There is a funny modern deck where you play , but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt?
9312
In Magic, at the start of the game, you draw 7 cards. How would you calculate the likelihood of drawing a specific card in your opening hand?  For example, let's say I have a 60 card deck, and I'm running 4 . What is the percent chance that I will have at least one Bird in my opening hand?
79402
Rules of the game with two players.  First player puts any number of coins in the first pot. Then second player, knowing that number, puts any amount of coins in the second pot.  Then they in turns (beginning with the first player) do one of these things: take any amount of coins from one pot OR take equal amount from both pots. Whoever cannot take a coin loses. Players always know how many coins there are in the pots.  Who wins and what is the perfect strategy?  I managed to solve it, but the solution is very complicated and is based on fact which I just randomly guessed, but cannot deduct without the full solution. So I am looking for a nice approach to this problem.  The easy part is determining that player two wins: if there is a smallest winning initial donation by the first player, then the first move of player one cannot be taking just from the second pot, and this fact leads to infinite count of winning positions for player two for bounded number in the first pot, which is impossible (since there cannot be two winning positions for player two with the same amount of coins in the first pot).
114545
So a friend of mine showed me a very interesting card trick that I don't quite understand.  You have 52 normal cards from any regular deck you can buy anywhere. The Ace is worth 1 while 2 through 10 are worth their number value. Jack is worth 11, Queen is worth 12 and King is worth 13.  You shuffle the cards any way you want. Then you take the first card from the top of the deck and put it down on the table face up. Now, if it's a king you put it back in the deck and try again. If it's anything else you read the value on the card. Lets say you picked a 3. Now you have to put down cards from the deck in your hand on top of the 3 until you have counted 13. So 13 - 3 = 10. So you place 10 cards from the deck face up on the 3 on the table. It does not matter what the cards you place on top have in terms of values.  Now you move on to make a new set of cards on the table after the same principle. Again you cannot start on a king. If you reach a point where you can no longer put down 13 cards you keep the cards in your hand.  So I carried this out and ended up with 5 sets of cards on the table.They started with 3, 6, ace, ace and 10 leaving me with 3 cards left on my hand.  Now you take three of the decks, turn them around so they lie face down and then you pick up the rest of the cards to your hand. Now you choose 2 decks of cards and make the first card turn around so that they are face up. I have a 3 and a 10 and I have 24 cards on my hand. With 3 and 10 facing up you put that together which is 13. You have to add an extra 10 to whatever number you get so it will become 23. Now I put down 23 cards from my hand which leaves me with one card. This means that the last card currently facing down of three sets of cards on the table will be an ace.  And it was.  Why does this work?  My friend told me this could be explained or at least proven to work 100 % of the time with a mathmatical formula.
63960
I did some calculation and found out, that my Living End deck (modern) might be better of with 66 cards maindeck. SB still contains of 15 cards. Depending on the match up, I'd like to cut my deck for game two down to the normal 60. The SB than would contain 15+6 = 21 cards. Is this legal?
65101
I just purchased the game . As per , the structure of the game is as follows:     Game has 55 round playing cards. Each card has eight randomly placed symbols. There are a total of 50 different symbols through the deck. The most fascinating feature of this game is any two cards selected will always have ONE (and only one) matching symbol to be found on both cards.   Is there a formula you can use to create a derivative of this game with different numbers of symbols displayed on each card.  Assuming the following variables:   S = total number of symbols C = total number of cards N = number of symbols per card   Can you mathematically demonstrate the minimum number of cards (C) and symbols (S) you need based on the number of symbols per card (N)?
119678
Suppose we had a normal 52 count deck of cards. If two cards are randomly picked, what is the probability that both are Hearts given at least one Heart was chosen.  I think the answer would be $ \frac{12}{51}$ because there are twelve Hearts left after one is chosen with 51 total cards, but I might be approaching this problem the wrong way. Any idea on how I should think about it, or what I did wrong?
79689
You start flipping cards over one at a time and holding them in one hand. When you have two cards of the same suit that are separated by two other cards (e.g. 4♣ 2♦ 7♥ 10♣), you discard the middle two. When two cards of the same value are separated by two other cards (e.g. 4♣ 2♦ 7♥ 4♥), you discard all four.  This game is neither Accordion nor Royal Marriage, but bears similarities to both. Somebody else  on Yahoo Answers five years ago and the response was unsatisfactory.  If someone can identify this game, I would also like to know what happens when you finish with two cards left: do you win, as in Royal Marriage, or not?
34443
If I control  and I activate his second ability and target  is the token copy a 4/4 or is it still equal to the number of cards in my opponents' graveyard?
