How do I calculate the odds of throwing a special die 50 times?
102501
The previous answer 50C30*(1/3)^30 is incorrect.  This is known as a set of Bernoulli trials (an experiment that can end in success or failure), and the probability of getting exactly 30 successes (landing on one particular side) is:\n\n50C30*(1/3)^30*(2/3)^20\n\nwhere 50C30 (read 50 choose 30) is the number of ways of selecting 30 objects from a group of 50 (the 30 objects here are the 30 flips of the 50 that yield the particular side).  It is computed:\n\n50C30 = 50!/(20!30!)
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.
56872
with dice, ones r worth 100,,3 ones are a thousand, 5's are worth 50,,.,.3 5's are worth 500, and so on,,,
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.
981433
I don't think there's anyone reference for this, at least not an authoritative one, although you might look at "True Odds" by Walsh.\n\nIf you wanted to know the probability of particular events happening, I think it'd be easier to look them up on-line individually.  For instance:\n\nhttp://golf.about.com/od/faqs/f/holeinoneodds.htm
547199
I'm assuming that you want the probability of drawing a heart followed by a queen.\nP(heart)=(13/52)=(1/4)\n\nAssume that the 1st card wasn't a queen, so (4/51) \n1st card is a queen, so (3/51)\nP(queen) = (4/51) or (3/51)\n\nor = add\n\nso P(queen) = 4/51 + 3/51 = 7/51\n\nP(heart) and P(queen) = 1/4 and 7/51\n\nand = multiply\n\nso P(heart/queen) = 1/4 x 7/51 = 7/204\n\nHope this helps\n\nTake care
360679
1, 2, 3, 4 are not greater than 4.  There are 6 numbers on a die.  4/6 or 66.6% chance that the number is not greater than 4.
1179755
50% For the first\n49.02% for two in a row\n48.98% for three in a row
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
698606
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)
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
242742
Remember, chance events are not dependent on the previous outcome.  The correct answer, as others have pointed out, is 50%.
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?
302246
if the order you draw them is important its a permutation.\n\nthe answer is basically; a 1 in 55 chance for the first card, a 1 in 54 chance for the second card (you've taken the first card out) and a 1 in 53 chance of getting the final card. all three events are independant of eachother. so the answer is 55 x 54 x 53 = 157410\nso 1 in 157410 chance.\n\nif the order you draw them is unimportant them its a combination.\nfor the first card there is a 3 in 55 chance\nthe second card has a 2 in 54 chance\nand the final card has a 1 in 53 chance\nthe answer is: (3/55)  x (2/54) x (1/53) = 1 in 8745
664305
For that to happen, 1 must come on each dice.\n\nThus P(Sum is 2) = P(1 appears)*P(1 appears)\n\n=(1/6)^2 = 1/36
857483
You could do something like this:\n\nSuppose that we play a game where there is a token placed on a board that has spaces to the left and to the right.  The game will consist of rolling a die and subsequently moving the token left on odd numbered rolls and moving it right on even numbered rolls - that is, on roll #1 move to the left & on roll #2, move to the right, etc...  The object of such a game might be to see how many rolls it takes to end up at the same space where the token started.\n\nNow suppose that we modify the game in the following way:  We introduce two different collered dice - one red and one green - that are used alternatively on each roll.  Every time the red die is rolled, you move the token in the direction indicated by whether it is an odd numbered roll or whether it is an even numbered roll.  However, whenever the green die is rolled, you move the token in the direction OPPPOSITE of the rule indicated by the odd/even numbered roll.  \n\nOne possible outcome of playing the modified game would be that you start out rolling a 5 on the red die - and since this would be roll #1, you would move the token 5 places to the right.  On the next roll, you roll a 4 on the green die.  Since this is roll #2, you would normally move the token 4 to the left.  However, the green die changes the direction.  Therefore, the token would move 4 to the right.  The effect would be the same as if a 4 were rolled on the red die.\n\nLastly, associate the operation of addition with that of the direction indicated by even numbered rolls and the operation of subtraction with that of the direction indicated by odd numbered rolls.  Also, associate the concept of positive numbers with that of rolling a red die and the concept of negative numbers with that of rolling a green die.\n\nI don't know if this is what you are looking for.  I hope it helps!
985314
its 1/36 chance of any two particular numbers coming up on each roll of the dice.  (1/6 is for one die)