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Everything posted by flamebirde

  1. Nope. Look more at the first line: what's the "mother's blessing"? I think Molly understood that.
  2. As pure in shade as mother's blessing true, And yet in form I'm still possessing, too; Partner both in math and drinking I was born of man's quick thinking
  3. Something that should be odd, but in this case is even, which makes it a symbol of good luck. hmm...
  4. I'm very much interested. I recognize the name, too! I expect great things, my friend.
  5. Question 1: Questions 2 and 3 are correct! Followup question: What are the chances I get the health from the 14th dig versus the 15th dig? Do they differ significantly?
  6. So the other day I was watching a speedrun of the Legend of Zelda: Ocarina of Time (a speedrun is a playthrough of a game with the intent of beating the game as fast as possible). In one particular part of the game, the player is forced to follow around a gravedigger as he digs up various holes. There is one particular outcome that is desired (the "jackpot" of the game, essentially): a permanent health upgrade. There are also three undesirable outcomes that only give out money: a green rupee (the least valuable prize, pretty much $1), a blue rupee (a fairly desirable prize, say about $5), and a red rupee (a very desirable prize, say $20). Here are the rules: The chances of digging up a green rupee is 40%, a blue rupee 30%, a red rupee 20%, and the health upgrade 10%. If the gravedigger has dug up eight green rupees already, and he would dig up a ninth green rupee this time around, he will instead dig up the health upgrade. If the gravedigger has dug up four blue rupees already and he would dig up a fifth this time around, he will instead dig up a green rupee. If the gravedigger has dug up two red rupees already and he would dig up a third this time around, he will instead dig up a blue rupee. As a result, the maximum number of attempts to dig up the health upgrade is fifteen. Example: four blue rupees and two red rupees have been dug up. If the gravedigger hit the 20% chance to dig up a red rupee on his seventh total attempt, he would instead dig up a blue rupee. However, four blue rupees have already been dug up, so he would actually dig up a green rupee. After digging up the health upgrade, the game is over. Three questions: One: what is the probability that it will take a player the maximum number of tries to dig up the most desirable outcome (the health upgrade)? Two: What is the expected average number of tries for the health upgrade? Three: There exists two methods to play this minigame. First is the method described above. Second is that after the first dig, the player exits and reenters the area. This resets the green/blue/red rupee counter, but also allows the player to try another dig far faster. This introduces the potential for an infinite number of tries (the world record currently sits at about 100, which is pretty unlucky to say the least). Say it takes a player ten seconds between digs using the first method, and five seconds between digs using the second method. On average, which one will get you the health upgrade the fastest? Disclaimer: I've got an answer to the first question and maybe the second, but I've got no clue for the third.
  7. Especially when you consider as well that at any point you've got an extra plate either to transfer to or to transfer from (although that does complicate things a bit, it also ensures that pairs such as {4,8} are solvable).
  8. But no matter what you'll always have at least one pair whose average is even. Try it: pick any three numbers such that the average of any two is odd. I'm fairly sure it's impossible, since between a+b, b+c, and a+c at least one is guaranteed to be even. Do you have another counterexample of two numbers whose average is even but can't be reached via doubling?
  9. Aw man, I was so confident I'd finally solved one of Bonanova's legendary puzzles. A start on thinking:
  10. Question: the process shown demonstrates a local maximum for the function, but is there a way to prove that (for instance) at n=1000 it doesn't suddenly reach a new maximum? (I guess what I'm trying to say is, is there a proof, or is the solution presented here just process of elimination?)
  11. flamebirde

    Westworld Mafia

    I didn't say anything because then I thought about the save's 50/50 to lose their vote the next day. When I went for the tie originally, I thought it would be a sure fire way to gain an advantage, but after the lynch I was left wondering about other possible vote manips. For the record, you had me fooled -- I figured Plasmid had to be a baddie considering everyone else had claimed, but then he claimed too and I saw the logic in his plan. I think Dolores may have been too powerful in the end. Having 2 different vote manips and a lynch save, on top of a kill and a save -- that rendered the baddie's vote manip practically useless in the endgame when roles were revealed. The "once per game" aspect of the actions never mattered because 1. I didn't use anything N1 and 2. The game was 2 nights long. The only way to neuter the role would be to block, which no sane person would do as discussed earlier (no one would use the block ability N1 for this exact reason: you might hit Dolores). But this may all be because of an inactive Indy: that was the only force opposing the goodies that could effectively counter a known Dolores (RID kill, and the block+save to a lesser extent).
  12. flamebirde

    Westworld Mafia

    this was one of the situations I had wanted when I went for the tie. Sure, it let the baddie control the lynch, but if he used his vote manip there then in the endgame he would lose since I had both still hidden up my sleeve (of course, that all depended on whether or not I lived). The other option was that the tie went through and no one died, leading to more time for the goodies to figure out the game. That was a good play by Plasmid at the end; it clearly worked out. Congrats to him and to all the goodies! Also, rip Aura who died twice in two consecutive phases
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