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  1. Yesterday
  2. Waiting, again II

    much more clearly stated than my babbling. I’m tickled that I’ve shown that it can be evaluated one flip at a time, based merely on the parity of contiguous Hs. here is my argument, expressed by plagiarizing your expression: Let e be the expected number of flips from the initial (even) state There are two states that are easy to analyze and cover all the possibilities: E is the initial state, it also represents the state of having seen an even number of H (including zero), since the beginning or the most recent T. O is the odd state, representing the fact of having seen an odd contiguous run of H. State E requires e more flips. In this state, H changes to state O, while T remains in state E State O requires o more flips. In this state, H changes to state E, while T terminates with a win. That allows us to write an expression for x as the sum of these terms, weighted by their respective probabilities, all 1/2. e = 1/2{1+e} + 1/2{1+o} o = 1/2{1+e} + 1/2 substitute o into e e = 1/2{1+e} + 1/2{1+1/2{1+e} + 1/2} = 1/2{1+e} + 1/2+1/4+e/4 + 1/4 = 3/2 + 3e/4 e/4 = 3/2 e = 6
  3. Born on a Wednesday

    (1) (2)
  4. Cubicle Stack #2

    Lol. I'm not sure what my score is anymore. Can you score this?
  5. Born on a Wednesday

    The probability that any one person selected at random was born on a Wednesday, is 0.1429 (~14 %). What is the probability that of any … (1) Seven persons chosen at random, exactly one was born on a Friday? (2) Five persons chosen at random, three were born on a Sunday? [Mr Moderator, if this question has appeared before, please remove it!]
  6. 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.
  7. Jelly beans join the clean plate club

    I can show you by counterexamples that they don't always land at the average (if, for example, their average is odd [say 4, 6], you'll never be able to double to the average). That's why I considered evaluating the difference between two as being divisible by 4 (instead of 2). Perhaps you'll succeed where I have failed.
  8. Jelly beans join the clean plate club

    I think I've gotten closer.
  9. Last week
  10. Cubicle Stack #2

    By the way, Thalia ...
  11. 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?
  12. 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?
  13. Waiting, again

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

    Yes, there can be a "bonus" row that contains 4 trees. Here's an adequate proof of the killer:
  15. Cubicle Stack #2

    Nevermind the CEX numbers. I really need to find a cube I can hold...
  16. Cubicle Stack #2

    I see what you mean about CEX although I'm not sure how to show that with the numbering system. That throws a wrench in my counting. . .
  17. Jelly beans join the clean plate club

    Aw man, I was so confident I'd finally solved one of Bonanova's legendary puzzles. A start on thinking:
  18. Largest unit sphere

    Nice, thanks!
  19. Cubicle Stack #2

    So 6/7 in the last post and check CEX?
  20. Waiting, again

    Gardner sets high standard in many ways. I was a child reading Childrens Activities and a few years later I was enjoying hexaflexagons and later mathematical games. I was kneeling behind you in worship. i enjoy the puzzles here, and sometimes I don’t understand something that is obvious to anyone else. I think I may have a touch of ambiguity flu. Keep on puzzling, Bonanova!
  21. Whodunit?

    My guess:
  22. 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.
  23. 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.
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