I’m not dead!

In fact, I have a new toy. A staple from my childhood that’s actually even more fun than I remember.

Pattern blocks are little wooden blocks (about 1 cm thick) that come in the following shapes:

• Equilateral triangle, side length 1 unit (green)
• Rhombus, angles 60-120-60-120, side length 1 unit (blue)
• Hexagon, side length 1 unit (yellow)
• Trapezoid, half of a hexagon (red)
• Square, side length 1 unit (orange)
• Rhombus, angles 30-150-30-150, side length 1 unit (white)

The only really fundamental shapes are the triangle, square, and white rhombus; the blue rhombus, trapezoid, and hexagon can all be built from triangles. Having them handy makes the patterns attractive, though.

(The link and the pattern therein does not do them justice. There is also a linked Java applet that really shows a tiny fraction of the neat things you can do, because of limitations as to where the pieces will snap to. Try using the white and green ones together and you’ll quickly see.)

Anyway, I highly recommend them as a gift for everyone you know, especially kids in grades 1-3 and adults that remind you of me.

If you get some, try this riddle: How many (different, rotationally symmetrical) ways can you make a regular dodecagon (12-sided polygon) of side length 1? I was a little surprised to find you can do it at all, but have found at least 3 different ways already. I guess you can do this riddle even without the blocks in front of you, if your spatial/geometrical imagination is much better than mine…

A variation of the Pirate Riddle from intrepid reader John Rhoadhouse, on honor of Talk Like A Pirate Day. Arrrr.

First, the pirates have a new constitution. The first pirate makes a proposal. regarding the distribution of treasure. If half or more of the pirates vote for a proposal, it succeeds. Otherwise the proposer is executed and the next pirate down the list makes a proposal. Pirates vote based on the following priorities:

1. Life–a pirate will vote for whatever will save his or her skin.
2. Treasure–the more, the better
3. Bloodlust–life and treasure being equal, pirates will vote in the way that will see as many of their fellows killed as possible.

So, the riddle: Our salty pirate crew of 10 has fallen upon hard times and has a meager treasure of only one gold coin, which cannot be cut or split up. How many pirates will die? In general, for n pirates, how many will die?

Hint: Solve the case where there is no treasure at all.

Three gunfighters, named A, B, and C, are having a rather contrived gunfight. Here are the rules:

• The fighters shoot in turn: first A, then B, then C, then (if more than one is still alive) back to A, and so on.
• On a turn, a fighter gets to shoot only once, at one opponent.
• A hits the target 1/3 of the time. B hits 2/3 of the time. C always hits.
• Fighters are allowed to shoot at the ground or otherwise intentionally not hit any target.
• If all fighters intentionally miss in a row, the last one to do so is eliminated. *

What is A’s best strategy?

Extra credit: Generalize for different accuracies.

*This rule is not usually given as a part of this riddle, but it is necessary to avoid degeneracy while working out the solution, and (as you will see if you try to generalize) is particularly important if all of the fighters have accuracy over 50%. I chose it arbitrarily; it could be replaced with any rule that keeps everyone from intentionally missing forever.

I am running a single-elimination tournament with 64 players. The seeding is random. Assume the teams are totally ordered in skill (no two have the same skill, and the “is more skillful than” relationship is transitive) and that the more skillful team wins every game.

The tournament determines the most skillful team, of course. After it’s over, how many games are needed to determine the second best team?

Consider a (fair) six-sided die with four sides red, two sides blue. That is, it has a 2/3 chance to roll red, 1/3 blue. We are playing a game as follows: One of us picks a sequence of two outcomes (like “blue, red”), and then the other picks a different sequence of two outcomes. We roll the die as many times as necessary until the most recent two rolls match one of our sequences, at which point that player wins.

(Example: you pick red, red and I pick blue, blue. The rolls come up 1. red, 2. blue, 3. red, 4. blue, 5. red, 6. red, at which point you win.)

Because I am magnanimous, I will allow you to pick the sequence first or second, as you like. What is your winning strategy?

Edit: I should credit the source, which is here. I recast the riddle because I feel that the die is more intuitive than a “biased coin flip” and to try to avoid what I feel is a major hint in the wording of the original.

Today, I was contemplating the idea of creating a Ra computer game. One hypothetical feature of this game would be to “auto-pass” a worthless auction. Turns out this feature won’t be seeing the light of day…

For both of the riddles below, you may use any play situation you can imagine, and you have complete knowledge of the game state; in particular, you can know exactly what tiles are still in the bag as well as each player’s face-down suns and score.

Riddle 1: The auction track contains one tile, Unrest (AKA Death of Civilizations). The sun in the middle is worse than any of your remaining face-up suns. You pull a Ra tile. Come up with a situation where you would want to bid on this auction. Make no assumptions about your opponents’ strategy–assume that whatever they do will be bad for you.

Riddle 2: The auction track contains nothing at all. The sun in the middle is worse than any of your remaining face-up suns. You pull a Ra tile. Come up with a situation where you would want to bid on this auction. For this one, you may make reasonable assumptions about opponents’ strategy; they may want, say, to maximize their own score or winning chances, not just hurt you.

Can be found here.

Tonight, I cancelled my World of Warcraft account. I look forward to having more time to read, work out, blog, design games, hang out with friends, play other computer games, swing dance, cook, go to concerts, watch the Daily Show, and more. The extra bonus is that I can probably get all of these activities and more into the newly-freed hours.

To inaugurate the new era, I present to you this riddle: An island is populated with chameleons in three colors: red, green, and blue. When two chameleons of different colors meet, they both became the third color. So if a green and a blue chameleon meet, they both become red.

At one time, there are 13 red chameleons, 15 green chameleons, and 17 blue. Is it possible for all the chameleons to ever be the same color? Why or why not?

This is an old riddle that many puzzle fans will have heard before. It’s not new to me, and I’m not a big fan of it.

But, a friend of mine was recently posed this brainteaser during a technical interview, and she hadn’t heard it before. As a public service, I now re-present to you this old riddle, that you may ace your next interview.

You have two jugs of water. One holds three gallons and one holds five gallons, though they are otherwise unmarked. You also have a sink that you can get water out of or dump excess water into. Your task is to measure out exactly four gallons of water.

For the record, I am opposed to using this kind of question on an interview; it is 90% “have you heard this question before”, 8% sangfroid, and no more than 2% of whatever mental ability the interviewer is hoping to measure.

This one is extremely crunchy and might be considered more of a math problem than a riddle. But it caught my interest. From the XKCD forums:
You are playing no-limit Texas Hold’em with one opponent. In this version, you can see each other’s hands.
You hold AK unsuited; your opponent has a pair of queens. After initial betting, each player has $100 in reserve and there is $100 in the pot. The flop comes down Q-J-10 with no flush possibilities, giving you the straight and your opponent three queens. You have first action.

Note: Currently you are winning, but no turn or river cards can improve your hand. Your opponent can win if one of the two hidden cards is a queen (for a four of a kind) or a 10 or jack (for the full house.)

What is your betting strategy?