What is the probability of getting either a strait or three of a kind with one deck of cards?
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
371410
Completely depends on the cards
286247
Ubergeek is right again, almost.  Half the cards are black, so prob that the first one is is 1/2.  After that, only 25 of the remaining 51 cards are, so prob that the second is black is 25/51, and prob both are is the product, 1/2 * 25/51 = 0.245098, roughly.  I used to teach this stuff.
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.
496704
Assuming that the 1, 2 and 3 are supposed to be consecutive:\n\nFirstly, for three throws of the die, the probability of throwing 1, 2 and 3 in order is (1/6)^3 = 1/216\n\nIf you throw the die 10 times then there are 8 independent chances of throwing 1, 2, 3. That's because if you have not thrown 1, 2, 3 by the time you get to the 9th throw, you cannot do it because you only have two throws left. \n\nTherefore the chance of throwing 1, 2, 3 with 10 throws of a die are (10 - 2) * (1/216) = 8/216 which is equivalent to 1/27.\n\nIf you prefer to express that as a decimal (not recommended!) then the answer is 0.0370 to 3 s f
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
727985
well the odds are TREE to ONE
438664
Some times I will use mulligans(may be 2-3 times a round).\nBUT DON'T CHEAT TOO MUCH!
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  :)
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
1137758
1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
719210
Really Good Question.\n\nThe answer is 3 socks.\n\nExplanation : Taking the case of least probability, suppose the first sock is brown and the second is grey, then the third has to be either brown or grey as there are only two colour sock. So the purpose is solved by drawing 3 socks.
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
648450
First get it straight. Algebra and probability are 2 very different things! Now:\n\n1. There are 12 marbels, so,\na) purple then red = (1/12)*(3/12)=1/48\nb) red then green = (3/12)*(3/12) = 1/16\nc) 2 greens = (3/12)*(3/12) = 1/16\nd) 2 reds = (3/12)*(3/12) = 1/16\ne) purple then green = (1/12)*(3/12)=1/48\nf) red then yellow = (3/12)*(5/12) = 5/48\n\n2. There are 9 total marbles, so:\na) purple then white = (3/9)*(1/8) = 1/24\nb) white then green = ((1/9)*(1/8) = 1/72\nc) 2 purples = (3/9)*(2/8) = 1/12\nd) 2 yellow = (4/9)*(3/8) = 1/6\ne) yellow then purple = (4/9)*(3/8) = 1/6\nf)yellow then purple = (4/9)*(3/8) = 1/6\ng) green then white = (1/9)*(1/8) = 1/72\n\n13. Red card is 1/2 and club is 1/4, add them = 3/4\n14. Dimond is 1/4 ans spade is 1/4, add them = 1/2\n15. face card is 12/52 and a 2 is 4/52, add them = 16/52\n16. A 7 is 4/52 and a 9 is 4/52, add them = 8/52\n17. Red card is 26/52 and a king of spades is 1/52 = 27/52\n18. Heart is 13/52 and quuen od diamonds is 1/52 = 14/52
992523
The probability they will not visit any of the three is (.5) x (.6) x (.65).  So the answer to the first question is 80.5 percent.\n\n.35 means there is a 35% probability.\n\nThese events are not mutually exclusive, since it is possible to visit more than one of the vacation spots in the same trip. We know this because the probabilities of the three events add up to more than 100%.
244013
First find the number of ways to have 5 spades in your hand.\nEach hand consists of 13 cards.  \nThat means 5 will be spades, and 8 will not be spades.\nThere are 13 spades in a deck and 39 cards that are not spades.\nThe number of ways of choosing 5 spades from 13 is\nC(13,5)\nThe number of ways of choosing 8 cards that are not spades from 39 is \nC(39,8)\nHence, the number of ways of having 5 spades is \nC(13,5)*C(39,8) = 1287*61523748 = 79181063676\nWe divide this number by the total possible number of 13 card hands, which is C(52,13) because you are choosing 13 cards from 52\nC(52,13) = 635013559600\nThus, the answer is\nC(13,5)*C(39,8)/C(52,13)\n= 79181063676/635013559600\n= .124691925832\nso there is approximately a .125 chance of being dealt 5 spades
249369
Is the term you are looking for "probability"?\n\nI hope this helps.