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 Subject: probablity with cards Category: Science > Math Asked by: frodo2366-ga List Price: \$10.00 Posted: 24 Jul 2006 20:01 PDT Expires: 23 Aug 2006 20:01 PDT Question ID: 749200
 ```I came upon this decades ago and has perplexed me ever since. It came from cards but can be more generic than that. Setup: 2 ramdomly suffled card decks (52 cards) Process: the top card from each deck is turned over simutaneously, one at a time, up to 52 times Question: What are the odds that on one of the 52 "flips" identical (suit and number) cards will match. A computor simulation would be relatively easy, but I'm looking for the math behind this. Suprisingly, "manual" simulation shows it to be close to 50%, but that might just be because at "52" that's the way it works out, other starting numbers (10, 25, 10,000...) may be quite different.```
 Subject: Re: probablity with cards Answered By: livioflores-ga on 24 Jul 2006 23:47 PDT Rated:
 ```Hi!! I found a well known problem, very similar in some way to this one, which asks for the probability of no one match: "Players A and B each have a well shuffled standard pack of cards, with no jokers. The players deal their cards one at a time, from the top of the deck, checking for an exact match. Player A wins if, once the packs are fully dealt, no matches are found. Player B wins if at least one match occurs. What is the probability that player A wins?" From "Nick's Mathematical Puzzles: 21 to 30": (see the number 26) http://www.qbyte.org/puzzles/puzzle03.html The answer to this problem is 1 - 1/1! + 1/2! - 1/3! + 1/4! - ... - 1/51! + 1/52!, which is within 1/53! of 1/e. You can see how to reach this figure at this page: "Nick's Mathematical Puzzles: Solution 26": (The solution to this problem have a nice surprise to you!!) http://www.qbyte.org/puzzles/p026s.html So you have that the probability of "no one match" is almost 1/e; then the probability of "at least one match on one of the 52 flips" is: P = 1-1/e = (e-1)/e = 0.63212 For additional references see: "Derangement -- from Wolfram MathWorld": http://mathworld.wolfram.com/Derangement.html I hope this helps you. Feel free to request for a clarification if you need it before rate this answer. Regards, livioflores-ga``` Request for Answer Clarification by frodo2366-ga on 25 Jul 2006 00:58 PDT ```I'm pretty good at math and understand series (taylors, etc) but a +/- alternating series doesn't make a lot of sense in this case.``` Clarification of Answer by livioflores-ga on 25 Jul 2006 07:27 PDT ```Hi!! The +/- alternating series have its origin, as the solution page tells us, on the Inclusion-Exclusion Principle. This principle is a little hard to understand, at least for me, but I found a very nice article that explains how to count derangements in a very easy way: "... We can do exactly the same sort of thing with the Yankee Stadium problem. If there are N seats, there are N! possible arrangements. We want to count the number of ?derangements??orderings where none of the people get the correct seat. To do this, begin with N! and subtract off the ones where there is a match. ... How many arrangements are there with person 1 in seat 1? Well, the rest of the N-1 people can be arranged in any of (N-1)! ways, so there are (N-1)! such arrangements. There are a similar number of arrangements with person 2 in seat 2, et cetera. So we?ll subtract all those off. But then we?ve subtracted many arrangements more than once. Some of the arrangements have the correct person in both seats 1 and 2, and we subtracted that arrangement twice. How many arrangements have person 1 and 2 in the correct seats? Well, there are N-2 remaining seats, and those can be arranged in any of (N-2)! ways. We need to add in all of those, then subtract off all the cases where there are at least 3 people in the right seats, then add in the cases with 4, et cetera. ..." From "Derangements" by Brian Conrey and Tom Davis: http://www.geometer.org/mathcircles/derange.pdf The above explanation is easier to understand if you read a couple of previous simple examples on the article; I strongly suggest you to read it completely, it will clarify almost all your doubts. As you can see there is an explanation for the alternate series, and if you continue reading the article you will find the nice and elegant solution: 1/e. For more formal reference about the Inclusion-Exclusion Principle see: "The Inclusion-Exclusion Principle": http://www.cut-the-knot.org/arithmetic/combinatorics/InclusionExclusion.shtml I hope this helps you. feel free to continue using the clarification feature if you still need it. Regards, livioflores-ga```
