What is the probability getting Straight in poker with a 52 card deck and a 5 hand where Ace is high or low?
97907
There are 10,200 (10,240 if straight flushes are counted) possible straights, so the odds of getting a straight given a five card hand are 32359:128, or approximately 253:1.
885606
I think this is only hard if you make it hard, but it's quite easy if you look at it in the right way. Since the order the cards are dealt in is arbitrary, you can assume that that the first card that was dealt was one of the two identical ones, and we can call the person who received it 'Alan' (note we are naming him AFTER he got the card, so we're not finding the probability of it going to a particular player). There are now 51 equally likely 'slots' that the other identical card can go in, and 12 of them are 'slots' that would give Alan the other identical card. So the probability is 12/51 .
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.
1056545
Find the probability of each hand and add them up.  The hands would be: straight, flush, full house, 4 of a kind (straight flush would be included in straights AND flushes, don't count them twice).\n\n\nThere are 52C5 = 2598960 total hands, that's the bottom number.\n\nThere are 4x4x4x4x4 = 1024 ways to get an A-2-3-4-5 straight\nThere are the same for each of the ten possible straights up to 10-J-Q-K-A so 10x1024 = 10240 possible straights.  40 of these would be flushes too.\n\nDo this for the other hands then add them up. Put the sum over the total.
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.
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.
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.
882956
Probability is 3 out of 5
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
998556
Find the probability of the complement occurring and then subtract that from 1.\nThe complement would be that an odd occurs 0 or 1 times.\nP(odd) = .5 and P(even) = .5\nP(0 odds) = C(10,0)(.5)^0 (.5)^10 = .0009765625\nP(1 odd) = C(10,1) (.5)^1 (.5)^9 = .009765625\n.0009765625 + .009765625 = .0107421875\n1 - .0107421875 = .9892578125
940139
26/52, 1/2 or 50%, half of a deck of cards are red, the other half black
74975
Which something?\n\nJust using roulette, a chance of winning on a single spin varies from 47.4% to 2.63% depending on which kind of bet you make: color, odd, even, a specific number, or a pair or quad of numbers, etc.\n\nFor some other event, to calculate the odds would require more information.
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)
566507
I hope you mean 52 cards.\n\nAnyway, to figure out this problem, you would need to find the number of combinations of 5 taken from 52. The equation for that is:\n52!/(52-5)!5!\nThis is equal to 52!/47!5!\nWhich is equal to: (48x49x50x51x52)/(2x3x4x5)\nThere are a total of 2,598,960 poker hands. \n\nIt's best to find the number of ways to get three kings, then two queens, then to multiply the two. \nThere are four possible kings and combinations of three. \n4!/(4-3)!3! This equals 4. \n\n4!/(4-2)!2! This equals 6. \n6x4=24\nThere are 24 possibilities of three kings and two queens. \n\n(c) is a little harder to explain. For every card Ace thru King there are 12 other cards it could make a full house with. The king can make 24 with a queen, but it can also make 24 with ace, two, three, four etc. To find the total number one must multiply 24 by 12. This equals 288. Now we must multiply that number by the total number of different kinds of cards in the deck. That would be 13. 288x13 = 3,744.\n\nThere are 3,744 ways to make a full house. \n\nTo find the probability, one must use the answer from question (a) and put the answer to question (c) over it. \n\n3744/2598960 =\n\n6/4165.\n\nOr an approximately 0.144 % chance of being dealt a full house.
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
476052
it's pretty complicated, and I'm too far removed from my math courses to keep it all straight.  But it varies based on what your point is, simply because some outcomes are more likely.\n\n4 has 3/36 odds of being the point\n5 has 4/36\n6 has 5/36\n8 has 5/36\n9 has 4/36\n10 has 3/36\n\nBut you can't roll 7.  So if we assume the point is 5, for each roll there is 32/36 odds that we don't roll 5, but 6/36 odds we roll a seven.  Reconcilling that involves equations that I've long since forgotten about.\n\nI think maybe [((8/9 - 1/6)^11)*(1/9)] would be the odds you'd roll 11 times without getting 7 or 5 and get a 5 on the 12th roll.  That'd be about .0214.  That you'd follow that up with a 7 or 11 would be .0214*(8/36), or .00535\n\nSite I link to be low discusses the outcomes a bit more formally than I'm cabaple of doing these days.
1009251
Ahh!  True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening.  But what are the odds that the seventh flip will be heads, too?