Imagine you have a deck of cards and want to be fairly sure that you draw each card once (with a perfectly fair, complete, and random deck on each draw, of course). You are drawing cards in groups of size n from the deck. What is the expected number of draws such that each card has been drawn at least once?  Similar to the , but not quite. How would one go about integrating the math for that algorithm with drawing multiple cards at the same time?  Edit: found some duplicate questions.
99775
Suppose I have a set of 100 integers. I randomly choose 10 of those, make a note of which ones I selected, and repeat the process. What is the expected number of times this process must be repeated before I see all 100 integers?  I'm also greatly interested in how this is calculated as I'm trying to increase the expected number of times this process is repeated by changing the set sizes.
73723
Is drawing multiple cards an atomic action, or do effects triggered by drawing a card trigger for each card drawn instead?  Consider , which reads:     Whenever a player draws a card, Spiteful Visions deals 1 damage to that player.   If a player were to be compelled to draw more than one card, would that player be dealt the damage all at once, or one at a time? For example, how many times would  need to be activated to prevent all damage?  Furthermore, if there exist additional effects that are triggered by damage dealt in this way, are such effects resolved together atomically or is each resultant effect resolved individually?    How about tutors, or other cards not using the word 'draw', eg:      If you would draw a card, look at the top three cards of your library instead. Put one of those cards into your hand and the rest on the bottom of your library in any order.
61101
I'm new to the game and trying to make my first constructed deck. My understanding of the card rule is I can only have four cards of one kind.  What does this precisely mean?  For instance, does this mean I can only have four rat cards total or four rat cards of several kinds? If two different cards do the same thing (like  and ), can I have four of each?
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Situation: You've built your at least 2 color deck, know exactly what number of lands you want in it, but don't know how many to put in of each basic-land type.   How do pro-players figure this out? Obviously it seems like a deck that is about 50-50 color-wise would have equal numbers of each basic-land for those colors, but what if the deck is 60-40, 70-30, 80-20, etcetera? Or perhaps its a tri-color deck with a 40-30-30 or a 50-40-10 color distribution? Obviously it will depend on the mana-curve and the cards in the deck, but I'm just looking for some general guidelines.  Not getting any land in your side-color is bad, but not getting any in your main color is even worse. How do you manage these problems to get the best chance of success?    , and  but I'm specifically interested in color distribution of lands. It would be awesome if answers could address both 40 and 60 card decks. I'm talking only about basic lands, as I believe addressing the entire issue of "What lands should I put in my deck?" is far too big for one question.
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A box contains $n$ identical balls numbered $1$ through $n$. Suppose $k$ balls are drawn in  succession.   (a) What is the probability that m is the largest number drawn?   (b) What is the probability that the largest number drawn is less than or equal to $m$?  I don't know how to solve this problem? Could you help me? For (a), I only know that the denominator will be $\binom{n}{k}$. How to find the numerator for both cases?
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If each suit is represented then we will have two cards from one suit and one card each from the remaining suits. So I am counting the ways this can happen like so -  ${52 \choose 1}$ ways of selecting the first card.  ${12 \choose 1}$ ways of selecting a card from the same suit as the first.  ${13 \choose 1}$ ways of selecting a card for each of the 3 remaining suits.  So I have -  $${52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}$$  Then I have the sample space as $52 P 5$ as we are counting with regard to order on the numerator so we need to do the same for the denominator.  So I have $$\frac{{52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}}{52P5}$$  Does that look correct?
71629
Recently I tried to solve a problem that based on coupons collector problem. Let it be sth like this:  If a package has one of 50 random baseball cards, how many packages do you need to buy to get a complete set? (or sth like this, doesnt matter)  If I need every card one time, so it makes one set that is easy, explaination is on wikipedia and I already understood it. But what if we consider to collect every card k times? (To collect k sets of cards)  How can I tried to solve this problem? I found sth about Chernoff's bound (http://www.math.ucla.edu/~pak/courses/pg/l10.pdf) but I dont get it actually if it is a solution of this problem. I need to estimate E(X) and Var(X) for k sets of cards.  Could anyone give me a hint how to solve this problem? Thanks for all answers:)  Greetings,
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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 ${26*26 = 676}$
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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???
