I think this one can be compared to both Pack Rat in terms of difficult to remove, and Assemble the Legion in terms of power. With that in mind, I think this card is a bit too powerful. Yes, you can wrath it away (you can also use -1/-1 like Loss from Profit//Loss and Golgari Charm), but Assemble the Legion has a casting cost of 5, and it is immensely problematic when it hits the board, so I would expect this to be a bit more troublesome at 3. It's definitely close though, difficult to say.
Is there value in trust? Yes.
Is there value in faith? Sometimes.
Is there value in superstition? No.
We're fortunate enough to live in an age where vast arrays of information on science and history are freely available. It's become pretty obvious that religion is closer to superstition than faith.
I heard that it's good, but I wonder what people think like about it. Could you tell me?
It's difficult to explain. Aside from the awesome soundtrack, art, and animation, the story is very mysterious but it also manages to be very actiony at times. The way the "mystery" and "reveal" aspects are presented are just very impactful. I'd just suggest watching the first episode. If that won't hook you I don't think this anime is for you.
Shingeki No Kyojin has been on a steady downslide since the 5th or 6th episode and Eren is literally the worst protagonist in anime history(outside of the guy from One Outs)
From what I've seen the reception of Shingeki no Kyoujin has been pretty much consistent, if not better as the series has been progressing.
The mana thing makes me think an MTG anime wouldn't go so well. I vaguely remember the show Duel Masters where they kept throwing down mana for multiple turns and it was extremely boring. Imagine the episode where the guy gets manascrewed and gets beat down by RDW. Maybe it would work, but I think it would be a mistake not to make the anime about the adventures of a newly minted Planeswalker. Anime is a great medium to portray cool magic battles.
I think the Yugioh show is getting a lot of unnecessary hate. The original is obvious outrageous but has nostalgia points. GX was okay, especially the Japanese 4th season. Yugioh 5Ds was the one I was sad to see go, as I thought the Japanese version was excellent. Yugioh Zexal was so bad in the English Dub I think it got canceled, and the Japanese doesn't make it much better. Obviously none of these shows are deep, but they're fun to watch nevertheless.
The flavor text seems to read: "The wills of mortals shift as the tide ebbs and flows." In case this wasn't mentioned already for those interested. <.<
I'm thinking of an integer n where 0 ≤ n ≤ 15. You get to ask me seven yes-or-no questions to determine the number, with the following restrictions:
- I can answer the questions in whichever order I choose. Therefore, each question must stand on its own and not depend on the answer of another question.
- I am allowed to lie when answering one of the questions. I can also choose not to lie on any of the questions.
What seven questions can you ask so you can always determine the number?
Does this one only look tedious to solve and prove or is it easier than it looks?
Two magicians perform the following card trick: Magician A gives an ordinary 52 card deck to a member of the audience, asking them to select 5 cards (at random or deliberately). Magician A looks at the 5 cards, and returns one to the audience member. He then re-orders the remaining 4 cards and places them face-down in a neat pile on the table (all cards facing the same way). Magician B takes the pile and looks at the four cards, and immediately guesses the fifth card in the audience member's hand. There is no communication between A and B and the trick can be repeated with a 100% success rate. Clearly the magicians have some sort of system- How does it work?
There are probably quite a few ways to answer the 5 card trick if we're talking in a practical sense. Here's an answer I came up with.
You need 2 different systems: One for determining the value of a card and one for determining the suit of a card.
Finding out the suit of the card is the easy part. When the magician places the pile on the table, they can agree the 4 quadrants of the table (top left, top right, bottom left, bottom right) to correspond to the 4 different suits. By looking at which quadrant the pile is placed, the suit can be determined. Another idea I had was the first magician putting their hands in their pockets, as there are 4 ways this can be done (left hand in left pocket, right hand in right pocket, both hands in both pockets, no hands in either pocket).
For the value of the card, the system is as follows. The magicians have numbered every card from 1 to 52. For example: AH=1, 2H=2, 3H, 3, ... , KH=13, AC=14, 2C=15, ... ,etc
Then, a pattern could be agreed upon depending on the ordering of the cards. For example, The magician is ordering the cards AH, 5C, KH, 7S. Firstly, they mentally number the cards from lowest to highest: AH=1, KH=2, 5C=3, 7S=4.
