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4526

General Discussion / Re: Logic problems

« on: January 14, 2013, 06:35:47 am »

I guess you cannot give a clue to those of us that are clueless...?

I cannot fathom how:
-Infinite people discuss a strategy;
-Devil puts hats on people according to the strategy they have agreed upon;
-Without any means to communicate (since they have to shout simultaneously, which negates the possibility that waiting "t" seconds before shouting means anything) after the hats are on, they have to guess the colour of their hat;

can somehow lead to a finite amount of errors. Afterall, the impossibility to communicate once the hats are on means that they cannot react to the devil's strategy, while the devil can react to their strategy, which would mean that their best shot at surviving would be to guess randomly...

4527

General Discussion / Re: DominionStrategy Forum Members Map!

« on: January 14, 2013, 06:30:21 am »

I'm planning to visit Barcelona in March.  Any advice? 

Nice! How long will you stay here? General advice, or places to visit?

4528

General Discussion / Re: DominionStrategy Forum Members Map!

« on: January 13, 2013, 07:10:14 am »

I was surprised to be the only one in Barcelona, but turns out I'm the only one in the whole Iberian Peninsula.
We've got nice warm weather here, I guess that's why people would rather be outside. I am so lonely... [cue Lonely by Akon... yes, I am a very sad person]

4529

Dominion General Discussion / Re: How much of dominion skill is luck?

« on: January 13, 2013, 06:16:51 am »

In this particular case, more of the issue is that you perceive Gold as being a step ahead of the 5$ cards, and often it's not.

Though, sometimes, in base set, it is.  If you get expansion, Dominion rapidly becomes a more skill-based game.

I wouldn't go as far as saying that "Gold often isn't a step ahead of the 5$ cards". Aside from cursers, there aren't that many 5$ cards that you would rather have than a gold, pre-third shuffle, since it's "usually" too early to have a combo built. A gold all but guarantees that you will get that 5$ card that you wanted before your next reshuffle, but a 5$ card doesn't guarantee a gold, and there aren't that many strategies that don't rely on getting one or two of them.

4530

General Discussion / Re: Logic problems

« on: January 13, 2013, 04:29:29 am »

I'd put the order of priorities of pirates this way:
-Survive
-Earn as much money as possible
-Kill as many pirates as possible

So, in the case that pirates are perfectly trustworthy, a follow-up to Titandrake's post:

[EDIT: actually, it doesn't work; I'm still thinking on it, but I forgot that in a 4 crew ship, any deal that doesn't involve the captain requires all other pirates to agree, so that they represent the majority]
-With 4 pirates, if no promise is made, the second pirate will accept whatever the captain offers, 0 coins included, or otherwise he will die, as Titandrake said.
However, the third pirate can strike a pact with the second pirate: if he votes against the captain, he will accept a 0, 100, 0 split afterwards, which the second pirate and fourth pirate would normally accept, since that way they kill one extra pirate; but the fourth pirate can also offer a 0,0,100 split, which is equivalent for the second captain; as such, they will both rise their offers until they both reach 99 coins for the second pirate and one coin for the relevant pirate, so that doesn't seem like a very good split for the third or fourth pirates. Assuming (*), this means a 99, 1, 0 split.
Thus, the captain (a bit nervous since everybody is discussing his death) will instead try to earn the loyalty of the third or fourth pirate, and one coin for the fourth pirate should suffice , since otherwise he would earn nothing. But! Whatever 100-n, 0, 0 n split the captain proposes, the second pirate can offer the same to either the third or fourth pirate, and they will accept, since they are bloodthirsty. As such, the captain, trying to survive, will offer the 100 coins to either the third or fourth pirate, and the second pirate, seeing that he will earn nothing, will also offer 100 coins to either the third or fourth pirate, as he is bloodthirsty.
Probably before reaching that point, the captain and the second in command will realize that it may be better to reach an agreement between them, but that's prisoner dilemma: if the captain offers a 100-n, n, 0, 0 split, the second pirate could accept, or try to get a better deal with either the third or fourth pirate, which is always possible, since a 0, n, 0, 100-n split seems better for him (he gets to kill the captain); but that reasoning leads to a 0, 0, 0, 100 split (according to the previous paragraph), which is worse than the previous 100-n, n, 0, 0 split...

Adding extra rules about how and when the pirates can strike a deal would make this quite easier... or solvable...

*I would add that, if a pirate has the choice among more than one deal that are equivalent to him (according to the three laws of buccaneering above), he will choose the one proposed by the highest ranking pirate. It can shave some lines of discussion for higher number of pirates.

4531

General Discussion / Re: Logic problems

« on: January 12, 2013, 10:35:05 am »

Captain keeps 97 for himself and gives 1 to each of 3, 5 and 7?

Yep! Obviously, if the pirates could leave the ship, I'm sure that wouldn't be the solution

Does the captain get a vote? I forget.

Figured this one out. Yeah, cap gets a vote too, checked wiki for exact setup. Also, they want as much money as they can get (greedy), and if they can kill some other pirate while still receiving the same amount of money, the will do so (bloodthirsty).

Ah, you beat me

You can solve the problem too if the captain doesn't get a vote; it's slightly harder, and gets you a different result, but the way you get the solution is the same.

