bridge hand is made up of 13 cards.   A royal flush (5 highest cards of a single suit) in poker.
211936
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
650618
Go to the California lottery website. I think Virgina does. It says you can play up to 20 draws in advance.
539410
How many sides are there on each die?
667245
Don't know, I have only ever won ladies on poker games!
416337
He already knows you were right and neither needs nor wants to hear it from you. You've "won" already. Leave it at that.\n\n As Kenny Rogers once said in his famous song The Gambler, \n\n"You got to know when to hold em, know when to fold em, \nKnow when to walk away and know when to run."\n\nWell, misschatea, you're already holding a Royal Flush. This ain't the time to draw another card.
309515
a)P(full house) = P(XXYYY or XYXYY or XYYXY or ...), where each combination keeps order in mind.\n\nnow\n\nP(XXYYY) = P(22333) + P(22444) + P(22666) + ...\nP(22333) = (4/44)(3/43)(4/42)(3/41)(2/40) , which you would find is the same as P(22444) and P(22666) and whatever combination you can think of.  There are 11*10 = 110 ways to select the two different types of cards(11 because 2, 3, 4, 6, 7, 8, 9, J, Q, K, and A), So\n\nP(XXYYY) = 110(4/44)(3/43)(4/42)(3/41)\n*(2/40)\n\nYou would see that P(XXYYY) = P(XYXYY) = P(XYYXY) =..., and there are 5C2 = 5!/(2!3!) = 10 ways to arrange 2 X's and 3 Y's, so\n\nP(Full house) = 1100(4/44)(3/43)(4/42)(3/41)\n*(2/40) = 165/67880 = 0.00243\n\nb) P(straight) = 0 because a straight has to have either a 5 or 10 in it.\n\nc) P(flush) = P(All H's) + P(All S's) + P(All C's) + P(All D's)\n\nP(All H's) = (11/44)(10/43)(9/42)(8/41)\n*(7/40)\n\nAnd you will see that P(All H's) = P(All S's) = P(All C's) = P(All D's), so\n\nP(flush) = 4(11/44)(10/43)(9/42)(8/41)\n*(7/40) = 3/1763 = 0.0017\n\nSorry about having to line break sometimes; Yahoo Answers apparently doesn't like when you have a whole bunch of numbers all in a row.\n\nedit: swalker5037, I'm trying to figure out why our answers differ slightly.  One thing I notice is that in a) you are computing P(PTTT|P), meaning the probability that the final four cards are PTTT given that the first is P.  Now in order to find P(PPTTT), you need to compute P(PTTT|P)P(P), which I'm not sure you did.  However if that were the case then your answer would be much more different from mine.  In c), we have the same reasoning, but we have slightly different answers.\n\nAnyway, close enough.
233739
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
586534
money can be made in blackjack.  a bunch of students from mit teamed up and made a nice chunk of cash in vegas on blackjack.  interesting story.  cardcounting gives you an advantage on the house but its also prohibited in the casinos.  check out the link i provided, its pretty sweet.
31877
Here are some code snippets that might be useful.\nNote: I wrote these down, did _not_ compile them, or test them.\n\nThe deck of cards is a array of 52 unsigned integers,\nThey can take a value of 0-52.\n\n//Create an array to hold the deck:\n\nunsigned int Deck[52];\n\nfor(unsigned int i = 0; 1 < 52; i++)\n{\n   Deck[i] = i;\n}\n\n// Shuffle the deck\nsrand(); // Initialize the random number generator\n\nunsigned int temp;\n\n// The shuffle works this way. First round, every card gets a chance\n// to be in slot 51. Second round, each of the remaining cards gets a \n// chance to be in slot 50, third round etc.\n\n\n// Note: This method allows an array value to be swapped with itself, i.e. not moved, which is what you want to get all possible shufflings of the deck.\nfor(unsigned int j = 52; 0< j; j--)\n{\n   // Generate an index in the range 0-j\n   unsigned int k = rand( void ) % (j);\n\n   temp = Deck[k];\n   Deck[k] = Deck[j-1];\n   Deck[j-1] = temp;\n}\n\n///////////////////////////////////\n\nunsigned int GetFaceValueOfCard(unsigned int n)\n{\n  return Deck[n] % 13 + 1;    // Returns a value 1-13\n}\n\n// If you want to get fancy:\n\nchar GetSuitOfCard(unsigned int n)\n{\n   char Suit[4] = "HDCS";\n   return Suit[n/13];\n}
25444
Yes it is a game you play with 2 or more people. Everyone shakes theirs hand and counts to 5 or whatever then show your hands flat for paper, fist for rock, first two finger like sissors and one finger for a stick. Which ever is stronger wins. Like rock breaks sissors and so on. Hope this helps:)
150818
http://images.search.yahoo.com/search/images?p=poker%20hand&fr=FP-tab-web-t&toggle=1&cop=&ei=UTF-8
182108
every time you hit the ball, you add a stroke.  if you go out of bounds or in the water, you add 2 strokes.  if the par for a hole (the amount of strokes you should be able to get the ball in the cup) is 5, and you hit 4 strokes, you get a -1 for the hole.  if you hit it in in 5 strokes you get par.  1 over par is a bogey.  (6 strokes on a par 5)  2 strokes over par is a double bogey.  if you get one under par, its a birdie.  2 under par is a eagle.
375777
for the checkers... if they eached played 5 games that would mean there were 10 games. it never says they played each other. so the answer would be 5
531071
someone cheated at the card game, presumably the dead guy.he's holding bicycle 52. bicycle is a brand of playing cards.
107188
The answer is 3/26.  There are 6 red face cards out of 52 total cards.
453386
21 over 40\n1 over 5\n83 over 132\n8 over 9
465612
The answer is obviously 24.\n\nThe only time that swapping the hands can result in the same hand position is when the hands are at the same position (e.g. 6:30), or when they are at opposite positions (e.g. 6:00).  Since the hour hand can only take on 12 different values, the answer is 2*12=24.