There are 24 different ways to order 4 different objects, so an order can always be agreed for the value of a card, let's say they agreed on this system: 1234 = A, 1243 = 2, 1324 = 3, 1342 = 4, 1423 = 5, 1432 = 6, 2134 = 7, 2143 = 8, 2341 = 9, 2314 = 10, 2431 = J, 2413 = Q, 3124 = K
This would let the other magician know the value of the card by looking at how the cards were ordered.
I usually prefer manga first, subbed anime second, and dubbed anime third. Manga is just very fast and has the most accurate story telling (for obvious reasons). Subbed anime will almost always have good voice actors and accurate story. Even with excellent voice actors and accurate story telling, dubs can have awkward moments due to cultural differences or just nuances in language.
Argh, Redthirst's answer is better. We were so fixated on 3's offer that we forgot to think harder about 4's offer. Well, more like a slight inaccuracy.
3 is not the axis of power. 4 is. 4's offer of 1 to 1, 1 to 2, and 998 to himself will definitely win if offered, which means 5's offer can be 2 to 2, 1 to 3, and 997 to himself. 2 and 3 will be in favor because it is strictly better than 4's offer.
---
Argh on the eyes problem as well. I'm better at outside-the-box thinking than pure logical thinking, so my previous answer was really dumb (Fun Fact: Another idea I had was pulling an eye out and checking the eye colour). I think I have a better answer for the problem this time:
From your perspective, there are 99 blue-eyed people and 101 non-blue-eyed people. You think to yourself, what if I'm not a blue-eyed person? Then to a blue-eyed person, there are 98 blue-eyed people and 102 not. That person thinks to themselves "what if I'm not blue-eyed?", then to another blue-eyed person, there are 97 blue-eyed and 103 not. Each hypothetical person down the chain thinks this until it comes down to 1 blue-eyed person. If it were 1, the blue-eyed person realizes that there are no other blue-eyed people, and disappears the 1st night. If there are 2, the blue-eyed people realize that the other didn't disappear because there is another, and they disappear the 2nd night. If there are 3, then you know there are 2 blue-eyed people, and you wait until after the 2nd night to confirm you're the third. If there are 4, you wait for the tension to resolve at the 3rd night to confirm a 4th. If there are 5, you wait for the tension to resolve with the 4 others on the 4th night before confirming you're the 5th and disappearing on the 5th night. Extrapolate and you realize the only reason all 99 blue-eyed people haven't all disappeared by the 99th night is because there is a 100th blue-eyed person.
On the 100th night, all the blue-eyed people disappear.
Also, I think have an answer in regards to the Pirate riddle.
2 must agree to any offer from 3 (as mentioned above) to avoid being killed (all of 2's offers can be declined, even if it's a 1000 to 0 split because the pirates are bloodthirsty*). 3 will never agree to an offer from 4 or 5 (knowing that his own offer of 1000 coins to 3 will always win). 3's offer will always win, therefore, 1 and 2 will agree to any offer that gives them at least 1 coin.
For 4 to be successful with his proposal, he has to make 2 people agree to his offer (therefore we must make pirate 1 and 2 agree to the proposal). His best offer is 1 coin to 1, 1 coin to 2, and 998 coins to 4. Both 1 and 2 agree because this is functionally superior to 3's offer (which we've established will always win).
Therefore, 5's offer needs to be 2 coins to 1, 2 coins to 2, and 996 coins to 5. 1 and 2 agree because this is better than 4's offer, giving the proposal a 3/5 majority.
*For this reason any offer that is functionally identical to an offer by a higher pirate will always be declined, because it means 1 more person dies before the money is distributed, despite that person not gaining any gold. For example, if the correct offer by 3 was 400 to 1, 400 to 2, and 200 to 3, even if 4 makes the exact same offer, it will be declined because it results in 1 extra death. However, if 4 offers 401 to 1, 401 to 2, and 198 to himself, this will win because we know the pirates are greedier than they are bloodthirsty.
