In Yugioh, you start out with a five card hand. You have a 40-60 card deck. I run a 60 card Exodia deck, and twice I have drawn all five pieces of Exodia in my starting hand. For those who don’t play, having all five pieces in your hand at one time causes you to immediately win the game. I would like to know what the actual odds of that happening once, let alone twice.
121653
In Yu-Gi-Oh!, there are some cards that allow to the duelist to win a duel automatically.   In this case, I'm focusing my question on . To win a duel using Exodia, you must have the five cards (parts of Exodia) in your hand.   These cards are (in no particular order):          I'm wondering:  How to calculate the probability to get all cards of Exodia The Forbidden One based on:   Parts of Exodia taken from the Deck during each Draw Phase with no additional card effects.   A Yu-Gi-Oh! deck has:   40 cards in total (35 monster, spell, trap cards and 5 Exodia cards). There are only five Exodia cards (no copies). Draw one card per turn.      I posted a  about this subject. The accepted answer should have a   to  of the . I'm looking for a detailed answer that explains how to make those calculations and the obtained results .
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,
22119
I was playing Uno, and the card was a yellow 4.  As I didn't have a yellow nor a 4 in my hand.  I played the draw 4 card.  Now should I have been challenged, I would have won the challenge.  But my friend behind me in the rotation, jumped in with another draw 4, making it a +8.  Now that he has played a +4, I am no longer challegeable, but would you be able to challenge him?  I don't think you would be able to, since it wasn't his turn so technically he didn't have to play anything, plus it was a jump in.  So can you challenge a jump in draw 4 card played?
69437
I'm completely new to Magic the Gathering. I've read the basic rulebook pretty thoroughly and have a pretty good grasp of the basic rules, but it's unclear to me as to when you can tap lands to generate mana.  First off,  mana is supposed to dry up after each phase in a turn. Can I untap my land cards and then tap them again in the next phase if I need more mana to power a spell? Or does a land card remain tapped until my next turn like everything else?  Second, the rules mention that I can play instants and activate abilities during my opponent's turn. Does that then mean that I can tap my land cards during my opponent's turn to power those spells?  Third, does my opponent's untap step for their turn only apply to their cards? Or can I untap mine as well? Not closely related to land cards, but when I can tap and untap my land cards would make a huge difference as to how I play.  Hope these questions aren't painfully obvious, but I've just started playing the game. Thanks for your time!
18723
Happy Sunday, I was wondering how to calculate how many combinations could be rolled in the game of Yahtzee in getting a large straight. A large straight is when you roll and the dice com out 1,2,3,4,5 or 2,3,4,5,6. Can any one give me some guidance???
33117
Your (honest) opponent choose a random number from 1 to 13 inclusive. You have to guess the number, and you win if the guess is correct. If not, your opponent either reduces the number chosen by one or increases it by 1, and you guess again.  The question is, what is the minimum # of attempts necessary to guarantee a win for you.  I am not able to get a handle on the problem.  Also, (a new variant just thought of), how many guesses should be allowed for a fair or "nearest to fair" game ?
89244
I'm constructing a Deck and I want to make all the mana Temple Cards. More Specifically I want them to be , , and . Are these mana cards limited to only four per deck like regular cards or can you put an unlimited amount of them in your deck?
110281
We were confused by the + action cards.  It says you have to start at the top of the card and work your way down. so when do you play a second action card and are you allowed to play another action card if you havent done everything on the first action card.  Also, can you total up all of the extra coins on all of your action cards and combine them for one buy?
102719
My friends and I are relatively new to MtG. At some point, one of them introduced us to the idea of mana weaving, and we'd do this to declump our decks prior to a very thorough shuffling. However, at the Theros pre-release, one of us was asked not to do this, and we learned it was frowned upon and ceased to do it.  I just now came across , and it sounds like mana weaving's only bad if you use it to leave your deck in an even distribution, and it's OK if your deck is then thoroughly shuffled. In hindsight, the judge may have simply thought my friend wouldn't have shuffled afterwards, but at the time I came under the impression it was frowned upon even if you did shuffle afterwards.  So: is mana weaving my deck OK, if I shuffle it very thoroughly afterwards?
