kerichu
Joined: Mar 22, 2010
• 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
Joined: Dec 1, 2009
• Posts: 1910
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.
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Croupier
Joined: Nov 15, 2009
• 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.
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cardshark
Joined: Nov 30, 2009
• 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
Joined: Nov 12, 2009
• 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.
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cardshark
Joined: Nov 30, 2009
• Posts: 239
March 23rd, 2010 at 6:51:57 AM permalink
Nevermind, I read the question wrong!
DJTeddyBear
Joined: Nov 2, 2009
• Posts: 10065
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.
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boymimbo
Joined: Nov 12, 2009
• Posts: 5988
March 23rd, 2010 at 7:34:14 AM permalink
Thanks. Corrected.
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statman
Joined: Sep 25, 2011
• 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?
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CrystalMath
Joined: May 10, 2011