1. Joined
    04 Aug '01
    Moves
    2408
    14 Jun '06 19:23
    Suppose we have a triangle with vertices A,B,C and a slug that travels around its perimeter. Whenever the slug reaches a vertex, it has a 1/2 chance of moving toward each of the remaining two vertices.

    Suppose the slug starts out at vertex A. After the slug has moved 10 triangle side lengths, what is the probability that the slug will find himself back at point A? After moving 20 side lengths, what is the probability that he will be at point C?
  2. Joined
    28 Nov '05
    Moves
    24334
    14 Jun '06 21:142 edits
    after 10 lengths I make it 171/512

    after 20 lengths I make it 174763/524288
  3. Joined
    04 Aug '01
    Moves
    2408
    14 Jun '06 22:14
    Originally posted by aging blitzer
    after 10 lengths I make it 171/512

    after 20 lengths I make it 174763/524288
    I agree with your first answer -- well done.

    For the second question, I got something very slightly different. I get that your answer would be the probability for the slug to end up at A after 20 moves -- not C. I could have made a mistake, but notice that your second answer is greater than 1/3, which at first glance should be a strike against its being correct.
  4. Joined
    28 Nov '05
    Moves
    24334
    15 Jun '06 07:261 edit
    as you surmised, I didn't read it right.

    to be at C after 20
    349525/1048576

    edit:
    either 1 or 2 points will have a greater than 1/3 probability.
    after an odd number of moves A < 1/3, B > 1/3, C > 1/3
    after an even number of moves A > 1/3, B < 1/3, C < 1/3
    so, yes, after 20 the probability of C should be < 1/3
  5. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
    Moves
    20443
    15 Jun '06 19:34
    Originally posted by aging blitzer
    as you surmised, I didn't read it right.

    to be at C after 20
    349525/1048576

    edit:
    either 1 or 2 points will have a greater than 1/3 probability.
    after an odd number of moves A < 1/3, B > 1/3, C > 1/3
    after an even number of moves A > 1/3, B < 1/3, C < 1/3
    so, yes, after 20 the probability of C should be < 1/3
    Let An, Bn, and Cn denote the chance of being in A, B and C respectively after n moves.

    A0 = 1, B0 = C0 = 0

    An = B(n-1)

    Bn = 1/2 * (1 - An)

    Cn = Bn

    I currently have no program to check the answers given 🙂
  6. Joined
    04 Aug '01
    Moves
    2408
    15 Jun '06 20:49
    Originally posted by aging blitzer
    as you surmised, I didn't read it right.

    to be at C after 20
    349525/1048576

    edit:
    either 1 or 2 points will have a greater than 1/3 probability.
    after an odd number of moves A < 1/3, B > 1/3, C > 1/3
    after an even number of moves A > 1/3, B < 1/3, C < 1/3
    so, yes, after 20 the probability of C should be < 1/3
    Well done.
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