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Posts posted by HoustonHokie

  1. (1a) solved

    You could use as many blue lines as you want (if all of them are intersecting in black point and if some red points remain). One blue line seems most elegant to me. Just wandering, why you chose to use two blue lines.

    There's something about the dimensionality that suggests why it's impossible to paint with four colors. In the simple (elegant) solution, color 1 is painted 0-dimensionally, color 2 is painted with 1 dimension, color 3 has 2 dimensions. If you paint with a 4th color, it would require another dimension to keep from violating the 3-colored line rule. But since you're limited to a 2-D plane instead of 3-D space, you can't do it.

  2. i'd pick game 2.

    on your first play, you're at mod 3 = 1. So 90% to win and increase $1. Then mod 3 = 2. Again 90% to win and increase $1. Now mod 3 = 0 so 99% to lose and decrease $1.

    Working it out, the chances to lose 2 games in a row when mod 3 = 1 or 2 is 1% which is the same chance to win 1 game when mod 3 = 0. So it seems like you're likely to bounce around between $1,000,002 and $1,000,000. That doesn't seem too bad to me.

  3. the guys in the ambulance because you read the word "AMBULANCE" when you turned around. "AMBULANCE" is always mirrored on the front of the vehicle so you can read it in your rear view mirror.

  4. Did some drawing and I came up with 31.51%. No time to show reasoning now, but that's my best guess. Area looks like a square made with 4 arc segments, each radius 1, with centers on corners of the square.

  5. That reducing the step size to infinitessimal values for a constant size circle

    is the same as keeping the step size equal to unity and increasing the size of the circle without bound.

    Is the probability still 1, looking at it that way?

    For bonanova's question (either form), the probability is 0. As either the distance walked by the ant approaches 0 or the circle size approaches infinity, the probability of escape approaches 0. It's just not gonna happen...

  6. "volume" implies a third dimension which is not present in the OP. Being 2-dimensional elements, both have 0 volume.

    But, assuming you meant "area" instead, I'd have to agree with sks. Maximizing the perimeter of the two shapes is an interesting task.

    For the 6-gon, I think I'd use the 4 corners of the square, the place a 5th point inside the square very close to one of the corners, and the 6th point on the edge of the square very close to the diagonal opposite corner of the square from the 5th point. In this way, the perimeter would be very nearly 4 + 2√2, which is similar to rigandy's perimeter, but the area is much larger than his thin capital N. The area would be very nearly 1.

    For the 7-gon, I think I'd use the 4 corners of the square again and 2 points just inside 2 of the square's corners, diagonally opposite from each other. The 7tgh point would be on the perimeter of the square, very close to one of the other 2 corners. In this way, the perimeter maxes out at nearly 6 + √2. In so doing, the 7-gon's area is only about 0.5

  7. ...and they finish exactly when they began, at 8:00, having completed 0 handshakes.

    If no 2 people are exactly the same height, every possible handshake must contain a person who is shorter.

  8. hmmm...

    3.14159*(162.815763^2 -162.814995^2) = 3.14159*(26508.972681 -26508.722597) = 3.14159*(.250084) = .785661

    oh, yeah. need to check my formulas - I combined area and perimeter into a formula that doesn't give any correct answer...

  9. I get the radii of inscribed and circumscribed circles to be 162.814995 and 162.815763, respectively. Calculating the difference in areas, I get 1.5713, so yes, the area is greater than 1.

    edit: more decimals

  10. If you know which position you are in line, it's pretty easy to determine your probability of getting the right seat.

    Assume passengers are numbered 1 to N, with passenger 1 being the guy who lost his boarding pass.

    If you define P(n) as the probability that passenger n gets his correct seat, then:

    P(1) = 1 / N

    P(2) = (N - 1) / N

    P(3) = (N - 2) / (N - 1)


    P(N) = 1 / 2 (which was demonstrated in the previous puzzle).