 frodo2366-ga rated this answer: and gave an additional tip of: \$5.00 ```Thank you, you have done an exceptional research job. I will still have to "work this out in my head". You have given excellent "starting points". A good "answer" does not always equal "instant understanding". You have done your job, now I have to do mine, ;-) Thanks again for a very good reply.```

 Subject: Re: probablity with cards From: livioflores-ga on 26 Jul 2006 08:04 PDT
 ```Thank you for the good rating and the generous tip!! A suggestion to understand this is to mannually do the derangement counting for the case of four cards using the method of the text on the clarification, it will take you a time but it will be very helpful. Kind regards, livioflores-ga```
 Subject: Re: probablity with cards From: eion-ga on 26 Jul 2006 09:34 PDT
 ```ok, i hit this with a bit of logic today while sitting in the loo... the probability has to be 1/52 every time. so about 1/52 of you picking 2 cards from 2 seperate decks the same. think on it this way . first card - first deck. you have a 100% cance of it being the card you want.. th card is just you marker, what you are looking for in the next pack it has no odds ok first deck done second deck 52 cards, you have 1/52 chance of your card being on top.. we keep on going 10 cards in both decks, first card odds dont count..still 2nd card - 2nd deck , you have a 1/52 chance the card is at that point in the pack.. the previous cards dont count.. nether do the ones left cause you are not after a specifc card... IF you said what are the odds of me picking the ace of spades as the first card in this pack and the first card in the second you would say 1/52*1/52...if you strik out, the 1/51*1/51 etc.. but here we dont care what the card it it is just what we are looking for.. in he next pick , and the odds of it being were we are upto in the pack being #1 or #52 is 1 chance in 52 cards. maybe this helps maybe it confuses ppl, who knows..```
 Subject: Re: probablity with cards From: frodo2366-ga on 26 Jul 2006 19:21 PDT
 ```Hi eion-ga, I think you misunderstood the question. We are not looking for a specific "card" just any value that matches in both decks upon a simutaneous flip of the cards in both decks. Anyway, hope you all are salvitating to take the above answer and pocket your friends money!```
 Subject: Re: probablity with cards From: livioflores-ga on 26 Jul 2006 20:50 PDT
 ```Dear eion: What happened with the probability of the event that the card on the top of the second deck was displayed before on the first deck in a previous hand, for example: 1st hand: Deck 1: 10 of hearts Deck 2: 5 of hearts 2nd hand: Deck 1: 5 of hearts Deck 2: xxx In the way you are thinking the problem there is zero chance to get a match not 1/51. Think about this and how this affect the result you got.```
 Subject: Re: probablity with cards From: eion-ga on 27 Jul 2006 03:34 PDT
 ```i understand you are not looking for a specific card that is why i stated the odds on the first deck have no impact.. the fact that the card you turn over at any time on the first deck is an unknown means you cant factor in it already having been discarded on the second deck.. you have infact 1/52 that the card is sitting in the deck where you are up to...no matter how you look at it.. yes you can factor in the chance that you may have removed an unknown card from pile already but these odds would remain inline with the fact that you may have already removed it from the first pile..... maybe i need to draw a big picture...```
 Subject: Re: probablity with cards From: frodo2366-ga on 29 Jul 2006 21:23 PDT
 ```Hi eion-ga, Don't know if you still don't get the question, but as the math answer behind the question suggests any thing over 6 different cards approaches the probability answer very closely. Therefore, try it manually with 2 sets of A-6 of spades a number of times, go to A-k if you are not convinced, still not convinced, try all 52 from 2 decks. It takes longer but as the answer suggests the results over time are the same. Of course this is probability, so you have to do the manual experiment a number of times.```