Assume Black 10, Blue 10, Red 10 and Joker are on the board as a set of four: The Joker, at this point has to be an orange 10. Questions:  Can you take one of the three non-joker 10's from the set to use elsewhere on the board? Leaving a three tile set. This would have the effect of now making that joker one of two colors, instead of just orange. Can you split up all four 10's keeping the joker an orange 10 elsewhere on the board?
122074
I have a group of tiles made up of 3 fours and a joker:     4 4 4 Joker   Now, if a player removes the joker, do they have to replace it with another 4?
7424
In how many ways can we choose k items from n distinct items put in a circle.   Then further extension of the question, if I want to find out the number of ways to choose k items from n distinct items in a circle such that no 2 are together.  Thanks
12605
I have the following problem from book "Introduction to Probability", p.32     A certain casino uses 10 standard decks of cards mixed together into one big deck, which   we will call a superdeck. Thus, the superdeck has 52 · 10 = 520 cards, with 10 copies   of each card. How many different 10-card hands can be dealt from the superdeck? The   order of the cards does not matter, nor does it matter which of the original 10 decks   the cards came from. Express your answer as a binomial coefficient.   Hint: Bose-Einstein.   My solution:  Because the number of cards of each type in the superdeck (10) is not less than the size of the hand (10), and thus not limiting, it's the same as sampling with replacement where the order does not matter,  so the number of possible 10-card hands would be $\binom{52+10-1}{10}$.  Is my thinking correct?
51880
A recent quandary came up with some friends in a game of Ticket to Ride.  If, in the last turn, you have the cards to play a route longer than the number of trains that you have left, can you do so?  Or are you limited by your number of trains?
47301
I'm working on a problem where each number is chosen randomly from 1-9. Given $n$ numbers chosen in this manner, we multiply all of these together. I'm looking for the probability that this product is divisible by 10.  I reasoned that we only get a multiple of 10 when we multiply 5 and an even number together. So if either all $n$ numbers are odd or if none of the $n$ numbers is 5, then the product will not be divisible by 10, and we can take 1 - P(Not Divisible by 10).   From here, I'm not sure how to come up with a numerical answer when there is an ambiguous number of $n$ randomly chosen numbers. Is there some way to get there? Or is this reasoning flawed?
3104
I'm stuck on what how to continue. I know I'm missing a few steps but this is what I have so far. Thank you in advance!   ¬Cube(b) → Small(b) Small(c) → (Small(d) ∨ Small(e))  Small(d) → ¬Small(c) Cube(b) → ¬Small(e)   Small(c)   ¬Small(b)   ¬Cube(b) Small(b) (→ Elim 1,7) ⊥ (⊥intro 6, 8)    10.Cube(b) (¬Intro 7-9)   ¬Small(e) (→ Elim 4,10)     Goal: Small(c) → Small(b)
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 ${26*26 = 676}$
52652
I don’t know how to handle this 3 players case, may anyone give a hand? Thanks!     When playing card game, there are 3 players playing 32 cards, each   receive 10 cards, 2 remind on the table. Overall are 4 aces and 4   kings in the game. How likely is P (E | D)? D = "The table remains 2   aces" E = "I (at least) get one king".   I guess I should have steps as follows.   P(E) = P(1King)+P(2K)+P(3K)+P(4K) P(D) = (4/32) * (4/32) = 1/256 P (E | D) = P(E) + P(D)   Equations may be used:  Counting combinations      The number of combinations of n objects taken r at a time is     nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!                 Number of favorable outcomes P(I) =     Total number of possible outcome
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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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How do I calculate the number of options for $k$ different balls in $n$ cells, with at least $1$ ball in each cell?   Lets say $6$ different balls in $3$ different cells, with at least $1$ ball in each cell.   like this:  $$x_1+x_2+x_3 = 6$$  I would like to see an explanation please and not just the answer.  Thanks!
7371
If I draw three cards at random (without replacement) from a standard 52-card deck, what is the probability that two of the cards will be black and one of them will be red?  Thanks!
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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?
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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?
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If, after drawing one of the five cards from the face up train cards, and after replacing that card, the second card is drawn from the top of the deck and is a locomotive card, is the player allowed to keep that card?     Likewise, if both cards are drawn from the the top of the stack and the second card is a locomotive card, is the player allowed to keep that card?
119318
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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There are nine pairs of gloves each of different colors in how many ways can we select six gloves such that each is of different color? I tried like this : First number of ways in which we can select 6 colors out of nine it is $^9C_6$  ways. Then we can arrange it in 6! ways and also we can select either of two gloves out of the 6 in $2^6$ ways. Hence required number of ways should be $^9C_6$  .$2^6$.6!
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Let n be an even integer. How many different ways can I choose subsets of the set {1,2...,n} if they all have to have odd size, but the intersection of any two of them has to have even size?  Please help me answer this question or lead me in the right direction. Thank you :)