    If you don't know which position you are in line, then your individual expectation of getting your correct seat is the average of P(2) -> P(N), which approaches 1 for large values of N. For N = 100, the expectation is 0.9577 for the random passenger. Call this value P(avg).

    The expected total number of passengers seated correctly is 1 / N + P(avg) * (N - 1). The 1 / N term describes the first passenger and the P(avg) * (N - 1) describes all the rest. For the case where there's 100 passengers, 94.82 are expected to be seated correctly. Again, as N gets large, the ratio of the expected number of passengers seated correctly to N approaches 1. Intuitively, this makes sense, because even though the first passenger is increasingly likely to take the wrong seat as N increases, he is also increasingly likely not to take "your" seat, especially if you're near the front of the line.

  11. How's this: Eagles are $15 a piece. Therefore, the dude bought: 3 Eagles for 45 dollars, 41 dollars (at $41) and 56 quarters for $14, totaling 100 coins for 100 dollars. How's that?

    The number of quarters (56) equals the number of dollars (41) plus an ODD multiple (5) of the number of Eagles (3). But the question said EVEN. Don't think that counts. If an ODD multiple is allowed, there are quite a few solutions possible...

  12. Unless I'm missing something, I get 2 solutions to the original problem:

    If we define the following:

    Q = # quarters

    D = # dollars

    E = # Eagles

    V = Value of an Eagle

    and write the solution as (Q,D,E,V) then my solutions are

    (80, 8, 12, 6) which is araver's solution and

    (80, 16, 4, 16)

    So Eagle could be worth $6 or $16

    For an odd value of V I get

    (64, 32, 4, 13)

    Eagle is worth $13

    edit: clarified target solution

  13. In order for a card to be the highest rank in the left pile, two things have to happen:

    1. The card has to be in the left pile

    2. All cards with a higher rank have to be in the right pile.

    So, assuming that cards are randomly ordered in the deck, the probability that a card is in the right pile is:

    p(R,right) = (R - 1) / 51 where R is the rank of the card in question

    p(R,left) = 1 - p(R,right)

    Then it's just a matter of combining probabilities of event occurring simultaneously.


    p(1,highest in the left) = p(1,left) * p(2,right) * p(3,right) * ... * p(50,right)

    p(2,highest in the left) = p(2,left) * p(3,right) * p(4,right) * ... * p(50,right)

    and so on. Similarly

    p(50,lowest in the right) = p(50,right) * p(49,left) * p(48,left) * ... * p(1,left)

    p(49,lowest in the right) = p(49,right) * p(48,left) * p(47,left) * ... * p(1,left)

    Tabulating those products, you will see that they reach their maximums at:

    p(8,lowest in the right) = 0.0892

    p(45,highest in the left) = 0.0892

  14. Actually, I don't think the strategy is flawed at all. I set up a simulation to run 10000 games with each strategy. With consistent $1 bets, you break even. With the doubled bets, you expect to win $0.50 every game. So after 10000 games, my simulation was up $5000 most of the time. All you need is enough seed money to cover a string of bad flips.

    A better strategy would be to triple your bet if you lose. Or go even higher... The results vary pretty widely, but after 10000 games using tripled bets, my simulations were up anywhere from $100,000 to $10,000,000!

    I don't look for this game in Vegas anytime soon. But let me know if you're going to start a casino with it - I'll come over and play a couple hundred times!

  15. that Davis did it. What if the 7th statement, when translated to truth, becomes "The killer did not arrive before I did." That leaves open the possibility that Davis could have been the killer (i.e., the killer arrived at the same time that Davis did).

  16. I'm guessing that you noticed that the value written on the board was even. If so, there's no way the answer could be out by one. It could have been out by 2, 4, 6, etc. but could not have been out by 1.

    Since you were in computer science class, the value of 2^32 was probably written in binary instead of decimal. So with your quick glance you counted the number of 0s written on the board. Confirming that there were, in fact, 32 0s, you knew that the correct representation 2^32 was on the board.

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