Wikipedia:Reference desk/Archives/Mathematics/2009 November 3

= November 3 =

Probability of getting 3 of a kind in a 5 card poker hand.
Yes this is a hw problem for me but I've tried it and keep getting stuck. The way I think it should be solved is 49 choose 2 / 52 choose 5. The 49 choose 2 represents the favorable outcomes --- there are 49 other cards besides the 3 of a kind and you want to select 2 out of these. The total number of hands would be 52 choose 5. I'm thinking that the 49 choose 2 doesn't completly make sense because there are 4 of each kind though. The answer I've been given is .2. If I solve it the way I described above I get an answer thats wrong by a factor of about 50. I'm thinking I should multiply the answer by 52 -- I get a solution that's close, but I can't figured why I would do that. 66.133.196.152 (talk) 00:09, 3 November 2009 (UTC)
 * This is how I'd do it. First compute how many different three-ace hands there are, considering both that there are several ways to choose the three aces from the four in the deck and that there are several ways to choose the two remaining cards from the 48 non-aces. Then simply multiply that by 13 to get the total number of three-of-a-kind hands. — JAO • T • C 00:35, 3 November 2009 (UTC)
 * Keep in mind you probably only want the probability of three of a kind, excluding any better hands (like a full house, or four of a kind). You can first pick the rank (13 ways), then 3 suits for that rank (4 ways), and then you have to pick the other two cards to ensure you don't get a full house or four of a kind. Particularly, those last two cards had better have ranks different from each other and from the rank chosen for the three of a kind. An answer of .2 is definitely wrong; if you've played poker for any amount of time, you'd notice that. The answer would be closer to .02, though you should try to get an exact probability.Nm420 (talk) 03:08, 3 November 2009 (UTC)

Consider the hand with three aces and two other cards. We split the pack into two piles: the four aces and the 48 non-aces. There are (4 &times; 3 &times; 2)/3! = 4 ways of choosing three aces from four, and (48 &times; 47)/2! = 1,128 ways of choosing two cards from the 48 non-aces. That gives a total of 4 &times; 1,128 = 4,512 five card hands made up of three aces and two non-aces. To avoid a full house the two non-aces cannot make a pair. Since there are four 2s in the non-ace pile there are (4 &times; 3)/2! = 6 ways of making a pair of 2s. Likewise, there are 6 ways of making a pair of 3s, etc. So there are 6 &times; 12 = 72 ways of making a pair with the two cards drawn from the non-ace pile. Therefore, there are 4,512 &minus; 72 = 4,440 five card hands with exactly trip As. We can do the same for trip 3s, trip 4s, &hellip;, trip Ks. This makes 13 &times; 4,440 = 57,720 hands with exactly three of a kind. There are (52 &times; 51 &times; 50 &times; 49 &times; 48)/5! = 2,598,960 five card hands. The odds of getting exactly three of a kind from a five card hand are then 1,629-to-37 against (about 44-to-1 against). The probability is then 37/1,666 ≈ 0.02. Dr Dec (Talk)   12:47, 3 November 2009 (UTC)
 * The above solution is close, but not quite. The subtraction is being made at the wrong place. In general, how I like to solve these types of problems, is to use only multiplication of binomial coefficients (when possible of course), and of course state the "real-world" choice being made with each. Nm420 (talk) 14:45, 3 November 2009 (UTC)
 * Oh yeah, silly me. We have to take out the possibility of the pairs right from the beginning. There are 4 ways of choosing three aces from the aces pile, and there are 1,128 &minus; 72 = 1,056 choices of two non-paired cards from the non-ace pile. That gives a total of 4 &times; 1,056 = 4,224 five card hands made up of three aces and two non-paired, non-ace cards. Doing this for three 2s, three 3s, &hellip;, three Ks gives 13 &times; 4,224 = 54,912 hands with exactly three of a kind. There are still 2,598,960 five card hands. This gives revised odds of 4,077-to-88 against (about 46-to-1 against). The probability is then 88/4165 ≈ 0.02. The Poker Probability article gives a full list of odds. Dr Dec  (Talk)   15:08, 3 November 2009 (UTC)

Mensuration and geometry
Could you please explain me the difference between mensuration and geometry with examplesKasiraoj (talk) 11:49, 3 November 2009 (UTC)


 * "Mensuration" just means measurement. You could be measuring anything - time, temperature, electric current, luminosity etc. - so many measurment techniques have nothing to do with geometry. On the other hand, if you are measuring a length, an area, a volume, or related concepts such as density, speed or acceleration, then you may be using geometric techniques. For example, to measure the width of a river, you can use surveying techniques, which are an application of geometry and trigonometry. Gandalf61 (talk) 12:01, 3 November 2009 (UTC)