unreality

Mekal: the pack was shuffled adrenalgirly: yes, the fact that the FIRST black ace is the first one flipped & seen changes it, and marks an ideal point in the pack that you should predict beforehand... and the goal is to discover this point and find whether or not you should accept the bet, and if you DO accept the bet, what your prediction is

Yep, that's correct

The odds are not 2/52 edit to clarify: There is a point in the deck where the odds are in fact 51/1326 (which is 2/52), but not all points are the same... your goal is to find the ideal point in the deck, and then determine what the odds are and thus if you would take the bet

I have just made a truly amazing program... I call it 'the Awesome' btw, we still need 2 or 3 more people: 1) unreality 2) Frost 3) dms172 4) Mekal 5) FOTH 6) 7) 8) possibly unreality #2 PM everyone you know

Nope, sorry

Your friend staggers up to you with a grin on his face. "Want to play a game?" he asks nonchalantly. "Just ten bucks. If you win, you get three hundred dollars back!" You know him pretty well, so you're reluctant. "What's the game?" Your friend pulls out a deck and shuffles it absentmindedly, then hands it to you. You inspect it; it's a standard deck, 52 cards, no jokers, all cards look the same and feel the same from the back. You give it a couple good shuffles and hand it back. "You saw the two black aces in there, right?" your friend asks. "Of course." "Well, here's the game. You're going to take the deck, and one at a time, flip over the top card, placing it on the table face-up. Easy enough?" "Sounds good," you say. "Well here's the catch. You're going to predict when the first black ace will turn up. Anywhere you want to pick in the deck... 1 being the top card, 52 being the bottom. Remember that it's been shuffled and re-shuffled, so each card has a 1/52 chance. If you accept the bet, you pay in $10, but get $300 if you win." Do you accept the bet?

Male - Azure ;D (a form of blue, the color of the sky)

Nobody so far I'll rephrase the OP if you want: Swarmbots are invading! They work by building copies of themselves. Swarmbots come in two genders: drones and producers. In one minute, a drone can assemble a single producer. However, a producer, given the same amount of time, will assemble two Swarmbots: a drone and a producer. (1) The swarm begins with one drone... assuming that all swarmbots work at their optimal speeds and work continuously, how would you define their growth (ie, the new number of swarmbots based on previous generation(s))? In each generation, how would you define the breakdown of how many drones and how many producers make up that generation? Why does it work this way? (2a) After x generations, how would you define the TOTAL number of swarmbots in ALL generations (the total population of swarmbots), assuming they receive no casualties? (2b) Now a swarmbot is guaranteed to die right after seeing its "great-grandchildren" get assembled. After x generations, how would you define the TOTAL number of swarmbots? (3) Prove (or disprove) that, given that no casualties or "swarmbot deaths" occur, the final two generations will always have a bigger total of swarmbots than ALL preceding generations... ie, say there are 10 generations. Call the total population of the first 8 generations 'x', and the total population of the last two generations 'y'. Prove (or disprove) that y>x (independent of any specific generation number)

Yeah - the unfortunate thing is that one of programs has rand() in it, so I need to change it Hehe

No other attemptees?

In testing, I'm having some second thoughts about the rand() function. It's okay to have the programs placed at random spots in the VNA, as that's the whole point - the programs are at unknown locations relative to each other, so you can't just bomb your opponent from the beginning But the rand() function is different. Not being able to reconstruct the same game twice is one minor reason not to have it, but also it adds more randomness to a game which is based on survival & battle strategy. Any program that uses COPY/SHIFT to launch BOMBs around the VNA using the rand() function is just putting its eggs in the 'luck' basket. For this reason I'm going to take away the rand() function... which means I have to rethink one of my best programs, but that's a good thing ;D This game isn't a game of chance, it's a game where the best program wins btw: for all games in VNA 2.0, luck will never be a factor. Which means that a battle between two programs will actually consist of multiple games, all at new random starting locations, in case there was one lucky scenario based on starting loc. A program must win at least 7 games AND at least 2 more games than the opponent for it to win the battle

I used random.org to determine the bracket order ;D If we don't get 8 we can have a buy-in round or something

Yes but you can't recursively use your 'total growth rate' formula to define the next step, since it doesn't distinguish between C & B for the next round.

Don't worry, you're nowhere near as impatient as Brandonb himself :P

I know you didn't say that exactly, but that's how I interpreted it I get it now - the 28 days in each month isn't a metaphor, it's an exact example (ie, all insects have 2 legs.... and 4 more ). Thanks for clarifying

Swarmbots are invading! They work by building copies of themselves. Swarmbots come in two genders: drones and producers. In one minute, a drone can assemble a single producer. However, a producer, given the same amount of time, will assemble two Swarmbots: a drone and a producer. (1) The swarm begins with one drone... assuming that all swarmbots work at their optimal speeds and work continuously, at what rate do they grow? In each generation, how would you define the total number of swarmbots in that generation and the breakdown of how many drones and how many producers make up that generation? Why does it work this way? (2a) After x generations, how would you define the TOTAL number of swarmbots in ALL generations, assuming they receive no casualties? (2b) Now a swarmbot is guaranteed to die right after seeing its "great-grandchildren" get assembled. After x generations, how would you define the TOTAL number of swarmbots? (3) Prove (or disprove) that, given that no casualties or "swarmbot deaths" occur, the final two generations will always have a bigger total of swarmbots than ALL preceding generations

that's what the Others forum is for Make a new topic

Ah I know what it is The "sorta" 2 legs says it: preying mantis

the OP says 15...

haha, awesome Last year my health teacher had a poster with a huge list of these ;D

Sounds fun 1. AAAsn888s 2. Star Tiger 3. Reaymond 4. Peace*Out 5. Izzy 6. akaslickster 7. unreality

Yeah I try to do that too as much as possible, but there's always things that evade me

hahahahahahahahahahAHAHAHAHAHA! These are hilarious