might roll
    1,4,4,4,1,2,3,6,3,6,5,6,1,5.. hold your breath as that makes 47.. and then toss a 4 and go one over and lose.

    Questions -
    * what are the chances of meeting some fairly big number exactly?
    * for which number are the odds best?
    * for which number are the odds worst?
  2. 13 Oct '11 10:20
    I haven't fully tested this but I reckon the actual number doesn't matter for any number greater than 6.

    I believe the probability of landing on exactly any chosen number greater than 6 using a 6-sided is 24.4667%.

    I haven't provided my workings on purpose so others can have a go at solving this.

    Hidden clue:
    It involves using the numbers 1, 3, 6, 10 & 15. The answer can be expressed mathematically. Intuition says the odds should be 1/6 but the starting point relative to the chosen number is crucial.
  3. 13 Oct '11 10:48 / 8 edits
    Probability for a 6 is higher:

    P(6) =
    1/6 + (Throwing a 6 first)
    1/6^2 + (Throwing a 5 first)
    1/6^2 + 1/6^3 (Throwing a 4 first)
    1/6^2 + 2/6^3 + 1/6^4 (Throwing a 3 first)
    1/6^2 + 3/6^3 + 3/6^4 + 1/6^5 (Throwing a 2 first)
    1/6^2 + 4/6^3 + 6/6^4 + 4/6^5 + 1/6^6 (Throwing a 1 first)
    (I think)
    which is roughly 36%

    and I'd say 6 is the best number and 1 is the worst.

    Edit: I agree, for n>6 P(n) should have the same probability.

    Edit: Probability for P(n) (n>6) should be (P(1) + ... + P(6)) /6

    Hence why P(6) is the best and P(1) is the worst.
  4. 13 Oct '11 10:59 / 3 edits
    I've just been thinking about the best number to pick to try to land on. I reckon this could be any integer that can be expressed in the form 3.5y + 2.5, where y is any odd number greater than 2. If the average value thrown is 3.5 then we want to start with a remainder of 6 to give the greatest probability of landing exactly on the number.

    Edit : have you mapped out a probability tree? Your number seems a little high. I solved this for 2, 3 & 4 sided die before I came up with a general equation that I tested on 5 and 6. The proof (for me) was calculating the probability of failing using a probability tree.

    Edit 2 : I agree P(6) > P(1) hence the 3.5y+2.5 method above.
  5. 13 Oct '11 11:16 / 2 edits
    Originally posted by andrew93
    I've just been thinking about the best number to pick to try to land on. I reckon this could be any integer that can be expressed in the form 3.5y + 2.5, where y is any odd number greater than 2. If the average value thrown is 3.5 then we want to start with a remainder of 6 to give the greatest probability of landing exactly on the number.

    Edit : have ...[text shortened]... ling using a probability tree.

    Edit 2 : I agree P(6) > P(1) hence the 3.5y+2.5 method above.
    I kind of used brute force and a few scribbles of a probability tree and stumbled onto to the pattern. The probability was higher than I expected, but I don't see how its wrong.

    Edit: Proving my conjecture for a two sided dice:
    P(1) = 1/2
    P(2) = 1/2+1/4
    P(n>2) = 1/2(P(1) + P(2)) depending whether or not we land 1 or 2 away from n.
  6. Standard member talzamir
    Art, not a Toil
    13 Oct '11 11:56 / 4 edits
    Indeed, P(1)=1/6 is the lowest, and P(6)=16807/46656 (or 36% ) the highest overall, and I even agree on the iterative format, which I expanded into

    P(n) = 0 for n= {-5, -4, -3, -2, -1}
    P(0) = 1
    P(n) = (P(n-6)+P(n-5)+P(n-4)+P(n-3)+P(n-2)+P(n-1))/6 for all n > 0.

    I wonder if there is a way to express that in a non-iterative form.. probably is as that can be done for the Fibonacci sequence, with square roots of five and stuff.. but I don't know what it is. I did however look into the probabilities, and they are not all equal.
  7. 13 Oct '11 12:32 / 1 edit
    Originally posted by talzamir
    Indeed, P(1)=1/6 is the lowest, and P(6)=16807/46656 (or 36% ) the highest overall, and I even agree on the iterative format, which I expanded into

    P(n) = 0 for n= {-5, -4, -3, -2, -1}
    P(0) = 1
    P(n) = (P(n-6)+P(n-5)+P(n-4)+P(n-3)+P(n-2)+P(n-1))/6 for all n > 0.

