Probability
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.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/
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
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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.
October 14th, 2011 at 6:43:26 PM
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Yeah, I got the same as miplet, too.
A falling knife has no handle.
October 14th, 2011 at 8:26:47 PM
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Quote: statman
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?
1.106e-63 is very small number, so I started looking at your formulas. I think I must be missing something. Sum(x=1 to 13, x^4)=1+16+81+256+625+1296+2401+4096+6561+10000+14641+20736+28561=89271, and 52^4=7311616. 89271/7311616=.012209, which is a very long way from 0.000000000000000000000000000000000000000000000000000000000000001106781796.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
October 14th, 2011 at 8:44:04 PM
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Quote: statmanQuote: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.
. ..
Is that what you guys got?
There are also 4! or 24 ways to arrange those 4 cards which I believe is where your error is.
Man Babes #AxelFabulous
October 14th, 2011 at 10:32:14 PM
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Quote: statmanQuote: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?
There are three errors in your formula:
1) It's a product, not a sum
2) Miplet already pointed this out -- there are 4! ways to arrange each 4-card hand.
3) You quoted 52! as the denominator but then wrote 52^4 in your formula.
Fixing all these yields
PRODUCT(x = 1 to 13, 4!*x^4)/52!
which gives the correct answer, 1.63387571E-11 or 1 in 61204166001.13
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 16th, 2011 at 3:41:14 PM
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Quote: MathExtremistFixing all these yields
PRODUCT(x = 1 to 13, 4!*x^4)/52!
which gives the correct answer, 1.63387571E-11 or 1 in 61204166001.13
Going by suits alone, 4!^13 / (52!/13!^4) .
Then i began to wonder, that for every ordering of the deck, there is at least one means of dealing which shall make each of those a solution. And intuitively, that some of these solutions might have more possibly-corresponding means of dealing than others. Then rendering this problem precisely-insoluble, without first observing the order of the cards before the means of dealing has been revealed, and the above number to merely an average over all means of dealing.
Even were the problem well-defined here... in practice, there is still the comparatively-likely chance of dealing (clockwise, the usual way,) from a new, "mistakenly" unshuffled "boxed rack" of cards (with the jokers removed).
Why bet at all, if you can be sure?
Anyway, what constitutes a "good bet"? - The best slots-game in town; a sucker's edge; or some gray-area blackjack-stunts?
(P.S. God doesn't even have to exist to be God.)
October 20th, 2011 at 1:52:39 AM
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Thank you for the correction, but I don't quite understand why the order of the cards in each hand is important.
A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant
October 29th, 2011 at 8:13:11 PM
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Quote: statman... but I don't quite understand why the order of the cards in each hand is important.
Eg, if the suits run in an order like spade, heart, diamond, club, on down through the deck, then you can (and might) deal out the cards in both most-typical manners to arrive at the solution: by a set of four to each the players; or singly, clockwise, one card at a time to each. Perhaps someone would check on that, as i just now ran it through my head quickly only once.
Besides that, understanding the differences from methods of shuffling to those of dealing can have implications in computer-program efficiencies, if not in also which is programed.
Why bet at all, if you can be sure?
Anyway, what constitutes a "good bet"? - The best slots-game in town; a sucker's edge; or some gray-area blackjack-stunts?
(P.S. God doesn't even have to exist to be God.)
October 30th, 2011 at 4:42:42 AM
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This event (getting 4 hands of 13 cards all of the same suit, known as 4 perfect hands in bridge) actually happened in the 1950's. The 4 ladies who got dealt the perfect bridge hand appeared as guests on both "What's My Line?" and "I've Got a Secret".
