Jump to content
BrainDen.com - Brain Teasers


  • Content count

  • Joined

  • Last visited

  • Days Won


Everything posted by bonanova

  1. Digging Probabilities

    My understanding of the puzzle If that's all true, then Q1: Q2: Q3:
  2. Waiting, again II

    I find probability questions interesting, because they often defy intuition. Particularly for me are those that involve waiting times. Other than the basic idea of an event of probability p needing on average 1/p trials to occur. But here's one not that trivial, yet still fairly easy to solve -- with the right approach. On average, how many times do you need to flip a fair coin before you have seen a (continuous) run of an odd number of heads followed by a tail? For example, T T H H H H T H H H T took 11 flips.
  3. Waiting, again II

    @CaptainEd - Nice solve. Here is a solution i was aware of. I think the two are similar or equivalent, parsed out into a different set of states.
  4. Waiting, again II

    Not sure if it was clear that the run of odd number of heads are contiguous, as in the example, or if I'm misunderstanding your algorithm. Can you add some words here and there?
  5. Waiting, again

    You're at a carnival and two people offer you money for getting six fair dice collectively to show all six numbers. Each charges you $1 per roll. Peter pays you $20 when you succeed, while Paul will pay you $50. There's another difference. Paul lets you roll all six dice each time, but Peter makes you roll just one die each time. Do you stop and play? If so, with whom?
  6. Waiting, again

    And Children's Activities had some cool features on the last page - cartoon, riddle or puzzle - as I recall.
  7. Whodunit?

    Yes, there can be a "bonus" row that contains 4 trees. Here's an adequate proof of the killer:
  8. Largest unit sphere

    This puzzle is inspired by the seventh of Rocdocmac's Difficult Sequences: In n dimensions, where a point is an n-tuple of coordinate values { x1, x2, x3, ... , ,xn }, the unit sphere is the locus of points for which x12+ x22 + x32 + ... + xn2 = 1. In one dimension this is a pair of points that encloses a volume (length) of 2 and comprises a surface area of 0. In two dimensions it's a circle that encloses a volume (area) of pi and comprises a surface area (circumference) of 2 pi. So the enclosed volume and surface area both start out, at least, as increasing functions of n. What happens as n continues to increase? Puzzle: Is there a value of n for which the volume reaches a maximum? Bonus question: What about the surface area?
  9. Waiting, again

    @CaptainEd - OMG no. Awhile ago I next-to-worshiped Martin Gardner (who wrote the math games column in Sci American for so many years) because he worded his puzzles perfectly, simply and clearly. His, unlike mine, (try tho I may) never needed editing. When I wrap prose around mine to make them perhaps interesting or, sometimes, to camouflage the solution, stuff gets added that has often has to be clarified later. My bad on this one.
  10. Whodunit?

    Clarification: Dick asserts that he had been out running, and that one of his three brothers has just lied. Inspector just called in and needs a final answer ... Fame awaits the brave.
  11. Jelly beans join the clean plate club

    Sorry guys, "fuller" should have read "at least as full." Examples always help, so here is an example. a b c 7 9 12 <- 14 2 12 -----> 2 2 24 -> 0 4 24 So { 7 9 12 } is a starting point where a plate can be emptied. Can any { a < b < c } lead to an empty plate? A Yes answer needs proof; a No answer just needs a counter example.
  12. Whodunit?

  13. Whodunit?

  14. Hurry up and Wait

    A discrete event (like rolling a fair die and wanting a 3 to appear) has a probability p of success (1/6 in this case.) The first roll is likely to fail, so let's keep rolling the die until we do get a 3, Then stop and write down the number of rolls that it took. Let's repeat the experiment a large number of times, each time recording the required number of rolls. So we have a bunch of 1s (the number of times 3 appeared on the first roll,) 2s (the number of times a 3 appeared on the second roll,) and so forth. What number will most appear most often?
  15. Whodunit?

  16. Waiting, again

    There are six dice. They are marked A. B, C, D, E and F in some order. Initially no dice are on the table. Peter's game: You pay $1 and roll die A or die B or die C or die D or die E or die F. You get to choose which die to roll. Paul's game: You pay $1 and roll die A and die B and die C and die D and die E and die F. You must roll all of them. Winning condition: All dice are on the table and collectively they show 1, 2, 3, 4, 5 and 6 dots.
  17. Whodunit?

    To clarify: (and I'll add it to the OP) Dick: I went out and ran 3 miles in the woods, and I've figured out that one of my 3 (living) brothers is lying.
  18. Whodunit?

