There are 100 types of cards, each type with a number from 1 to 100 on it. Each minute I am given a random card. Random means that the number of my card has an equal chance of being 1,2,3,...,100. I stop when I have cards of each number, that is, I stop collecting once I have each number on at least one card. How many minutes is this process expected to take? That is, find the expected value of the number of minutes I have to wait.  E.g. I could potentially get duplicates so this method will take a lot more than 100 minutes. Once I get 99 different cards, I will stop as soon as I get the other one which I haven't got yet.  Thanks for helping everyone!
580
What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die?  The order in which the numbers appear does not matter.  I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation.  We were given the answer $(1-(\frac56)^n)^6 = .5$ or $n = 12.152$  Can someone please explain this to me, possibly with a link to a general topic?
18104
I've just played a game with my kids that basically boils down to: whoever rolls every number at least once on a 6-sided die wins.  I won, eventually, and the others finished 1-2 turns later. Now I'm wondering: what is the expectation of the length of the game?  I know that the expectation of the number of rolls till you hit a specific number is  $\sum_{n=1}^\infty n\frac{1}{6}(\frac{5}{6})^{n-1}=6$.  However, I have two questions:   How many times to you have to roll a six-sided die until you get every number at least once?  Among four independent trials (i.e. with four players), what is the expectation of the maximum number of rolls needed? [note: it's maximum, not minimum, because at their age, it's more about finishing than about getting there first for my kids]   I can simulate the result, but I wonder how I would go about calculating it analytically.    Here's a Monte Carlo simulation in Matlab  mx=zeros(1000000,1); for i=1:1000000,    %# assume it's never going to take us &gt;100 rolls    r=randi(6,100,1);    %# since R2013a, unique returns the first occurrence    %# for earlier versions, take the minimum of x    %# and subtract it from the total array length    [~,x]=unique(r);     mx(i,1)=max(x); end  %# make sure we haven't violated an assumption assert(numel(x)==6)  %# find the expected value for the coupon collector problem expectationForOneRun = mean(mx)  %# find the expected number of rolls as a maximum of four independent players maxExpectationForFourRuns = mean( max( reshape( mx, 4, []), [], 1) )  expectationForOneRun =    14.7014 (SEM 0.006)  maxExpectationForFourRuns =    21.4815 (SEM 0.01)
120466
I have a set of documents numbered 1 to n. If I arrange the documents randomly, what is the chance of selecting the last k pages in the correct order if I select them at random from the documents?   Because order is important, I use the following formula:  $\displaystyle\frac{n!}{(n-k)!}$  From this I calculate how many ways there are to select k pages from n pages in the correct order. I need to have the last k pages. Which I can do in only one way. So is the chance   1/$\displaystyle\frac{n!}{(n-k)!}$?   It seems unlikely. It seems to me spontaneously that the chance should be smaller.
83982
I've seen many variations of this problem but I can't find a good, thorough explanation on how to solve it. I'm not just looking for a solution, but a step-by-step explanation on how to derive the solution.  So the problem at hand is:   You have m balls and n bins. Consider throwing each ball into a bin uniformly and at random.    What is the expected number of bins that are empty, in terms of m and n? What is the expected number of bins that contain exactly 1 ball, in terms of m and n?   How would I approach solving this problem?  Thanks!
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.
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...
116543
A deck of cards contains $52$ cards. There are $13$ different kinds of cards, with four cards of each kind. (Among the terms commonly used instead of “kind” are “rank,” “face value,” “denomination,” and “value.”) These kinds are twos, threes, fours, fives, sixes, sevens, eights, nines, tens, jacks, queens, kings, and aces. There are also four suits: spades, clubs, hearts, and diamonds, each containing $13$ cards, with one card of each kind in a suit. A hand consists of five cards.  How many possibilities are there for the following poker hands?:  (1) Full house = $3$ of one kind and $2$ of another kind  (2) Straight = $5$ consecutive kinds (note that Ace can be high or low)   (3) $4$ of a kind  (4) Straight flush = five cards of the same suit of consecutive kinds  I calculate answers for each above, but I am not totally sure about my answer. Can someone take a look? Thanks in advance!  For (1), $C(13, 2) \cdot 2 \cdot 4^5$ since we choose $2$ kinds from $13$ kinds, and $3$ cards can have one of the kind, while the other $2$ cards can have the other kind. Since there are two choices for which kind to be for the three kinds, we multiply by $2$. Each card can be one of four suits, so we multiply $4^5$.  For (2), there are $10$ consecutive kinds. Each cards can be one of $4$ suits. Thus, $4^5 \cdot 10$.  For (3), I think there are $13$ kinds in total, so the answer is $13$.   For (4), there are $10$ consecutive kinds and four suits totally. So $10 \cdot 4 = 40$.
