Find the probability of the following card hands from a 52-card deck. In poker, aces are either high or low. A
467993
There are only 4 different ways to have a royal flush.\nThere are C(52,5) different 5 card hands.\n4/C(52,5)\n= 4/2598960\n= 1/649740\n= .000001539077
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.
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
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.
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.
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.
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.
577657
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
940139
26/52, 1/2 or 50%, half of a deck of cards are red, the other half black
277670
probability is the # of favorable outcomes over the # of possible outcomes:\n\nprobability=\n# of favorable outcomes\n------------------------------\n# of possible outcomes\n\nfavorable outcomes would be the # of possible resultis in your favor\n\nexample:\ni have 5 cards- a joker, a jack, a queen, a king, and a ace. there is a 1 in 5 chance (1/5) that you will pull a jack...there is a 3 in 5 chance (3/5) that you will pull a royal card (jack, queen, king)\n\nin both examples, the first number (1 and 3) are favorable outcomes
465846
Depends on the game played.\n\nAnd if you mean pre or post draw, here is the predraw\nHE PROBABILITY OF BEING DEALT SPECIFIC HANDS BEFORE THE DRAW\n(52-card deck - without Joker)\n\nThe probability that you will make... Expressed in percent (%) is... The odds against it are... Number of possible combos\nStraight Flush 0.00* 64,973 to 1 40\nFour-of-a-kind 0.02 4,164 to 1 624\nFull House 0.14 693 to 1 3,744\nFlush 0.20 508 to 1 5,108\nStraight 0.39 254 to 1 10,200\nThree-of-a-kind 2.11 46.3 to 1 54,912\nTwo-Pair 4.75 20.0 to 1 123,552\nPair of A,K,Q,J\n   (Openers)  \n13.00  \n6.69 to 1  \n337,920\nShorts (10's - 2's) 29.25 2.42 to 1 760,320
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%.
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
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
714109
When there are 80 numbers with equal chance the probability is 1 / 80 for each number.\nThe chance of 20 out of 40 numbers is 1 out of \n40 x 39 x 38 x... x 22 x 21 = 40! / 20! = 3 x 10^29\nSince you have all the 80 numbers, the chance is 4 x 10^27.\nThe very, very, very most probability is that you waist your money again\n\nTh
830442
52 Cards in a Playingcard-deck
412043
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.