March 22nd, 2010 at 9:43:30 PM
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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
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Quote:kerichu13 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

March 23rd, 2010 at 4:09:16 AM
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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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March 23rd, 2010 at 6:36:59 AM
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Quote:miplet1 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.

March 23rd, 2010 at 6:47:33 AM
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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.

[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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You want the truth! You can't handle the truth!

March 23rd, 2010 at 6:51:57 AM
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Nevermind, I read the question wrong!

March 23rd, 2010 at 7:08:59 AM
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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.

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. 🤗

March 23rd, 2010 at 7:34:14 AM
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Thanks. Corrected.

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You want the truth! You can't handle the truth!

October 14th, 2011 at 4:43:31 PM
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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

October 14th, 2011 at 5:02:52 PM
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I calculate 1:61204166001, which is what Miplet said.

I heart Crystal Math.