39252
I'm working on a card game, which uses a non-standard deck of cards. Since I'm still tweaking the layout of the deck, I've been using variables as follows:  Hand size: $H$ Number of suits: $S$ Number of ranks: $R$ Number of copies of each card: $C$    Thus, the total number of unique cards in the deck is $S*R$, and the absolute total number of cards in the deck is $S*R*C$.  Since there are duplicates for each card, I'm trying to find the number of unique hands of size $H$ that are possible for given $S$, $R$, $C$, and $H$. If I'm remembering correctly, ${S*R*C\choose H}$ would over count the number of unique hands in this case.  How would I handle this calculation with duplicate cards?  EDIT: As an aside, how would calculations of unique number of winning hands be different with duplicate cards? I'm thinking of having general types of winning hands, such as $N$-flushes, $N$-straights, $N$-of-a-kinds, etc (restricted to $1 \le N \le H$, naturally).
120623
Say that both players each had 2 , 2 , and a . (Or any other way to get unlimited mana, and a Nantuko Shade).  Player 1 attacks with his , and player 2 blocks with his . Before damage is assigned, each has chances to pump up their  as much as they want. They can't just declare them to be "infinite," but each can always one-up the other player. What happens?  It seems there would be a loop of events, but would one player eventually be forced to break the loop as per rule 714.3? Is it considered the same game state after each player pumps their creature a bunch as it was before?  This was discussed at length on the Draw3Cards Magic Q&amp;A site (which is no longer available), but no conclusion was reached. Some felt that the game would simply be a draw, others felt that the active player would be forced to make a different decision to break the loop. Others felt that you would just have to call a judge and see what he decides.
26565
4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?  Here's how I tried to solve:  For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.  For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.  Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.  Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$  But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.  My answer differs by a factor of $4!$. What did I do wrong?
29271
I am having some trouble solving the following.  You draw cards from a standard 52-card deck without replacement until you get an ace. You draw from what is remaining until you get a 2. You continue on with 3. What is the expected number you will be on after the entire deck runs out?  It was natural to let   $T_i = \text{first position of card whose value is }i$ $U_i = \text{last position of card whose value is }i$   So the problem essentially amounts to figuring out the probability that you will be on $k$ when the deck runs out, namely:  $$Pr(T_1&lt;\cdots&lt;T_k \cap U_{k+1} &lt; T_k)$$  I can see that  $$Pr(T_1&lt;\cdots&lt;T_k) = 1/k!  \\  \text{and}  \\ Pr(U_{k+1} &lt; T_k) = 1/70$$  but could not get any further...
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.
714
What is wrong with my method of find the probability of a 2 pair?  For the first card we have 52 choices, one this is picked( call it A) we have 3 choices(different suite than A but this same value) for the second card, ( lets call this card B). Now we have AB, to we have to throw away 2 more cards so we don't have 3 of a kind or a four of a kind. Therefore one our third card we have 48 choices and on our fourth card we have 3 choices again similar to above.  On our last card we only have 44 choices left. So the probablity of getting a 2 pair should be (52*3*48*3*44)/(52 choose 5).  But this isn't correct.     Note: I edited this to "2 pair"; the original version was wrong.
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
110394
Assuming you enter the spells area of effect and end your turn inside this area on the same turn. Does it trigger twice?     A creature must also make this saving throw when it enters the spell’s area for the first time on a turn or ends its turn there.   Not a native speaker, so I am not sure whether this means you make a saving throw when entering and when ending your turn in the area, or that you have to make one saving throw when one of this happens and only once.  It seems a bit overpowered when it triggers twice, dealing 8d10 damage is ok for a 5th level one-target spell, not a lasting multi-target AoE spell (DMG p.284)