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  3. flamebirde

    Mining Gold

    whoops. In that case, I say refuse the offer. Assuming that the rest of the puzzle (i.e. the vehicle) is just story fluff and that the $1000 price tag of the rock is after processing, my answer remains the same. His share isn't even worth $100, let alone $200.
  4. bonanova

    Poisonous apples

    The OP does not say how many apples there are. It says the proportion that are poisonous. Question: was that the intent?
  5. rocdocmac

    Mining Gold

    A different approach ...
  6. rocdocmac

    Crypto-anagram

  7. Thalia

    Crypto-anagram

    Can you give an example of a one word policy? Assuming a male politician from the phrasing "his policies". Any one here more politically inclined?
  8. plasmid

    A Lovely Little Riddle

    I was thinking the same as Thalia. Let's see if this works. @plainglazed @Wilson @DudleyDude @fabpig Master Shakee hath returned.
  9. Skinnybitz

    Story riddle.

    A friend is trying to figure this out any help or ideas on how to tackle this would be great The time is NOW said Hugh Jackman: but was it: 6:00 am or 6:00 pm or was it 6:15 for all he knew it could have been 930, but probably closer to 6:15. Although, if it was 9:30, the 6:15 train would not arrive until 9:45 if it even came at all. The crazy 6:15 train would NOT arrive at 9:45! At 3:00 he noticed on the movie he was watching with Nicole Kidman, something about a witch casting a spell at a quarter to one. Or was it the Devil himself mister 666. That spell was cast as 12:45 the six stones were placed at the zero hour and Heath Ledger was trapped in a bubble. It took 1215 angels 645 hours to set him free! The Zero hour was the name of a play I directed with 6 wonderful actors. 3 of them were male and 9 of them were female. That does not add up. Sorry if I am confusing the issue, I am not sure if it was 6 after all. In the Zero hour 3 of the actors were male and 9 were female. I have done it again. Maybe there were more than 315 males .. there may have been 9 females though, yes of that I am certain!
  10. plasmid

    Poisonous apples

    I agree with Flamebirde's answer. But I would add a bit more to it.
  11. Donald Cartmill

    Poisonous apples

    the odds of eating a poison apple for the 1st bowl = 3/5 + 3/4 = 12/20 + 15/20 = 27/20 = !.35 chances of eating the P/A 2nd bowl =2/5 + 2/4= 2/3 = 24 /60 +30/60 + 40/60 = 84/60 = 1.4 CHANCES Therefore you would eat from the 1st bowl with a 0.05 better chance of avoiding the P?A
  12. Thalia

    Mining Gold

    fb- Joe is trying to sell you his share. You are paying him the $200... if you accept his offer. A question
  13. Last week
  14. BMAD

    Mining Gold

    Mining gold in a particular region is hard work. The metal only appears in 1% of rocks in the mine. But your friend Old Joe created a detector he’s been perfecting for months and it is finally ready. To your astonishment it always detect gold if gold is present. Otherwise it will have a 90% accuracy rate in detecting that a particular rock does not have gold. Working with Old Joe, You guys scan a large rock and determine that it gives a positive result. In loading it up, Old Joe realizes that both of you can't fit into the vehicle. He offers to sell his share to you for $200. You know that a rock of gold that size is worth easily $1000. Is that a fair price? Assume the vehicle remains with the proper owner.
  15. rocdocmac

    Crypto-anagram

  16. Thalia

    Crypto-anagram

    Half way?
  17. flamebirde

    Poisonous apples

    assuming five apples in each bowl,
  18. BMAD

    Poisonous apples

    There are two bowls that you and a challenger must eat from. After flipping a coin you were selected to pick the bowl that each would eat from. In the first bowl there are three out of five poisonous apples. In the second bowl, there are two out of five poisonous apples. Whoever eats from the first bowl must eat two apples at random from the bowl. Whoever eats from the second bowl must eat three random apples from the second bowl. Which bowl should you pick to eat?
  19. rocdocmac

    One Girl - One Boy

    Geez! ... How do these "fossil" questions keep surfacing?
  20. rocdocmac

    Sequence puzzle

    Deleted!
  21. kristy1992

    Sequence puzzle

    Thank you. He did figure it out after I posted.
  22. rocdocmac

    Calc-quickie

  23. Shakeepuddn

    The Executioner's Riddle

    Good guess . . . Right track. Go lower : )
  24. plasmid

    The Executioner's Riddle

    I'll count being in any way confused for Shakee as high praise. As for this one
  25. bonanova

    The triangle puzzle

    You've probably seen this puzzle. There are 15 holes in triangular array. (See sketch below.) The game begins with pegs in 14 of the holes. The play is to jump pegs over adjacent pegs, removing the "jumped" pegs afterward, as in checkers. The jump is made in a straight line. To make a jump, you need a contiguous group consisting of { peg1, peg2, hole } in a straight line. Peg1 ends up in the hole, and peg2 is removed. The object is to make 13 legal jumps and end up with a single peg. This happens about 6% of the time. That is, about 94% of the time you get a configuration, with more than one peg remaining, that permits no further legal jumps. In some games the peg must end up in the original empty hole, and that happens only about 3% of the time. So, it's not a trivial puzzle. This puzzle asks for something different, and easier: Lose as badly as possible. That is, select a location for the empty hole, and then find a sequence of moves that leaves the greatest number of pegs on the board where there are no more legal jumps. It's simple enough to play, even without the game, by marking hole locations on a sheet of paper and using pennies. As already stated, there are 15 holes. There are also 36 possible jumps. For convenience in writing sequences of jumps, they can be numbered, as follows: Number the jumps like this: and the holes ---------------------------> o ------------- 1 So Jump #1 means the / \ like this: peg in hole #1 jumps 1 2 ----------> 2 3 over the peg in hole #2 into the empty 4 5 6 hole #4. o o / \ / \ 7 8 9 10 Jump #18 is peg 7 3 4 5 6 over peg 8, into 7 13 11 12 13 14 15 hole 9. / \ o-8 o 14-o Holes 4, 6, 13 / \ / \ / \ begin 4 jumps; 9 10 11 12 15 16 the others 17 19 21 23 begin two. / / \ \ o-18 o-20 22-o 24-o There are 36 jumps. 25 27 29 30 33 35 / / \ / \ \ o-26 o-28 31-o-32 34-o 36-o With symmetries taken into account, the holes have four equivalence classes: Corners (1, 11, 15) Adjacent to corners (2, 3, 7, 10, 12, 14) Edge centers (4, 6, 13) Centers (5, 8, 9) This means that there are just four distinct places for the empty hole to start a game: { 1 2 4 5 }. All other holes are symmetrically equivalent to one of these. Just to be sure the numbering above is understood, here is a winning game of the normal type. Start with pegs in every hole except #1. (The top hole is empty.) Then make these jumps: { 7 14 2 17 23 27 34 26 30 6 35 14 7 }. If done correctly, the original hole #1 contains the final peg. Enjoy.
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