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.

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

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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.

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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

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.

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

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..

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!

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.

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...

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.