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Probability
| March 22nd, 2010 at 9:43:30 PM permalink | |
| kerichu Member since: Mar 22, 2010 Threads: 1 Posts: 1 | 13 players are each given 4 cards from a 52 card deck. what is the probability that each player has one card of each suit? |
| March 23rd, 2010 at 12:24:43 AM permalink | |
| miplet Member since: Dec 1, 2009 Threads: 3 Posts: 552 |
1 in about 61,204,166,001 assuming I don't have any typos and/or brainos. |
| March 23rd, 2010 at 4:09:16 AM permalink | |
| Croupier Member since: Nov 15, 2009 Threads: 54 Posts: 1095 | Wow, that number is a lot bigger than I thought it would be. Goes to show how little I know about actual probability, and maths in general. [This space is intentionally left blank] |
| March 23rd, 2010 at 6:36:59 AM permalink | |
| cardshark Member since: Nov 30, 2009 Threads: 6 Posts: 212 |
Nevermind, it was me who read the question wrong! I agree with miplet. |
| March 23rd, 2010 at 6:47:33 AM permalink | |
| boymimbo Member since: Nov 12, 2009 Threads: 12 Posts: 2533 | The odds that ALL players have exactly one card from each suit is: [Edited for correction to numbers - see below] 52/52 * 39/51 * 26/50 * 13/49 * 48/48 * 36/47 * 24/46 * 12/45 * 44/44 * 33/43 * 22/42* 11/41 .... * 4/4 * 3/3 * 2/2 * 1/1 = 1.63388 E-11 = 1 in 61,204,166,001 as Miplet calcuated. My guess is that the odds that ANY one player has exactly one card from each suit is 1-((1-(39/51*26/50*13/49))^13) = .765276891 = 76.5% But I think the actual answer will require combinatorial analysis. -----
You want the truth! You can't handle the truth! |
| March 23rd, 2010 at 6:51:57 AM permalink | |
| cardshark Member since: Nov 30, 2009 Threads: 6 Posts: 212 | Nevermind, I read the question wrong! |
| March 23rd, 2010 at 7:08:59 AM permalink | |
| DJTeddyBear Member since: Nov 2, 2009 Threads: 105 Posts: 5727 | 52/52 * 39/51 * 26/50 * 13/49 * 48/48 * 36/47 * 24/46 * 12/45 * ... * 4/4 * 3/3 * 2/2 * 1/1 I.E. The first card can be anything. The second can be any of the other 3 suits. The third can be any of the other 2 suits, the fourth card can be any of the last suit. Continue for the other 12 players. Superstitions are silly, childish, irrational rituals, born out of fear of the unknown.
But how much does it cost to knock on wood? |
| March 23rd, 2010 at 7:34:14 AM permalink | |
| boymimbo Member since: Nov 12, 2009 Threads: 12 Posts: 2533 | Thanks. Corrected. -----
You want the truth! You can't handle the truth! |
| October 14th, 2011 at 4:43:31 PM permalink | |
| statman Member since: Sep 25, 2011 Threads: 12 Posts: 95 |
In the first hand, the diamond could be any one of 13 cards; the same for the club, heart, spade, so the number of ways the first hand could be formed is 13^4. The second hand could be formed in 12^4 ways. For the third hand get 11^4 ... For the next to last hand get 2^4 For the last hand it is 1 so the number of ways of forming the hands is SUM(x = 1 to 13, x^4) The number of ways 52 cards could be dealt any which way is 52!, so the final formula is SUM(x = 1 to 13, x^4)/52^4 I get 1.1067817962E-63 Is that what you guys got? A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant |
| October 14th, 2011 at 5:02:52 PM permalink | |
| CrystalMath Member since: May 10, 2011 Threads: 3 Posts: 476 | I calculate 1:61204166001, which is what Miplet said. I heart Crystal Math. |
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