brengle@hplabsc.UUCP (Tim Brengle) (12/12/85)
Upon finding out that I claimed to have been a mathematician once upon a time, a co-worker asked me a probability question. Now, I never did all that well in probability or combinatorics, so I am passing this on in hopes that someone out there might be able to answer it or give me an appropriate pointer: You have two decks of normal playing cards, say red-backed and blue-backed. You shuffle each individually, and place them on the table. Turn the top card of each deck face up and place it on the table. Repeat the turning process until all cards have been turned face up. What is the probability that at least one pair of cards turned at the same time are exactly the same (except for the color of their backs)? I think that the problem is equivalent to having a single deck of cards numbered from 1 to 52 and determining the probability that at least one card occurs at its numbered position. Thanks in advance, Tim Brengle
tan@ihlpg.UUCP (Bill Tanenbaum) (12/14/85)
> Upon finding out that I claimed to have been a mathematician once upon a > time, a co-worker asked me a probability question. Now, I never did all > that well in probability or combinatorics, so I am passing this on in > hopes that someone out there might be able to answer it or give me an > appropriate pointer: > > You have two decks of normal playing cards, say red-backed and blue-backed. > You shuffle each individually, and place them on the table. Turn the top > card of each deck face up and place it on the table. Repeat the turning > process until all cards have been turned face up. > > What is the probability that at least one pair of cards turned at the same > time are exactly the same (except for the color of their backs)? > > I think that the problem is equivalent to having a single deck of cards > numbered from 1 to 52 and determining the probability that at least one > card occurs at its numbered position. > > Thanks in advance, > Tim Brengle ------------ The probability is extremely close to 1 - 1/e, or about 63%. The probability that one or more cards will be in their proper position in an N card deck is greater than (1 - 1/e) for odd N, and less than (1 - 1/e) for even N. In either case, the probability approaches the limit of (1 - 1/e) as N increases. -- Bill Tanenbaum - AT&T Bell Labs - Naperville IL ihnp4!ihlpg!tan
stevev@tekchips.UUCP (Steve Vegdahl) (12/17/85)
> > You have two decks of normal playing cards, say red-backed and blue-backed. > > You shuffle each individually, and place them on the table. Turn the top > > card of each deck face up and place it on the table. Repeat the turning > > process until all cards have been turned face up. > > > > What is the probability that at least one pair of cards turned at the same > > time are exactly the same (except for the color of their backs)? > The probability is extremely close to 1 - 1/e, or about 63%. The > probability that one or more cards will be in their proper > position in an N card deck is greater than (1 - 1/e) for odd N, > and less than (1 - 1/e) for even N. In either case, the probability > approaches the limit of (1 - 1/e) as N increases. In particular, the probability is within 1/N! of (1 - 1/e). For N = 52, this means that the (1 - 1/e) approximation is accurate to 67 decimal digits. Steve Vegdahl Computer Research Lab. Tektronix, Inc. Beaverton, Oregon
norman@batcomputer.TN.CORNELL.EDU (Norman Ramsey) (03/05/86)
Doubtless everyone remembers the old chestnut about how many people you have to have in a room before it becomes likely that two of them share the same birthday (I believe that it's something like 23 people for probability 1/2, and that with fifty people it approaches certainty). Well, the other day I got to wondering, with N people in the room, what is that probability that two of them share the same eyeglass prescription. (Note that I don't want to count two people who don't need glasses, since they can't swap glasses, and that's no fun). To define "the same" I'm willing to say "within a tenth of a diopter" or some such nonsense. How about it. I know this isn't net.optometry, but does anyone know where I can find the appropriate statistics? Or does anybody know the answer? -- Norman Ramsey norman@tcgould.tn.cornell.edu Pianist at Large