4532

General Discussion / Re: Logic problems

« on: January 12, 2013, 06:04:43 am »

My favorite puzzles have already been posted (blue eyes, 100 prisoners), so I'll just post a somewhat easy one:

7 pirates have just ended plundering the Caribbean, so they get together to discuss how they will share the 100 gold coin booty. Now, these aren't your normal kind of pirates, they are blood-thirsty, unlimitedly greedy, perfectly rational pirates, and they are ranked by fighting prowess.

Thus, the captain has to propose how to share the booty, knowing that if more than half of his crew starts a mutiny, he will get killed, and the 2nd in command will take his place (who will in turn have to propose how to share the booty, eventually being replaced by the third in command, etc.).

How much gold will each pirate receive?

4533

General Discussion / Re: Probability paradoxes

« on: January 10, 2013, 06:49:28 am »

I have one sibling.  At least one of us is a boy.  What is the probability that the other is also a boy?

In this case, knowing that at least one of you two is a boy leaves three possibilities for you/sibling:

boy/boy
boy/girl
girl/boy

And now we have 2/3 male sibling.

Maybe I'm just being dense, but how do we know that these three possibilities are equally probable? The knowledge that there is at least one boy affects the possible outcomes (girl/girl isn't possible), so why wouldn't it also affect the probability of each one?

4534

Dominion General Discussion / Re: Dirty way to decide on platinum/ shelter use

« on: January 09, 2013, 06:20:08 pm »

If there is one DA card and one prosperity card, the odds of getting both shelters and colonies are:
-1/100 when shuffling a second time;
-1/90 when using the top/bottom technique.

Not really intuitive, but that's how it is.

4535

Other Games / Re: So Pokemon X and Y.

« on: January 09, 2013, 06:03:21 pm »

I agree that a type change would be nice, but Fighting/Psychic/Dark isn't balanced because Dark is not just resistant to Psychic but fully immune.  A workable triangle would be Fighting/Rock/Flying.  Two more are Fire/Rock/Steel and Grass/Poison/Ground, but those dip into the original three.  There aren't any triangles other than these, I think.

Here's the list of all the imperfect triangles, id.est., A is very effective against B, B is very effective against C, C is very effective against A. Took me a while to find it out, I'm sure it has been done already, but oh well:
-fire/grass/water
-fire/ice/ground
-fire/steel/rock
-fire/grass/rock
-fire/grass/ground
-water/ground/electric (immunity)
-grass/ground/poison
-grass/rock/ice
-ice/flying/fighting
-grass/rock/flying
-ice/ground/rock
-ice/ground/steel
-grass/rock/flying
-grass/rock/bug
-fighting/rock/flying
-fighting/dark/psychic (immunity)

If you also want to have A is very resistant against B, B is very resistant against C, C is very resistant against A, then you are left with only:
-fire/grass/water
-fire/steel/rock
-grass/ground/poison
-fighting/rock/flying

So, if you don't want to repeat any type, fighting-rock-flying it is!

EDIT: this gets more difficult with dual types. If I was bored enough, I could write a small script to find perfect dual-type triangles, but you know...

4536

Puzzles and Challenges / Re: Ruined Palace

« on: December 16, 2012, 12:39:33 pm »

Actually, you need 88 plays of abandoned mine, and 7 plays of ruined market, and enough king's court/throne room/ruined library to support them.

Chaining King Courts provides 2*n-1 effective plays of King Courts for one action. King courting a throne room provides 6 plays of the card in question, but requires three copies of the card and the throne room, instead of one; so, its only benefit is allowing a division of the multiplied actions, but it's just as good/bad from a card draw perspective. Maybe you can dig this to find a solution with more throne rooms and less king courts?*

Anyway, we need one king courted ruined market and two throne roomed; 28 king courted abandoned mines, and two throne roomed. That means two king courted throne rooms (and we are left with two uses of throne room to spare), so 48 (16 of which are king courts) cards to draw. Two king courted ruined libraries provide six cards, and costs three cards (the king court, with one use left for the next one; and the two ruined libraries), for a net gain of three cards. Since we have to draw 43 cards (48, minus the 5 you start with), and we had two spare throne rooms (the first applied on a ruined library, the second on a king court applied on two libraries), the end result is the need to draw 39 cards, so that means another 13 (+1) king courts, and 26 (+3) ruined libraries, and no card draw left.

End result:
-30 king courts
-2 throne rooms
-29 ruined libraries
-28 abandoned mines
-3 ruined markets

And no spare buy/coin/card draw.
*Fast thinking: normally, once your king court chain is going, you would do king court (ruined library; ruined library; king court (abandoned mine; abandoned mine; REPEAT), which draws itself (6 cards) and provides 6 coin for two king courts.
Instead, you could do king court (ruined library; ruined library; king court (ruined library; ruined library; king court (ruined library; ruined library; king court (throne room (abandoned mine; abandoned mine; abandoned mine); throne room (abandone mine; abandoned mine; abandoned mine); REPEAT))); which draws itself (18 cards), and provides 12 coins for four king courts and two throne rooms, so you're worse than in the first case;

So I'm pretty sure I got the best answer.

Man, I sure love procrastinating