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jupiter
Member since Jan306

Jan0306, 12:00 PM (EST) 

"probability"

I am thinking of our UK bank creditcard PINs, which have 4 digits. Obviously many PINs are repeated. Say a bank has a million customers My question is: What is the chance that if the PINs are assigned randomly, all the possible 4digit numbers 00009999 will be used at least once?To simplify the problem, let us consider, say, only n customers, with n<10000, and only 2digit PINs 0099. Is there a generalised answer to this kind of problem? CDA (from United Kingdom) CDA 

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alexb
Charter Member
1756 posts 
Jan0306, 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 2digit PINs 0099. 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 xystrings of length n contain only 1 letter? 

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jupiter
guest

Jan0406, 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 the problem 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 get at 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. 

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Pierre Charland
Member since Dec2205

Jan1706, 01:40 PM (EST) 

3. "RE: probability"
In response to message #0

Basically, this is the <Coupon Collector Problem>, If a complete set has K coupons, how many coupon to buy (on average) until a complete Nset 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 2digit, then K=100, then H(K)=5.18737751763.., K*H(K)=518.737751763.. If 4digit, then K=10000, then H(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. Wilf AlphaChapMtl 

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jupiter
guest

Jan2906, 02:13 PM (EST) 

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

Thanks ! Your formula passes OK. I played the game with K=100 by generating random numbers. After some 500 games, the average worked 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 500 such games. Have not tried it for K=10,000. It would consume too much machine time and resources. Thanks again. 

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