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A pair of dice is rolled repeatedly until each outcome (2 through 12) has occurred at least once. What is the expected number of rolls necessary for this to occur?  Notes: This is not very deep conceptually, but because of the unequal probabilities for the outcomes, it seems that the calculations involved are terribly messy.  It must have been done already (dice have been studied for centuries!) but I can't find a discussion in any book, or on line. Can anybody give a reference?
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Alice and bob are playing cards. for each round of the game, player can win 1 card if he or she won the round, otherwise, he or she loses $1$. It's a $30$ rounds game, and player can hold negative amount of cards which he or she owes to the other player. the game result will be presented by series of $30$ characters, built from $A,B$.  A- if Alice won the round.  B- if Bob won the round.  What is the number of possible sequences (game results) under the following restriction: Bob started with 2 cards, finished with no cards and during the competition Bob never owed Alice cards.    if I wasn't clear enough, please tell me and I'll fix it.
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When playing with progressive Draw 2 and Draw 4 and (say that there are 2 players playing) and    Player 1 goes out with a Draw 2  Player 2 puts down a Draw 2    Does player 1 have to pick up 4 cards?
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Edit: Sorry for not providing more on this question the first time. I tried to solve it. When choosing 13 cards from a deck of 52 cards, what is the probability of choosing precisely a card with a number 6,7,8 of the same suit? For example a 6,7,8 of heart  or a 6,7,8 of diamonds, or a 6,7,8 of spades, or a 6,7,8 of clubs. All three cards must be the numbers of 6,7,8 and only of a single suit. These must be among the hand of  13 cards. My first thoughts on this question that I only have 4 possibilities of this happening for the entire deck of 52 cards. So I thought $(_4 C_1)(_{13}C _{3})$ will account for the possible number of decks that will include cards of 6,7,8 of all the same suit out of 13 cards in the hand. And $(_{52} C_{13})$ gives me all the possible combinations of 13 cards from 52 cards.  So this probability that I got would be: $\frac{(_4 C_1)(_{13}C _{3})}{(_{52} C_{13})}\approx 5*10^{-7}$. However I feel something is off.
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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 ?
67947
I'm in a computer algorithms course and have a question about basic probability. My math background includes no more than discrete math and a little calculus, so this probability question left me completely stumped. I checked for similar questions but didn't see any. If this question has been answered already I apologize in advance but I don't even know the correct terms to search to find an answer.  The question goes like this:   Suppose you're playing a set of games and the first to win 10 games  wins the set. Each player has an even chance (1/2) of winning a  game. What is the probability that the first player will win if she  has already won i games and the second player has already won j  games? How do you calculate this for different values of i and j?    Please word the explanation for a non-maths student. I need to provide a recursive formulation for a function that computes this probability then solve it using dynamic programming. No worries on recursion or dynamic programming. I can solve that once I understand how to solve for probability.
107625
An urn has $4$ balls of $4$ different colours; red, blue, green, and   yellow. I pick one ball at random at first and if it is red, I paint   it blue and return it to the urn. If it is blue, I paint it green. If   it is green, I paint it yellow. If it is yellow, I paint it red. What   is the expected number of trials to get all $4$ balls of the same   colour?   Reminder:  $$\color{red}{red}\to \color{blue}{blue}$$ $$\color{blue}{blue}\to \color{green}{green}$$ $$\color{green}{green}\to \color{yellow}{yellow}$$ $$\color{yellow}{yellow}\to \color{red}{red}$$  I am really stuck with this problem. Help!
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My problem is: suppose I have an urn containing balls of $n = 10000000$ (i.e., $10^7$) different colors, with $1000$ balls of each color (so the total number of balls is $1000n = 10^{10}$). Suppose I draw $100000000$ (i.e. $10^8 = 10n$) balls. My question is:  how do I calculate the probability that I have drawn at least 90% (in this case $1000000$ or $10^6$) of the different colors?  Many thanks in advance