78298
Am I allowed to look up a card during a game, if it's being played with serious Rules Enforcement? Either to read the full card from the Gatherer, or to try to find the exact card I'm vaguely thinking of.  Two possible scenarios:   An opponent has revealed a card previously, and I want to remember the exact details. (Something like the exact wording of a sorcery, mana cost, keywords on a creature, subtypes, etc. etc.) I have a "name a card and remove it from the opponent's deck" type card, and I want to, say, ask google for "Blue Gearhulk" to remember that it's that that one is called "Torrential Gearhulk".   Obviously, in casual or semi-casual games neither of these are a problem. Any reasonable opponent will just re-reveal the known card (assuming it IS still guaranteed to be known) or will accept "I name the Blue Gearhulk".  Also obviously, it seems unlikely that I (or anyone else) would get to the stage of playing at the REL without knowing the cards too thoroughly for this to be relevant.  But in principle... is it permitted? Or not? (and if not, is there a good reason, or just "we decided 'No'"?)
68651
If you have one property and another wild card, can you put them together as a set and charge rent on them, and then place the wild card separately, all on your turn? (to save your set from a deal breaker card)
78221
Yesterday my housemates and I were playing card games and someone popped this question. We tried to solve the problem, but we couldn't figure it out. This morning I woke up and I am still wondering how to solve it. Will you please help me out?
112480
Suppose that a pair of cards are selected from a standard deck of 52 cards. And since there are 26 black cards and 26 red cards, so I'm wondering if my thinking is right?   ${26!/25! * 26!/25!}$ so it would be ${26*26 = 676}$
36968
Does a player need to complete a full turn (i.e. 4 actions, pickup, infect) if during the action portion the final disease was cured? I only ask because there was only 1 player card left in the deck and I wouldn't have been able to complete a full turn, which in the rules, signifies that players have lost. However, I knew that I could cure the final disease in one action.
76744
Assume you have 52 cards in a deck, and you have to draw 10 cards. Assuming you have to put back the card after each draw, what is the probability that you will not draw any repeated cards?  Here's what I was thinking...   Draw 1: $\frac{52}{52}$ since first card is allowed to be any card. Draw 2: $\frac{51}{52}$ since you don't want to draw the first card, you're now allowed to draw 51 of the remaining cards.  ... Draw 9: $\frac{44}{52}$ since you don't want to draw the last 8 cards, you're now allowed to draw 44 of the remaining cards. Draw 10: $\frac{43}{52}$ since you don't want to draw the last 9 cards, you're now allowed to draw 43 of the remaining cards.   So $\frac{52}{52} \times \frac{51}{52} \times ... \times \frac{44}{52} \times \frac{43}{52} = 0.397$ or $39.7\%$ chance of not getting a repeat.  Just wanted to check if this is the correct approach to the problem or not.
18379
How many 5-cards poker hands are there containing at least 3 of the 4 suits?  I don't how know how to properly handle this question.  There 13 ranks of each card, and 4 suits  My Attempt:  Begin with the first 3: C(13,3) X 4 ways to choose suits  Then pick last 3 cards of anything:  C(49,2)  My solution: 1,345,344  Thanks,
99610
I'm not a pretty much fun of Poker, but I'd like to study that game.  What are the probabilities of getting a Straight flush in a  game considering this factors?    Number of playersHow are cards dealtWho is the first player
31456
The following is the problem put to me:  A 5-card poker hand is dealt from a well shuffled regular 52-card playing card deck. Find the probability that the hand is a Flush (5 nonconsecutive cards each of the same suit).   I am completely unfamiliar with poker, and just learning the principles of probability. I would appreciate some help understanding the problem, and figuring out how to proceed from there.  For example: when it says "5 nonconsecutive cards," does it mean no single card can be consecutive to any other, or does it only mean that they cannot all 5 be consecutive to one another? (Same goes for other combinations- are 4 in a row disallowed? 3 in a row? 2? Are 2 and Ace considered consecutive?)  For the sake of beginning the problem, I'm going to assume none of them can be consecutive to any other, and that the 2 and the Ace are not considered to be consecutive. If I'm wrong I would appreciate someone correcting me.  Here's how I think I might begin $\ldots$  First, we're choosing cards of the same suit, so we have 13 of any one suit.  None of these can be next to each other, so we'd have to consider the different ways they could be alternating.  They could either be in the alternating slots: 2,4,6,8,10,Q,A , of which there are 7 possibilities.  Or in the alternating slots: 3,5,7,9,J,K , of which there are 6 possibilities.  So I figure, the answer to this would be:  $(7C5 * 6C5) / 52C5$  Can anyone either verify or correct my assumptions about the nature of the question, and point me in the right direction if my logic/solution is incorrect? Thank you very much.  P.S. I assume this question is asking ONLY about the probability of getting a flush, rather than a straight flush or a royal flush (which I believe is a "thing"), since it's asking about nonconsecutive cards specifically.