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).
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.
95556
The following is a question I've made myself, but I need help in solving it:  Suppose there are 100 balls in a box. 20 balls are blue, 30 balls are green and 50 balls are yellow. Now we randomly pick out 10 balls out of the box (one ball after the other) and we don't put the balls back in the box.  What's the probability of the 10th ball being picked having color blue?  I tried thinking of all the possibilities of the the 10th ball being blue, divided by all of the possible combinations. I tried for several hours and couldn't figure out neither of them. Help is appreciated!
8995
There are $n$ types of balls in an urn; $a$ balls of type $1$ , $b$ balls of type $2$ , $c$ balls of type $3$ and so on. Now balls are drawn until a ball of type $1$ is obtained with condition that if a ball of type $i$ is drawn at any moment ( $i \neq 1$ ) then all balls of type $i$ will be removed from the urn. What is the expected number of draws done? (In one draw, all balls of that particular type are removed)  My approach for $n = 3$ is like:  expectation value = $a/(a+b+c)+2(ba/(a+b+c)(a+c)+ca/(a+b+c)(a+b))+3(bca/(a+b+c)(a+c)a+cba/(a+b+c)(a+b)a)$  But it's messy , is there any general form for n type of balls ?
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!
18507
Suppose there is a fair six faced dice, and you roll it once at each round. The rule is as follows: 1). If you roll 1 or 2 at an odd-number round, you win the game and the game ends immediately. 2). If you roll 1 or 2 at an even-number round, you lose the game and it ends immediately. 3). If you roll 3 at any round, then it is a draw and the game ends immediately. If two consecutive $3$'s appear then the game ends immediately in a draw. 4). Otherwise the game goes on.   Now, what is the probability that you will win the game?  I attempted to compute the probability of winning at the 1,3,5,...-th round separately and sum them up. However, due to rule 3) it is ever more complicated to compute this for large round numbers.  Another attempt was to calculate the probability of stopping at each round and to get a closed form expression for this sequence, but this is also hard, because it depends backwards on the previous rounds in an endless manner.  Simply put, if we only consider 1) and 2), or only consider 3), the problem isn't hard, but a combination of the three really messes things up.  Can anyone help? Thanks in advance.  EDIT: I'm very sorry for the mistake when I transcribed rule 3), I have corrected it.
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,
8434
I have a very simple problem.  Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13.  If there were only 13 cards I could easily use the $\ \frac{!n}{n!}$ for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game     1st card: 4 - Continue   2nd Card: A - continue   3rd Card: K - Continue   4th Card: K - Continue   5th Card: 6 - Continue   6th Card: 9 - Continue   7th Card: 10 - Continue   8th Card: A - Continue   9th Card: J - Continue   10th Card: 3 - Continue   11th Card: 2 - Continue   12th Card: 8 - Continue   13th Card: A - Continue   1st card: 5 - Continue   2nd Card 2 LOSE   So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability?
37556
Each box of a certain breakfast cereal contains one of ten different coupons, each with the same probability. We win a prize if we manage to obtain a complete collection of all the different coupons. How long on average do we have to wait? (For example, suppose we draw the following coupons in order: 5, 2, 3, 7, 5, 1, 1, 4, 8, 4, 1, 9, 10, 2, 3, 3, 6. With the last coupon 6, we have completed our collection, and so we stop after 17 steps.)
76514
folks,  i am new to this forum and not a math expert. so please bear with me if am asking silly questions.   The question is "probability of getting 50 heads from tossing a coin 100 times".  So the answer for this is, I guess, ${100 \choose 50} (2 ^{-100})$.  So all am trying to get is easier way to calculate ${100 \choose 50}$, or another approach to the parent problem only.  Thanks all, appreciate that.  raj