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**aces**4 kings 4 queens 4 jacks 4 10s *20 cards* half...1 Answers · Science & Mathematics · 02/02/2009

Okay, so we've got an

**ace**turned up and four cards turned down... in the 4 down cards could include the other three**aces**. Each of these must be a combination of the**aces**...1 Answers · Science & Mathematics · 27/06/2013

...39;s how I reason: There are 4

**aces**in the deck, so there are 4 ways I... think the number of ways you can hold**Aces**and another pair is: 6 * 6 * 12. Now, for...2 Answers · Science & Mathematics · 06/08/2009

qa 12 ranks remain for forming pair excluding 3

**aces**Pr = 4c3*12c1*4c2 = 4*12*6 = 288 <---- qb (choose rank) *(choose...3 Answers · Science & Mathematics · 16/05/2013

P(all four

**aces**are in one half) = 1 * 25/51 * 24/50 * 23/49 = 92/833 = 0.1104 approximately Note this is for one half not one particular half of the deck1 Answers · Science & Mathematics · 23/01/2010

... been dealt, leaving 3

**aces**and 44 other cards to deal to the others... in both the "someone has 2 or 3**aces**" calculation AND the "how many deals...1 Answers · Science & Mathematics · 16/10/2010

P(picking at least 2

**aces**) = 1 - (4C0*48C13 + 4C1*48C12)/52C13 = 1...1 Answers · Science & Mathematics · 20/09/2008

P(all hearts) =(1/4)^5 P(all

**aces**) =(1/13)^5 P(all faces) =(3/13)^5 P(2h,1s,1c,1d) =(1...1 Answers · Science & Mathematics · 07/10/2007

... of 48 cards, 5 at a time, because

**aces**are excluded from a 52-card deck. ...48,2) = 4,512 For a = 4, we drew all 4**aces**out of the 4 available**aces**, and 1 card out of the...2 Answers · Science & Mathematics · 22/01/2011

P[red joker or red

**ace**] = 1 - P[not red joker nor red**ace**] = 1 - [ 51/54 * 50/53] = 312 /2862 = 312...red**ace**only] = 2/54* 51/53 *2 (sequences) = 204/D P[2 red**aces**] = 2/54*1/53 = 2/D P[red**ace**& red joker] = 2/54 * 1/53 *2 = 4/...2 Answers · Science & Mathematics · 07/03/2011

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