    @Molly Mae Bravo. I should give you a solve (and a gold star) for this answer. But the Inspector has a reputation to uphold -- he needs a conviction -- and he did appeal to us for help. So, let me repair my flawed puzzle by adding this phrase about the brothers (and I'll add it to the OP as well.) None of them lied and told the truth in a single day.
  19. Cubicle Stack

    OK, I agree with 22. Nice work. Where my thinking was wrong - I considered left- and right-hand knight moves to be in the same class. (I put all mirror images into the same class.) That's wrong, because mirror image solids (if they lack further symmetry) can in fact be distinct. Three small cubes? Maybe look at it at some point. rodomac's images provide an advanced start point. That said, it still means finding distinct ways to remove C, E, F and (possibly) B small blocks from 22 different images.
  20. Cubicle Stack

    Here's my argument for 20 distinct shapes. First, I'm in awe of rocdocmac's images! The meager sketch below indicates only the visible small cubes. It does not show the small cube at the center, which I refer to below as B. The ones that do show are labeled as Corner (C) Edge (E) Face (F) We agree that C, E, F and B are the four distinct classes of small cubes, so that removing just one small cube gives rise to four distinct shapes. C / \ E E / \ C F C | \ / | | E E | | \ / | E C E | | | | F | F | | | | C E C \ | / E | E \ | / C And now we're removing two small cubes and identifying the equivalence classes. If we first remove a C, we can remove another C three distinct ways we can remove a E three distinct ways we can remove a F two distinct ways we can remove a B one distinct way If we first remove a E, having already counted the EC case, we can remove another E four distinct ways we can remove a F three distinct ways we can remove a B one distinct way If we first remove a F, having already counted the FE and FC cases, we can remove another F two distinct ways we can remove a B one distinct way There is no BB case, so we're done. And the total is 20. In summary, CC 3 CE 3 EE 4 CF 2 EF 3 FF 2 CB 1 EB 1 FB 1 BB 0 The cases that seem to disagree both involve Edge-cube cases. Namely, Corner-Edge (CE). Some say 4, I say 3. Edge-Edge (EE). Some say 5, I say 4. Here are my enumerations: CE - having removed a Corner, what classes of Edge faces remain? (I claim three.) Three small cubes E that touch C. Six small cubes E that do not touch C, but do lie on the same big-cube face. (Think of a chess knight move.) Three small cubes E that touch C'. C' is the small cube diagonally opposite C. EE - having removed a E, what classes of other Es remain? (I claim four.) Four small cubes E that touch the first E at one of its corners. Four small cubes E that touch E' at one of its corners. E' is the small cube diagonally opposite the first E. Two small cubes E each of which, along with the first E, surround and touch a common F. The (one) final small cube, E'. Again, E-E' passes through B. My class descriptions exhaust the 12 (CE) and other-11 (EE) Edge small cubes are stated in a way intended to suggest (at least) that the classes are homogeneous. I'm eager to hear other class descriptions for these cases.
  21. Whodunit?

    In my post I meant to assert the Inspector tried but failed to make the sketch following the method suggested by rocdocmac. That is, the antecedent of "He" was meant to be "the Inspector." Bert, of course, had already succeded.
  22. Two can tango

    Two pennies can be placed on a table in such a way that every penny on the table touches (tangos with) exactly one other penny. Three pennies can be placed on a table in such a way that every penny on the table touches exactly two other pennies. What is the smallest number of pennies that can be placed on a table in such a way that every penny on the table touches exactly three other pennies? (All pennies lie flat on the table and tango with each other only at their edges.)
  23. Whodunit?

  24. Whodunit?

    Regarding (2), can you make a sketch for the Inspector? "Absolute truth-teller" includes "without mistakes," and he did say random. Remember also that ...
  25. Line splitting

    Draw a line segment, and place a point on it anywhere you like. Now place another point, but be sure the two points are in opposite halves of the line segment. Continue by adding a third point, but make sure they all are in different thirds of the line segment. And so on. After placing eight points you'd have something that looks like this. Here are eight points, each in a different eighth of the segment. But in what order were they placed, and in what part of the 8 intervals must they lie in order for each successive group of points to have comprised a legal arrangement? It may seem a trivial task: it's certainly trivial to describe. Have a go at it, with some graph paper or a calculator, and see how many points you can place. Let's standardize the problem as follows: Define the line segment as the number interval [0, 1000], and give your answers as integers. For example, for the pretty easy case of four points you might get {200, 800, 450, 700}. And so on. Have fun.