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Everything posted by HoustonHokie

  1. From time to time, I've played a dice game called Farkle. Maybe you've played it before. To start your turn, you roll 6 dice, and, if your roll scores (see below for scoring), you can do one of three things: 1. Add the score from all the scoring dice to your total for the turn and set those dice aside. Then, pick up the remaining dice and roll again. 2. If you score in more than one way (say you roll a 1 and a 5, or two 1s), you choosing only the scoring combinations you want to keep. You add the score from those dice to your score for the turn and set them aside. Then, pick up the remaining dice and roll again. [The difference between #1 and #2 is the rejection of some scoring combinations in favor of having more dice available for your next roll.] 3. Add the score from all the scoring dice to your total for the turn and stop rolling. These points become permanently attached to your score for the game. The twist to the game is what happens if you don't score on a particular roll. This is called a Farkle, and you lose all the points you have accumulated in that turn. You still have the points from previous turns, but anything you might have gained in previous rolls that turn is lost. Here's an example of the effect of a Farkle: Roll 1: Three 1s appear, adding 1000 to your score for the turn. You can stop rolling now and add 1000 to your score for the game. Or, you can roll again, but with only 3 dice this time. Roll 2: Farkle - no scoring dice appear. You turn ends, and you add nothing to your score for the game, not even the 1000 points you earned on the previous roll. If you manage to score with all 6 dice on one or multiple rolls, you get to roll all 6 again. A Farkle, however, will still cause you to lose all your points for the turn. Scoring: If you roll a 1, you get 100 points If you roll a 5, you get 50 points If you roll 3 of a kind, you get 1000, 200, 300, 400, 500, or 600 points (for rolls of three 1s, 2s, 3s, 4s, 5s, 6s respectively) If you roll 4 of a kind, you get twice the value of 3 of a kind If you roll 5 of a kind, you get twice the value of 4 of a kind If you roll 6 of a kind, you get twice the value of 5 of a kind If you roll 3 pair, you get 750 If you roll a straight (1 through 6), you get 1500 For the 3, 4, 5, 6 of a kind, 3 pair, or straight, you have to have all those dice appear on a single roll. For example, rolling a 1 on your first roll can't be followed by rolling 2, 3, 4, 5, 6 on your second roll to score a straight. Obviously, the goal is to score as quickly as possible. Most times, the game is played to a certain limit (say first player to reach 10,000 wins). It's easy to score - but the trick is knowing when to stop. Enough about the rules. Here's my questions: 1. Given a certain number of dice remaining to roll (1-6), what is your probable score for the next roll (assuming you keep all your scoring dice)? 2. Assuming you're required to keep all your scoring dice and keep rolling (ignore possibilities #2 & #3 above), what is the value of an average Farkle turn after the first roll? After the second? Third? Fourth? 3. When is it advantageous to apply option #2 above and not keep all your scoring dice? 4. Can you develop a rational method for stopping your turn (#3 above)?
  2. HoustonHokie

    If Cody picks the joker to begin with, does the host tell him when she goes to remove it from the deck?
  3. yeah, but I think that's cheating... need to stay inside the grid (although most puzzles need you to think outside the box, this one does not ). The blank square is the one just northeast of the 7 & 4 at the top. Famous puzzle, famous answer, so I'll refrain from posting. But I finally remembered where I'd seen Prof. Templeton's avatar before...
  4. Here's what I've never understood about the gossip ring in these type of puzzles: And here's the other thing I don't get:
  5. Looks like bushindo was saying the prisoners were limited to 49 bits of information. A name and an ordinal would be one bit each. I agree that it makes sense for them to alphabetize themselves prior to dividing, but I'm not sure how to divide or how beneficial it would be to only know part of the equation. The prisoners could only memorize their own ordinal plus the names and ordinals of 24 others. So, when they got to the jar room, how do they open the jars? Do they go 1-50, but if they come across a name they know that's higher than 50, they open 1-49 plus that jar? Or do they open all the jars matching ordinals they memorized, plus the first 25 others in line? I have no idea how to calculate the odds that names would get put where they belong. Say they know 3 and 6 from memory, and happen to come across 4 & 5 in the initial 50 jars. They could sort 3-6 properly in that case. But if they only came across 4, where would you put it? Presumably you'd want it in 4 or 5 (because it's between 3 & 6), but which? And what do you put in the other jar (4 or 5)? It would have to be something out of order, and that would be a big problem. It almost seems like the extra information is enough to confuse, but not to help. The more I think about it, the more I think you're limited to a sort and search procedure unless you know every name.
  6. But we can do better...
  7. Okay, just the line this time
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