    I wonder if there is a way to express that in a non-iterative form.. probably is as that ...[text shortened]... I don't know what it is. I did however look into the probabilities, and they are not all equal.
    Yeah, good point, I was over simplifying. Forgot to take into account that the odds of landing on 1 through 6 at some point where all different (ironic considering that was exactly what I just proved) so P(12) > P(11) > P(10) ... > P(7) (< P(6) ) I think.
  8. 14 Oct '11 05:48
    Originally posted by VelvetEars
    Edit: Proving my conjecture for a two sided dice:
    P(1) = 1/2
    P(2) = 1/2+1/4
    P(n>2) = 1/2(P(1) + P(2)) depending whether or not we land 1 or 2 away from n.[/b]
    This was how I worked it out for a 2-sided dice:

    We have a 50% probability (P) of landing 1 away from our desired number (n).
    Deal with landing 1 away first:
    1. P(n) = 50% (for landing 1 away) x 50% (to throw a 1) = 25%
    2. P(n+1) = 50% x 50% (to throw a 2) = 25%

    We also have a 50% probability of landing 2 away from n.
    The probabilities are:
    3. P(n) = 50% (for landing 2 away) x 50% (to throw a 2) = 25%
    4. P(n-1) = 50% x 50% (to throw a 2) = 25%. {this result is ignored and re-tested below}

    Given a result of n-1 as above we must throw again, and we have the following probabilities:
    5. P(n) = 50% (for landing 2 away) x 50% (for throwing a 1) x 50% (to throw a 1) = 12.5%
    6. P(n+1) = 50% x 50% x 50% (to throw a 2) = 12.5%

    P(n) = sum of parts 1 + 3 + 5 = 62.5%
    P(n+1) = sum of parts 2 + 6 = 37.5%
    Sum(P(n), P(n+1)) = 100%

    Note part 4 is not counted given it is not a final solution. But the sum of parts 1 to 4 is 100% (being the first 4 possible outcomes). Parts 5 and 6 are re-allocating the 25% that was not yet finished per part 4.

    This can be expressed as a non-iterative formula for a 6-sided dice (see my next post).
  9. 14 Oct '11 06:09 / 2 edits
    Originally posted by talzamir
    Indeed, P(1)=1/6 is the lowest, and P(6)=16807/46656 (or 36% ) the highest overall, and I even agree on the iterative format, which I expanded into

    P(n) = 0 for n= {-5, -4, -3, -2, -1}
    P(0) = 1
    P(n) = (P(n-6)+P(n-5)+P(n-4)+P(n-3)+P(n-2)+P(n-1))/6 for all n > 0.

    I wonder if there is a way to express that in a non-iterative form.. probably is as that ...[text shortened]... I don't know what it is. I did however look into the probabilities, and they are not all equal.
    I believe the mathematical formula for the 6-sided dice is the sum of:

    1) 6*(1/6)^2
    2) 15*(1/6)^3
    3) 10*(1/6)^4
    4) 6*(1/6)^5
    5) 3*(1/6)^6
    6) 1*(1/6)^7

    I worked this method out with a probability tree for 3, 4 and 5 sided dice until I saw the pattern : (ignoring the first line) notice the pattern of the sum of the digits starting from the bottom.

    The first term needs explaining - you have a 1/6 chance of landing y away from n, and then a 1/6 chance of throwing exactly y to land on n. This is essentially the same as claiming the first 1/6 probability no matter your starting point, i.e. 6*(1/6)^2 = 1/6.

    The power represents the number of throws needed to land exactly on n. Assuming 6 different possible starting points I believe the number of potential outcomes to test is 6^7 not 6^6.

    Regarding the last term : this represents the probability of starting 6 away from n and then throwing 6 consecutive 1's to land exactly on n. 6 throws with P= 1/6 multiplied by 1/6 probability of starting 6 away from n gives the formula (1/6)^7.

    The key to working this out is to work out the probability of starting 2 away from n, and then throwing two 1's. This is (1/6)^3. Notice this term is found 15 times in the solution, representing the 5 times it is used for starting 6 away, 4 times for starting 5 away, 3 for 4, 2 for 3 and 1 for 2 - the sum of which is 15. Maybe I haven't explained myself well but if I post a picture it makes sense (to me anyway!).

    Andrew
  10. 15 Oct '11 00:11
    My apologies - when I look back at VelvetEars post I misunderstood your equation. I see now you were giving a formula for P(6) - I had misinterpreted this as P(1-6) hence my comment regarding it being a bit high. I think we are both essentially saying the same thing although I was providing a solution irrespective of the starting point.