A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of drawing a face?
339616
There are 12 face cards.\nThere are 26 red cards.\nThere are 6 face cards that are red.\n12+26-6 = 32\n32/52\n= 8/13
633627
Sorry for repeating some of the previous answers.\n\n(1) The probability of getting no aces (~X) is:\n\nP(~X) = (48/52)*(47/51)*(46/50)*(45/49)*(44/48) = 35673/54145\n\nso:\n\nP(X) = 1-P(~X) = 18472/54145\n\n(2) I agree with mathgirl826. The last reply is almost correct except that there are 4*3 = 12 possibilities for the selection of the two suits (not 6). Once the suits are selected there are C(13,3)*C(13,2) hands, so:\n\nP(Y) = 12*C(13,3)*C(13,2)/C(52,5) = 429/4165\n\n(3) blahb31's analysis looks good, but maybe a slightly easier way to do it is to first calculate P(Y|~X), the probability of Y when no aces are drawn. It's just the same as P(Y) would be if the deck contained only 48 cards with 12 cards (no aces) in each suit:\n\nP(Y|(~X)) = 12*C(12,3)*C(12,2)/C(48,5) = 110/1081\n\nThen we can use (this follows from P(X and Y) = P(Y) - P((~X) and Y)):\n\nP(Y|X)\n= P(X and Y)/P(X)\n= (P(Y) - P(~X)*P(Y|(~X)))/P(X)\n\nAll of the values in the above expression have already been calculated:\n\n= (429/4165 - (35673/54145)*(110/1081))/(18472/54145)\n= 1947/18472\n\nThis is about 2% higher than P(Y). The event Y which indicates a relative lack of diversity in suits, is somewhat correlated with a decreased probability of duplication in rank values, and therefore a slightly higher probability of finding any given rank such as an ace.
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
507905
if the 1st card was replaced: (1/13)(1/13) = 1/169\n\nif the 2nd card was not replaced: \n given that the first card is an ace: (1/13)(3/17)=3/91\n  \n given that the 1st card is not an ace: (1/13)(4/51)=4/663
855670
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
1092672
Its 1:220\n\n(4/52)(3/51) = 1/221 which is the probability\nThus the odds are 1/220 or 1:220
422184
There are 5 ways of getting a 6 (1-5, 2-4, 3-3, 4-2, 5-1) and two ways of getting 11 (6-5, 5-6), so the probability is 7/36.
523765
1.  I'm assuming you mean "exactly one even card."\nP(1st card is even and 2nd card is odd and die is even) +\nP(1st card is odd and 2nd card is even and die is even)\n= (20/40)(20/39)(1/2) + (20/40)(20/39)(1/2) = 0.256\n\n2.  P(1st card is even and 2nd card is even and die is even)\n=  (20/40)(19/39)(1/2) = 0.123\n\n3.  Again, I'll assume you mean "exactly one card of 3 is drawn."\nP(1st card is 3 and 2nd card is not 3 and die is less than 3) + P(1st card is not 3 and 2nd card is 3 and die is less than 3)\n= (4/40)(36/39)(2/6) + (36/40)(4/39)(2/6) = 0.062\n\n4.  P(115 or 124 or 133 or \n142 or 151 or 214 or \n223 or 232 or 241 or \n313 or 322 or 331 or \n412 or 421 or 511) = \n(4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(3/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6)\n= 0.024\n\n5. P(111 or 112 or 121 or 211) =\n(4/40)(3/39)(1/6) + (4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6)\n= 0.0060
1152612
If you're accepting the thought that every hand is random and that the probability of winning any hand is 1/2, then you're correct.\n\nThis would be a binomial distribution.  The probability of winning at least one hand = 1 - (the probability of losing all four).\n1 - (1/2)^4 = 1 - 1/16 = 15/16.\n\nThe only fallacy to the actual process is that the probability of winning subsequent hands after the first isn't going to be exactly 1/2, as the cards that were used are gone from the deck and change the odds for the rest of the deck.  But if you're looking for a good ballpark figure, you're on the right track.
