Let's say, I have a deck of $m$ cards from which I will draw $n$ cards at random, watch them and put them back in the deck. I want to know; what is the change after $k$ draws, that I have seen all cards in the deck.  It is actually a variant to the problem described , however in this and  case the number of items per draw is only one.  simulation  I made a simulation where the decksize is 10 and the number of cards each draw is 4, the results are plotted below ().   personal attempt  I only got to the point to calculate the probability that a single card is not in a single draw:  $\displaystyle P = (1 - \frac{1}{m})(1 - \frac{1}{m-1})(1 - \frac{1}{m-2})(1 - \frac{1}{m-3}) = \sum_{i=1}^n (1 - \frac{1}{m-i+1}) $
85661
I'm playing the , which is an Internet adaption of the classic Panini albums that are usually published for the soccer world cup, European championship, and possibly other tournaments.  The album has placeholders for 424 different stickers. The purpose of the game is to collect all 424. The stickers come in packs of 5, which can be obtained through codes found online (or, in case of the classic printed album, bought at your local newsstand).  I make the following assumptions:   All stickers are published in the same quantity.  One pack of stickers does not contain duplicates.     How can I find out how many packs of stickers I need to acquire in order to be reasonably sure (let's say 90%) that I have all 424 unique stickers?
54353
I have this task about cards:     Consider choosing a card from a well-shuffled standard deck of 52   playing cards.           Suppose that, after the first extraction, the card   is not reinserted in the deck. What is the probability that the second   card is an ace?   Suppose that, after the first extraction, the card   is reinserted in the deck. What is the probability that the second   card is an ace?      For the first case, I've been thinking about $$\left(\frac{4*3}{52*51}\right) + \left(\frac{48*4}{52*51}\right)$$ Where I counted two examples: a)the ace was drawn already, and b) it wasn't. For the second part of the task, would it be $$\left(\frac{4*4}{52*52}\right) + \left(\frac{52*4}{52*52}\right)$$ I'm pretty sure that I'm assuming wrong, but I can't come up with anything else. Any help would be great!
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?
28043
There is a party with 20 people, and everyone writes their name down   on a piece of paper and puts it into a bag. The bag is mixed up, and   each person draws one piece of paper. If you draw the name of someone   else, you are considered to be in his "group". What is the expected   number of groups after everyone draws?   So basically if we have a loop where each person draws someone else's name, and the last person draws the first person in that list's name, we have a group. Not quite sure how to approach this problem.  Thanks for any help.
120455
I have been given $25$ objects numbered from $1$ to $25$ and a set $S = \{-2, -1, 0, 1, 2\}$. I have to choose a random number from the given set and add it to all the objects. I have to do this $5$ times. What will be the expected sum of all objects after all $5$ turns?
67864
Completely shuffle and then lay out a full deck of 52 cards in a single row (i.e. single row 52 columns). Do the same thing with a second deck of playing cards and place this deck in the second row(also has 52 columns). What is the expected total number of matches in all the columns?   A match means that the top card and the bottom card in the column are the same (same rank and suit).  I was trying to solve it by brute force but seems like it's not that easy to me.
73718
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
54265
This question has probably been asked before, but just to be clear here, I am NOT asking for the answer, I know the answer.  What i want to know is why my solution is not equivalent to the actual solution.  Three of a kind in a deck means you draw 3 cards of the same rank, and 2 other cards each of different ranks.  So. The solution in my book is the following: (13 C 1)(4 C 3) * (12 C 2)(4 C 1)(4 C 1)  The logic is the following:  1) Choose 1 out of the 13 ranks, then for the 4 cards in the chosen rank, pick 3.  2) Choose a combination of 2 ranks from the remaining 12, then for the 4 cards in each rank, choose 1.  Total combinations: 54912  My solution is the following:  ((13 C 1)(4 C 3)) * ((12 C 1)(4 C 1)) * ((11 C 1)(4 C 1))  1) My logic for the 3 of a kind is the same as the solution.   2) Choose 1 of the remaining 12 ranks, and 1 card from the 4 cards in the rank.  3) Choose 1 of the remaining 11 ranks, and 1 card from the 4 cards in the rank.  Total combinations: 109824  My solution says that the combinations are exactly double of the book's solution.  I'm just very confused as to why this is. If anyone could explain this to me I would be very appreciative, thank you.
113072
Here is a question I've often wondered about, but have never figured out a satisfactory answer for. Here are the rules for the solitaire game "clock patience." Deal out 12 piles of 4 cards each with an extra 4 card draw pile. (From a standard 52 card deck.) Turn over the first card in the draw pile, and place it under the pile corresponding to that card's number 1-12 interpreted as Ace through Queen. Whenever you get a king you place that on the side and draw another card from the pile. The game goes out if you turn over every card in the 12 piles, and the game ends if you get four kings before this happens. My question is what is the probability that this game goes out?  One thought I had is that the answer could be one in thirteen, the chances that the last card of a 52 card sequence is a king. Although this seems plausible, I doubt it's correct, mainly because I've played the game probably dozens of times since I was a kid, and have never gone out!  Any light that people could shed on this problem would be appreciated!
59151
Given the coupon collector's problem, the expected number of coupons is calculated as follows:  $E[X] = N \sum_{i=1}^N \frac{1}{i}$  This assumes we can draw one coupon at a time.  Let's assume one can draw a pack of size $m$. All coupons in a pack are independent, which means there may be duplicates in one pack. In each draw we are only interested in one coupon which we do not have yet. All other coupons are discarded for that draw. Drawing packs is repeated until we have all $N$ coupons.   How do I calculate the expected number of draws for that case?
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.
120886
Can someone explain intuitively why this is 52 C 5?  What does it mean that the "order doesn't matter?" If I wanted to pick 5 cards from a 52 card deck, why wouldn't I just do 52!/47! since there are 52 spots for the first card, 51 for the second, 50 for the third, etc.?  Any help would be greatly appreciated!
59351
When you discard cards at the beginning of the game, where do they go? The appearance is that they get shuffled back into the deck. However, after discarding key cards and not receiving them the whole game, I am now wondering if they get put in the bottom of the deck. Where do they go after being discarded?
26899
What is the probability that a shuffled standard deck of 52 cards has no two cards of the same rank together ?  I am unable to get a handle on this problem, and wonder whether there is an analytical solution ?
101379
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.
85632
Suppose there are $N$ cards, $1,2,\dots,N$.   We start drawing cards (with replacement), until each card has been drawn at least once (we stop when the last card is drawn for the first time).   Let $x$ be the number of cards that have been drawn exactly once.   What is the expected value of $x$?   The answer is supposed to be $1 + \frac{1}{2} +\dots+ \frac{1}{N}$.
98913
The probability of two dice summing to k is simple enough, make a diagram of the throws,  $$ \begin{array}{c}  &amp;|&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\\hline 1&amp;|&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7\\ 2&amp;|&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8\\ 3&amp;|&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9\\ 4&amp;|&amp;5&amp;6&amp;7&amp;8&amp;9&amp;\color{green}{10}\\ 5&amp;|&amp;6&amp;7&amp;8&amp;9&amp;\color{green}{10}&amp;11\\ 6&amp;|&amp;7&amp;8&amp;9&amp;\color{green}{10}&amp;11&amp;12\\ \end{array} $$  and the probability of a sum is just the length of the corresponding diagonal divided by $36$. For instance, $\mathrm{P}(\mathrm{sum} = 10) = \frac{3}{36} = \frac{1}{12}$. However, it's not as easy to draw a diagram for three dice or more, so how would a general formula look?