kerichu
kerichu
Joined: Mar 22, 2010
  • Threads: 1
  • Posts: 1
March 22nd, 2010 at 9:43:30 PM permalink
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?
miplet
miplet
Joined: Dec 1, 2009
  • Threads: 5
  • Posts: 1883
March 23rd, 2010 at 12:24:43 AM permalink
Quote: kerichu

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?


1 in about 61,204,166,001 assuming I don't have any typos and/or brainos.
“Man Babes” #AxelFabulous
Croupier
Croupier
Joined: Nov 15, 2009
  • Threads: 58
  • Posts: 1258
March 23rd, 2010 at 4:09:16 AM permalink
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]
cardshark
cardshark
Joined: Nov 30, 2009
  • Threads: 9
  • Posts: 239
March 23rd, 2010 at 6:36:59 AM permalink
Quote: miplet

1 in about 61,204,166,001 assuming I don't have any typos and/or brainos.



Nevermind, it was me who read the question wrong! I agree with miplet.
boymimbo
boymimbo
Joined: Nov 12, 2009
  • Threads: 16
  • Posts: 5988
March 23rd, 2010 at 6:47:33 AM permalink
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!
cardshark
cardshark
Joined: Nov 30, 2009
  • Threads: 9
  • Posts: 239
March 23rd, 2010 at 6:51:57 AM permalink
Nevermind, I read the question wrong!
DJTeddyBear
DJTeddyBear
Joined: Nov 2, 2009
  • Threads: 174
  • Posts: 10044
March 23rd, 2010 at 7:08:59 AM permalink
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? 😁 Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition. 🤗
boymimbo
boymimbo
Joined: Nov 12, 2009
  • Threads: 16
  • Posts: 5988
March 23rd, 2010 at 7:34:14 AM permalink
Thanks. Corrected.
----- You want the truth! You can't handle the truth!
statman
statman
Joined: Sep 25, 2011
  • Threads: 12
  • Posts: 95
October 14th, 2011 at 4:43:31 PM permalink
Quote:

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?


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
CrystalMath
CrystalMath
Joined: May 10, 2011
  • Threads: 8
  • Posts: 1745
October 14th, 2011 at 5:02:52 PM permalink
I calculate 1:61204166001, which is what Miplet said.
I heart Crystal Math.

  • Jump to: