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Everything posted by bonanova

  1. I'm not understanding something about two shifts building exactly one car. A whole car (in one shift) or two half-cars (in two shifts) seem to be the only cases. Maybe spoiler one other possibility as a means of explaining? Thanks.
  2. It must be symmetric about the NW-SE diagonal, so your figure show all the cases you computed. Nice, btw. Hint
  3. @plasmid Does the program imply a proof that it can always be done? Or is it a statement that no counterexample has yet been found? A proof could be a repeated procedure which after each application reduces the smallest number of beans on a plate. Does your algorithm always reduce the number on place C?
  4. @ThunderCloud Nailed it. @Izzy Honorable mention
  5. I think it boils down to that, but how to justify doing it?
  6. @Izzy Very close. @Donald Cartmill OP restricts what each prisoner is allowed to say: Each prisoner must guess the color of his own hat, without having seen it, by saying one of the three colors
  7. @Izzy As you say, the distribution is surprising. To be certain of this expected attendance at the smaller party, you might want to ...
  8. No spoiler needed: There are only three no-win-after-13-digs possibilities: BBBB BBB GGGG RR Win on dig 14 with G or R or H = 60%. Win on dig 15 with B = 40% BBBB BBBB GGG RR Win on dig 14 with R or H = 30%. Win on dig 15 with B or G = 70% BBBB BBBB GGGG R Win on dig 14 only with H = 10%. Win on dig 15 with B or G or R = 90%. Their relative occurrences were not saved in the simulations, so it's a bit uncertain how weight these cases when taking an average. But if they are equally likely, the relative dig-14 and dig-15 wins would be exactly 33.33...% and 66.66...% That is, Win 15 would be twice as likely as Win 14. For the simulations, the last few probability estimates are the least precise, because they are averages of fewer cases. In particular, the probability of going beyond 13 digs is only 5.6%, so that out of 2 million total cases, only about 112,000 14-digs or 15-digs cases were averaged. Those relative win probabilities are 35.7% and 64.3% respectively. I'll point out that the proportions of needed B G R and H (8 4 2 1) are similar to their occurring probabilities ( 40% 30% 20% 10% ). That partially justifies an equal-likelihood assumption. But the proportions do differ, somewhat. In particular, B is needed 8/15 of the time but occurs only 40/100 = 4/10 = 6/16 of the time. This fact might well make a missing-B-after-13-moves (Case 1) the most likely case of the three. That case has the highest win-14 probability. So we might expect an upward bias on the win-14 probability. The simulation suggests that is the case.
  9. One hundred prisoners stand in a straight line seeing those visible to them only from the back. You get the picture, back guy sees 99 others, front guy sees no one. They are fitted, one each, with a hat, whose color is uniformly randomly Red, White or Blue. Each prisoner must guess the color of his own hat, without having seen it, by saying one of the three colors, and he is executed if he is wrong. The guesses are made sequentially, from the back of the line to the front. The guesses are not identified as to their accuracy, and no prisoners are executed, until all 100 guesses are made. The prisoners may collaborate on a strategy, with the object of guaranteeing as many survivors as possible. (Their communication ends, of course, once the first hat is placed.) How many can be saved, in the worst case?
  10. Assuming your interest is in Method 1:
  11. My understanding of the puzzle If that's all true, then Q1: Q2: Q3:
  12. @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.
  13. 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?
  14. Peter and Paul, who are neighbors, each threw a party last Friday. Bad scheduling, to be sure, but that's life. Even worse, their guest lists were identical: all 100 of their friends were sent invitations to both parties. When guests arrived, the happy sounds of those already present could be heard through the two open doors, and the old phrase "the more the merrier" figured in their choice of which party to attend: If at any point there were a people present at Peter's party and b people present at Paul's party, the next guest would join Peter with probability a/(a+b) and join Paul with probability b/(a+b). To illustrate: When the first guest arrived only the two hosts were present. (a = b =1.) So that choice was a tossup, and let's say that the first guest chose Peter's party. (a = 2; b =1.) Now the second guest would follow suit, with probability 2/3, or choose Paul's party, with probability 1/3. And so on, until all 100 guests arrived. What is the expected number of guests at the less-attended party?
  15. And Children's Activities had some cool features on the last page - cartoon, riddle or puzzle - as I recall.
  16. Yes, there can be a "bonus" row that contains 4 trees. Here's an adequate proof of the killer:
  17. @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.
  18. 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.
  19. 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.
  20. On a table are three plates, containing a, b and c jelly beans, in some order, where a < b < c. At any time you may double the number of jelly beans on a plate, by transferring beans to it from a fuller (or equally populated) plate. After one such move, for example, the plates could have 2a, b-a, and c beans. Using a series of these moves, Is it possible to remove all the jelly beans from one of the plates?
  21. 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.
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