what is the probability of being dealt 4 kings in a 5 card hand?
83480
By the hypergeometric formula, it's\n\n(4 choose 4) * (48 choose 1) / (52 choose 5)\n\nIf you want to understand the above, here's what it means.  \n\nFirst, there are four possible kings, and you need all of them.  4 choose 4 counts how many ways 4 of the cards could be kings; it's a combination ("choose") instead of a permutation because we don't care about the order.  By the way, \n4 choose 4 = 4! / (0! * 4!) = 1.\n\nNow, what about the 5th card?  It could be anything other than a king.  So, we get to pick 1 card out of the 48 that aren't kings.  That's where (48 choose 1) comes from.  Which is 48! / (47! * 1!) = 48.\n\nSo 1 * 48 = 48 is the number of ways you can be dealt 4 kings.\n\nThe denominator is the total number of ways to get a 5 card hand, \n(52 choose 5) = 52! / (47! * 5!)\n= 52 * 51 * 50 *49 * 48 / (5 * 4 * 3 * 2 * 1)\n= 2598960\n\nSo the final answer is 48 / 2598960 = 0.0000184689\nor about 1 time out of every 50,000 hands that you're dealt.
933079
There are 2 black 2s in a deck out of 52. The odds of selecting a black 2 is 2/52. \nThe odds of NOT selecting a black 2 is therefore 50/52.
539344
I assume the question you are asking is how many possible five card hands contain at least one pair.\n\nThere are 13 face cards from which the pair can come.  Once you have chosen from among the 13, there are (4C2) ways to choose the two suits.  There are 52-4=48 remaining cards, and you need to choose 3, so there are (48C3) possibilities for the other three.\n\nThe answer is 13 x (4C2) x (48C3) possible five card hands containing at least one pair.
107013
Maybe there's a trick to this, but if the probability is\n5%, then if (for example) you opened 100 boxes, you would\nhave "gotten something" 5 times (ie. 5% of 100 = 5).\n\nThat seems like you'd get something, on average, once\nevery 20 boxes.
600121
Your logic is flawed. Say you always pick card #2. The first time, you'll have 14.2857...% (that's the same as saying one seventh) of the chances of winning the car.\n\nThe second time, with a new set of cards, your chances that the card #2 is the winner are again 14.2857...%, and so will every new time, because your chances are always 1 out of 7. You are picking one card out of seven cards that you are given.\n\nYour reasoning would be right if you could be assured that the car card will always be a different one (that for example, it will be in the first case #4, then #6, and so on). In that case, after the 7th. guess (if not earlier), you'd get the car. But the way you explained the game, nothing prevents the car from being always for example at card #5. If that is the case, always trying card #2 will make you fail every time.
785346
Yes, 26 red and 26 black.\n\n4 suits: spades (black) clubs (black) hearts (red) and diamonds (red).\n\nThere are four each of Queens, Kings, Jacks, 4s and Aces (one of each suit e.g Queen of Spades, Queen of hearts etc.).\n\nThere are 13 cards of each suit.  Ace, 2, 3, 4,5,6,7,8,9,10, Jack, Queen, King.  \n\n(Usually 2 is counted as low and Ace is high... this is the case in most types of poker as you mentioned and many other games - there are however some exceptions and some games in which a player may choose high or low).
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.
950716
The only way to fail to win at least one prize is to loose each time.\nThe probability that you will loose one time is 1 - 0.11 = 0.89.\nThe probability that you will loose 5 times is 0.89^5.\nSo the probability that you will win at least once, or fail to loose all 5 times, is 1 - 0.89^5 = 0.44
771495
The easy part:\n\nExpected number correct: 5 = (1/4) * 20\n\nThe less-easy part:\n\nThe number of correct answers is binomially distributed, with p = 0.25 and n = 20.  You want to know about the probability that the number of success (labeled 'k') is 10 or greater.\n\nFor a more details on binomial calculations, see:\n\nhttp://www.itl.nist.gov/div898/handbook/eda/section3/eda366i.htm\nhttp://people.msoe.edu/~jorgensen/MA262_Binomial1.pdf\n\nSkip down to the middle of page 4 in this one:\nhttp://academic.csuohio.edu/holcombj/projects/Discrete_distributions.pdf
801480
4 is the answer I get
728992
The probability that the Reds win all 5 games is (0.57)^5. There is only one way that this could happen.\n\nThe probability that the Reds win 4 and lose 1 is:\n\n5 * (0.57)^4*(1 - 0.57)\n\nThe probabilities represent the probability of the Reds winning 4 times, and the probability of the Cubs winning once. We multiply by 5 because there are 5 ways that it could happen: The Cubs could win game 1 OR game 2 OR game 3 OR game 4 OR game 5.\n\nThe probability that the Cubs win 4 and lose 1 is:\n5 * (0.57)(1 - 0.57)^4.\n\nAgain, we're multiplying by 5 because there are 5 different ways this event could occur.\n\nThe probability that the Cubs win all 5 games is (1 - 0.57)^4. There is only one way that this could happen.\n\nRespectively, these probabilities are:\n\n0.0601\n0.2270\n0.0974\n0.0147
462472
I assume you mean the lotto that is done through the ping pong ball machines.  This cannot be predicted due to the random selection of numbers.  For example if there are 50 numbers available and you need to pick 5, then there are 50 available for the first number then fourty nine for the second (because you already used one!)  Then fourty eight for the third...etc.  So mathematically your odds are 50*49*48*47*46 to 1.  That 254251200 to 1.
660290
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56.\n\nThe winning combo is amongst the nice hot lucky winning numbers I listed above. Good luck!!!
630810
a) probability of rolling a 2 is 1 out 6. for 5 balanced dice\n(1/6)(1/6)(1/6)(1/6)(1/6) = (1/7776)\n\nb) isnt this the same as part a?
54347
Kings play cards on flat green stools.\n\nKingdom\nPhylum\nClass\nOrder\nFamily\nGenus\nSpecies
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.
132305
Hearts is a game involving chance but also considerable skill. If someone were very proficient at keeping track of the cards that had been played, they could probably win 30+ times in a row. Because of this, I don't think it would be able to calculate a probability. But congrats, though, 13 times is tres awesome!