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jupiter
Member since Jan-3-06
Jan-03-06, 12:00 PM (EST)    "probability"

 I am thinking of our UK bank credit-card PINs, which have 4 digits.Obviously many PINs are repeated. Say a bank has a million customersMy question is: What is the chance that if the PINs are assigned randomly, all the possible 4-digit numbers 0000-9999 will be used at least once?To simplify the problem, let us consider, say, only n customers, with n<10000, and only 2-digit PINs 00-99.Is there a generalised answer to this kind of problem?CDA (from United Kingdom)CDA

alexb
Charter Member
1756 posts
Jan-03-06, 08:48 PM (EST)    1. "RE: probability"
In response to message #0

 >To simplify the problem, let us consider, say, only n >customers, with n<10000, and only 2-digit PINs 00-99. It's not actually a simplification.>Is there a generalised answer to this kind of problem? There is a generalized question that can be asked:An experiment may result in one of k equiprobable events. What is a probability that an event will not show up in n trials?In your original problem k = 10000. To really simplify the problem, take k = 2. As a result of a trial you may have, say, either x or y. n trials produce a sequence of x's and y's. The question now is equivalent to asking how many of xy-strings of length n contain only 1 letter? jupiter guest
Jan-04-06, 05:51 PM (EST)

2. "RE: probability"
In response to message #1

 On looking at a few notes I wrote many years ago, I find theproblem is, as you say, an extended form of a much simpler problem.Example: There are 6 faces to a die. How many throws needed to getat least one showing of every number 1 to 6?And given, say, 50 throws, what is prob of getting all 6 faces at least once?Trouble is that it'seems to involve things called Stirling Numbers, after which I gave up. But didn't (James) Stirling have a formula for large factorials? The arithmetic would get far too complicated for my original question, I guess, without some good approximations available. Thanks for putting me on the right track.

Pierre Charland
Member since Dec-22-05
Jan-17-06, 01:40 PM (EST)    3. "RE: probability"
In response to message #0

 Basically, this is the ,If a complete set has K coupons, how many coupon to buy (on average) until a complete N-set is collected ?The answer is K*H(K) = K*(1/1+1/2+1/3+..+1/K)Note: H(K) = 1/1+1/2+1/3+..+1/K) is called the Kth Harmonic number.If 2-digit, then K=100, thenH(K)=5.18737751763.., K*H(K)=518.737751763..If 4-digit, then K=10000, thenH(K)=9.78760603604.., K*H(K)=97876.0603604..This does not exactly answer your question, but relates to it.My reference are:--Problems and Snapshots from the World of Probability, p.85 by Gunnar Blom-- Generatingfunctionology, 4.2 p.132 in 1st edition by Herbert S. WilfAlphaChapMtl jupiter guest
Jan-29-06, 02:13 PM (EST)

4. "RE: probability"
In response to message #3

 Thanks !Your formula passes OK. I played the game with K=100 bygenerating random numbers. After some 500 games, the averageworked out very close to your figure of 518.73, so I took your formula as correct. With K=50 the average came to 224.38 after 500such games. Have not tried it for K=10,000. It would consume too much machine time and resources. Thanks again. mr_homm
Member since May-22-05
Jan-29-06, 10:24 PM (EST)    5. "RE: probability"
In response to message #4

 >>Have not tried it for K=10,000. It would consume too much >machine time and resources. >Speaking of machine time, of course the calculation of the formula itself is much faster than the simulation, but for large K the formula itself simplifies:The limit of the harmonic series differs from that of ln(K) by the Euler constant, gamma = 0.5772156649..., so the expected value for K=10000 would by very, very close to 10000*ln(10000) + gamma, or 40000*ln(10) + 0.577.... Since the first term is so large, gamma contributes little.--Stuart Anderson

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