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**Puzzles**

Colonel |
Posts: 2,579 Cash: 46,264 / 266,023 Group: Nobility Joined: 12/14/05 03:32 PM |

Here's how the game works, the first person to get the correct anwser to a puzzle gets to post thier own. If the puzzle remains unanwsered for a week, that person wins.

I'll start first with a really easy one, just to get it moving.

3 guys go into a hotel, and each chip in 10$ for a single room. They go up to the room, and the guy who took the money realizes that the room is only 25$ a night. He sends a cleaning woman up to return the 5$. Now, she knows that the guys have no idea how much money they should be getting back, so she takes 2$ out and gives the 3 guys back 3$. They split it up evenly. So, each guy first spent 10$, and then got 1$ back, so they each spent 9$ total. The cleaning lady has 2$. 9x3=27, plus 2=29. There was 30 originally. Where's the extra dollar.

I'll start first with a really easy one, just to get it moving.

3 guys go into a hotel, and each chip in 10$ for a single room. They go up to the room, and the guy who took the money realizes that the room is only 25$ a night. He sends a cleaning woman up to return the 5$. Now, she knows that the guys have no idea how much money they should be getting back, so she takes 2$ out and gives the 3 guys back 3$. They split it up evenly. So, each guy first spent 10$, and then got 1$ back, so they each spent 9$ total. The cleaning lady has 2$. 9x3=27, plus 2=29. There was 30 originally. Where's the extra dollar.

Commander in Chief |
Posts: 5,699 Cash: 2,145,476,571 / 2,147,483,647 Group: Representative Joined: 12/23/06 04:45 AM |

QUOTE (Jinghao @ Feb 20 2009, 12:55 PM)

Every person is referenced by a number, say 0, 1, ..., n-1. And given any person i, you can find everyone i hates (and hence is hated by). That's all you need.

Oh. Well in that case, pick person 0, and move everyone he hates over to the other island. If anyone fights, it's impossible, of course (any 3 way hate relationship ultimately ruins the whole thing if there are only two islands. By the pigeonhole principle, at least two people who hate each other would end up on the same island if such a relationship existed).

So I guess in that case, if person 1 is still on the original island, repeat the procedure, if not, skip him and move onto the next person until you've covered everyone. Since what you're doing is getting rid of everyone who hates each other on the first island, then you just have to check the second island for a pair that hates each other.

Colonel |
Posts: 2,579 Cash: 46,264 / 266,023 Group: Nobility Joined: 12/14/05 03:32 PM |

All you would need to do is find one example of where it wouldn't work, like a hate triangle. It would be fastest to start looking for this example by looking at whoever is the most hated person on the island, because if any of the people who hate him, hate one of the other people who hate him, it won't work.

Then, work on down the line of most hated people.

Then, work on down the line of most hated people.

Not Odd anymore |
Posts: 45,875 Cash: 1,915,578 / 1,817,041,051 Group: Administrator Joined: 7/10/02 09:48 PM |

QUOTE (Dek Kvar @ Feb 28 2009, 11:23 AM)

All you would need to do is find one example of where it wouldn't work, like a hate triangle. It would be fastest to start looking for this example by looking at whoever is the most hated person on the island, because if any of the people who hate him, hate one of the other people who hate him, it won't work.

Then, work on down the line of most hated people.

Then, work on down the line of most hated people.

For that case there will be a lot of redundancies. Is there a more efficient way?

And Sing, that's close...

Peasant |
Posts: 1 Cash: 1,645,400,000 / 2,147,483,647 Group: Newbie Joined: 1/09/10 01:28 AM |

What is the probability of winning the best of n consecutive games of Rock-Paper-Scissors?

Allow me to give an example, as this does not explain the details of the problem. Say we're looking at person A playing person B. For the first game (say n=1), A has a 1 in 3 chance for each of either winning (W), tying (T), or losing (L) the first game (because player B picks rock, paper, or scissors randomly, and person A has an equal chance of either picking the move which will beat B's move [rock beats scissors, scissors beats paper, and paper beats rock], tying it, or losing to it). If n=2 however, the overall chance now depends on the first game. If A won the first game and loses the second game, A and B are tied. If A won both games or won the first and tied the second, A wins the overall game. However, if A tied the first game, the overall outcome depends solely on the second game: a 1 in 3 chance of winning. If A lost the first game, there is only a 1 in 3 chance of tying (by winning the second game), and a 2 in 3 chance of losing (by either tying or losing the second game).

Here are all 9 possible outcomes for A:

Game

1, 2, End result

-- -- ------------

W, W, W

W, T, W

W, L, T

T, W, W

T, T, T

T, L, T

L, W, T

L, T, L

L, L, L

There are of course 3 outcomes in which A wins, and out of a total of 9 possible outcomes, that gives a probability of (3/9=1/3) a 1 in 3 chance of A winning best of 2 games. Once n=3 however, the pattern chances. Obviously the total number of outcomes for n games is 3^n, and so the interesting part is the number of winning outcomes, W_n. Using brute force and simply counting all possible outcomes as I just did above, I have found that for n=1,2,3,4, W_n=1,3,10,31.

Allow me to give an example, as this does not explain the details of the problem. Say we're looking at person A playing person B. For the first game (say n=1), A has a 1 in 3 chance for each of either winning (W), tying (T), or losing (L) the first game (because player B picks rock, paper, or scissors randomly, and person A has an equal chance of either picking the move which will beat B's move [rock beats scissors, scissors beats paper, and paper beats rock], tying it, or losing to it). If n=2 however, the overall chance now depends on the first game. If A won the first game and loses the second game, A and B are tied. If A won both games or won the first and tied the second, A wins the overall game. However, if A tied the first game, the overall outcome depends solely on the second game: a 1 in 3 chance of winning. If A lost the first game, there is only a 1 in 3 chance of tying (by winning the second game), and a 2 in 3 chance of losing (by either tying or losing the second game).

Here are all 9 possible outcomes for A:

Game

1, 2, End result

-- -- ------------

W, W, W

W, T, W

W, L, T

T, W, W

T, T, T

T, L, T

L, W, T

L, T, L

L, L, L

There are of course 3 outcomes in which A wins, and out of a total of 9 possible outcomes, that gives a probability of (3/9=1/3) a 1 in 3 chance of A winning best of 2 games. Once n=3 however, the pattern chances. Obviously the total number of outcomes for n games is 3^n, and so the interesting part is the number of winning outcomes, W_n. Using brute force and simply counting all possible outcomes as I just did above, I have found that for n=1,2,3,4, W_n=1,3,10,31.

Commander in Chief |
Posts: 5,699 Cash: 2,145,476,571 / 2,147,483,647 Group: Representative Joined: 12/23/06 04:45 AM |

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