Is there value in faith? Sometimes.
Is there value in superstition? No.
We're fortunate enough to live in an age where vast arrays of information on science and history are freely available. It's become pretty obvious that religion is closer to superstition than faith.
Blue can't kill creatures but it's allowed to "morph" creatures into something else i.e. Pongify and Rapid Hybridization.
It's difficult to explain. Aside from the awesome soundtrack, art, and animation, the story is very mysterious but it also manages to be very actiony at times. The way the "mystery" and "reveal" aspects are presented are just very impactful. I'd just suggest watching the first episode. If that won't hook you I don't think this anime is for you.
From what I've seen the reception of Shingeki no Kyoujin has been pretty much consistent, if not better as the series has been progressing.
Underworld Cerberus
1RR
[Creature - Hound]?
Double-strike
2/2
Heliod sounds like Optimus Prime.
I think the Yugioh show is getting a lot of unnecessary hate. The original is obvious outrageous but has nostalgia points. GX was okay, especially the Japanese 4th season. Yugioh 5Ds was the one I was sad to see go, as I thought the Japanese version was excellent. Yugioh Zexal was so bad in the English Dub I think it got canceled, and the Japanese doesn't make it much better. Obviously none of these shows are deep, but they're fun to watch nevertheless.
Does this one only look tedious to solve and prove or is it easier than it looks?
There are probably quite a few ways to answer the 5 card trick if we're talking in a practical sense. Here's an answer I came up with.
Finding out the suit of the card is the easy part. When the magician places the pile on the table, they can agree the 4 quadrants of the table (top left, top right, bottom left, bottom right) to correspond to the 4 different suits. By looking at which quadrant the pile is placed, the suit can be determined. Another idea I had was the first magician putting their hands in their pockets, as there are 4 ways this can be done (left hand in left pocket, right hand in right pocket, both hands in both pockets, no hands in either pocket).
For the value of the card, the system is as follows. The magicians have numbered every card from 1 to 52. For example: AH=1, 2H=2, 3H, 3, ... , KH=13, AC=14, 2C=15, ... ,etc
Then, a pattern could be agreed upon depending on the ordering of the cards. For example, The magician is ordering the cards AH, 5C, KH, 7S. Firstly, they mentally number the cards from lowest to highest: AH=1, KH=2, 5C=3, 7S=4.
There are 24 different ways to order 4 different objects, so an order can always be agreed for the value of a card, let's say they agreed on this system: 1234 = A, 1243 = 2, 1324 = 3, 1342 = 4, 1423 = 5, 1432 = 6, 2134 = 7, 2143 = 8, 2341 = 9, 2314 = 10, 2431 = J, 2413 = Q, 3124 = K
This would let the other magician know the value of the card by looking at how the cards were ordered.
---
Argh on the eyes problem as well. I'm better at outside-the-box thinking than pure logical thinking, so my previous answer was really dumb (Fun Fact: Another idea I had was pulling an eye out and checking the eye colour). I think I have a better answer for the problem this time:
On the 100th night, all the blue-eyed people disappear.
For 4 to be successful with his proposal, he has to make 2 people agree to his offer (therefore we must make pirate 1 and 2 agree to the proposal). His best offer is 1 coin to 1, 1 coin to 2, and 998 coins to 4. Both 1 and 2 agree because this is functionally superior to 3's offer (which we've established will always win).
Therefore, 5's offer needs to be 2 coins to 1, 2 coins to 2, and 996 coins to 5. 1 and 2 agree because this is better than 4's offer, giving the proposal a 3/5 majority.
*For this reason any offer that is functionally identical to an offer by a higher pirate will always be declined, because it means 1 more person dies before the money is distributed, despite that person not gaining any gold. For example, if the correct offer by 3 was 400 to 1, 400 to 2, and 200 to 3, even if 4 makes the exact same offer, it will be declined because it results in 1 extra death. However, if 4 offers 401 to 1, 401 to 2, and 198 to himself, this will win because we know the pirates are greedier than they are bloodthirsty.