[math-fun] RE: [mat, h-fun] "Random" Deck?
R. Wainright asks: << Recently, my daughter asked the following question: After shuffling an ordinary deck of cards, what is the probability of having none of the cards in their proper place with respect to rank only? That is, ace through king, ace through king, etc. (four times). In this problem the suit of each card does not matter. Any help or direction to a reference would be appreciated. Thank you.
For a deck of n cards, this probability is well-known to approach 1/e (where e = 2.7218281828... is the base of natural logarithms) as n -> oo. So this is already a pretty good approximation for n = 52. 1/e = .3628794411714423 to 16 decimal places. The answer turns out to be exactly 1 - 1/2! + 1/3! - 1/4! + ... - 1/52! where k! denotes k(k-1)(k-1)...3.2.1. According to my calculation, (which may be wrong in the last decimal place) this comes out to .3678794411714424 to 16 decimal places (agreeing with 1/e at least through the 15th decimal place) though with more work this probability can be expressed precisely as a fraction. --Dan Asimov
Dear Dan: Actually not. Notice that the question didn't require having the _suites_ in order. What Dan is referring to is the mixed-up hats problem -- in the old days, you checked your hat at a restaurant -- what is the probability that when a group of n people left, and the hats were picked randomly, that no person would get his own hat. This question is subtly different, in that no person would get his own hat _class_, where each class consists of 4 hats. At 02:13 PM 4/1/2005, Daniel Asimov wrote:
R. Wainright asks:
<< Recently, my daughter asked the following question: After shuffling an ordinary deck of cards, what is the probability of having none of the cards in their proper place with respect to rank only? That is, ace through king, ace through king, etc. (four times). In this problem the suit of each card does not matter.
Any help or direction to a reference would be appreciated. Thank you.
For a deck of n cards, this probability is well-known to approach 1/e (where e = 2.7218281828... is the base of natural logarithms) as n -> oo. So this is already a pretty good approximation for n = 52. 1/e = .3628794411714423 to 16 decimal places.
The answer turns out to be exactly 1 - 1/2! + 1/3! - 1/4! + ... - 1/52! where k! denotes k(k-1)(k-1)...3.2.1. According to my calculation, (which may be wrong in the last decimal place) this comes out to .3678794411714424 to 16 decimal places (agreeing with 1/e at least through the 15th decimal place) though with more work this probability can be expressed precisely as a fraction.
--Dan Asimov
Whence a rough approximation of the answer should be e^-4, or about 1.8%. (This isn't going to be as good an approximation as e^-1 is for the shuffled hats problem.) Paul At 3:29 PM -0800 4/1/05, Henry Baker wrote:
Dear Dan:
Actually not. Notice that the question didn't require having the _suites_ in order.
What Dan is referring to is the mixed-up hats problem -- in the old days, you checked your hat at a restaurant -- what is the probability that when a group of n people left, and the hats were picked randomly, that no person would get his own hat. This question is subtly different, in that no person would get his own hat _class_, where each class consists of 4 hats.
At 02:13 PM 4/1/2005, Daniel Asimov wrote:
R. Wainright asks:
<< Recently, my daughter asked the following question: After shuffling an ordinary deck of cards, what is the probability of having none of the cards in their proper place with respect to rank only? That is, ace through king, ace through king, etc. (four times). In this problem the suit of each card does not matter.
Any help or direction to a reference would be appreciated. Thank you.
For a deck of n cards, this probability is well-known to approach 1/e (where e = 2.7218281828... is the base of natural logarithms) as n -> oo. So this is already a pretty good approximation for n = 52. 1/e = .3628794411714423 to 16 decimal places.
The answer turns out to be exactly 1 - 1/2! + 1/3! - 1/4! + ... - 1/52! where k! denotes k(k-1)(k-1)...3.2.1. According to my calculation, (which may be wrong in the last decimal place) this comes out to .3678794411714424 to 16 decimal places (agreeing with 1/e at least through the 15th decimal place) though with more work this probability can be expressed precisely as a fraction.
--Dan Asimov
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=Dan Asimov The answer turns out to be exactly 1 - 1/2! + 1/3! - 1/4! + ... - 1/52! where k! denotes k(k-1)(k-1)...3.2.1.
I think you left off an initial 1/0! - ...
According to my calculation, (which may be wrong in the last decimal place) this comes out to .3678794411714424 to 16 decimal places (agreeing with 1/e at least through the 15th decimal place) though with more work this probability can be expressed precisely as a fraction.
Or a computer algebrar (here Macsyma): 193937806586896328746924473226467394949530139620339174273819171217 ------------------------------------------------------------------ 527177615496365219422618541545122659969212453861982208000000000000 = 0.3678794411714423215955237701614608674458... (after including that missing 0th term).
participants (4)
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Daniel Asimov -
Henry Baker -
Marc LeBrun -
Paul R. Pudaite