480448
there were a few games we played with cards. in one you use a deck of cards 1-10 no face cards. each person got a hand of cards. the target card was turned over from the stock pile, play went around they table, each person put a series of cards down to reach the number, if you could not play you  drew and the play passed. lets say the number this hand was 7, you can lay 5+2 or 3+4 or 9-2 or 9+10-8-4 you get the idea. you can try other sites for math type games also. you can do math bingo. you can make a board game and depending on the square they land on they do +,-,* and roll 2 die for the numbers to use. You can get 10 and 20 sides dice also, in game shops or book stores. you can do something like this with dominoes also.   See the UCSMP site below.
1058117
Think of the probability of picking 4 cards from somewhere in the deck and finding them all the same... it would be something like (52/52 x 3/51 x 2/50 x 1/49).  But whatever it is, it doesn't matter.  Just call this probability p.  Now does it matter where you draw the cards from?  If I were to pick the first 4 cards, the last 4 cards, a random collection of cards, alternating cards, etc. will the results be any different?  No, the probability is always the same probability p.\n\nIt's the same with the lottery.  While the sequence 1,2,3,4,5 might seem less likely to occur than something like 13, 17, 26, 37, 42, they actually have the same probability.\n\nHowever, you may notice that dealers always deal one card to each player, then repeat.  They don't deal 5 cards to one, then 5 cards to another, etc.  The reason has nothing to do with probability and everything to do with preventing cheating. It's much easier to perhaps keep a set of cards together while shuffling (with false shuffles, palming, etc.) and then make sure they get placed at the top of the deck.  Then if you dealt them to one person, they'd get all the desirable cards.  However, if you have to alternate, then it is near impossible to arrange the cards so that one person gets the desirable cards.
1113134
We'll need two things to calculate this: the number of ways of getting a hand with exactly one pair and the total number of hands.\n\n1) The total number of hands with one pair is of the form\naabcd.\nThe number of ways of choosing the rank of card "a" is 13.  And within that rank, there are 6 different ways to choose the 2 cards.\nFor choosing "b", we have 48 cards to choose from, basically those that are not of the same rank as "a".  For "c", there are 44 ways, and for "d" there are 40 ways.  \nHowever, we are multiple counting things here, because I could choose b = 5 of clubs, c = 3 of hearts, d = king of diamonds, but I could also have chosen b = 3 of hearts, c = king of diamonds, and d = 5 of clubs.  So to account for this, we have to divide by 3!.  This gives:\n6 * 13 * 48 * 44 * 40 /6\n= 1 098 240\n2) The total number of hands is simply \n52C5\n= 52*51*50*49*48/5!\n= 2 589 960.\n\nSo the probability of getting exactly one pair is\n1 098 240 / 2 589 960\n= 9152 / 21583\n\nThe odds are\n9152 : (21583-9152)\n9152:12431
435663
The first card you draw has a 54/55 chance it's not Saddam Hussein (I'm assuming he's the only Hussein in the deck). The second card has a 53/54 chance; the third 52/53 and so on.  When 14 of these are multiplied, everything cancels out to 41/55, or about 74.5%.
869566
12 in 51\n\nDraw one card any card.\nNow what is the chance that the second card is of the same suit.\nSay the first card is a heart.  There are 12 hearts left in a deck that now have 51 cards so 12/51.
939050
Assuming order matters on the first 5 numbers, there are 39 x 39 x 39 x 39 x 39 x 19 combinations of draws.  Only one matches with your selection making the odds\n1 in 1,714,259,781
279356
This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)
307770
Shuffling is pretty complex as you probably mean a riffle shuffle or something like that.  But let's assume you're shuffling by spreading out all the cards on the table, and randomly picking them.\n\nAssuming you want the cards in a specific order, then for each card, you'd have to randomly select the "right" card.  For the first card, you have 52 to choose from, and a 1/52 chance of getting it right.  For the second card, you now have 51 to choose from and thus a 1/51 chance of getting it right...on down to the last card.\n\nSo your odds are:\n\n1/52 * 1/51 * 1/50 * 1/49 * ... * 1/3 * 1/2 * 1/1\n\nor 1/52! or about 1/8*10^69.  To put that in perspective, lets say all 6 billion people on earth are shuffling one trillion times a second (man is that fast shuffling!) for a billion years, you're still not even in the ball park of being close (1/10^30) of having one person get it right.