Math (probablility) Problem

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BackInTex
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Math (probablility) Problem

#1 Post by BackInTex » Thu Sep 08, 2022 2:57 pm

I need the formula to calulate x where x is the probablity that a randomly generated number between and including 0 and 9999 already exists in a population of y distinct numbers between 0 and 9999.

Real world explanation:

We issue company credit cards to employees. They are tracked/identified in the system by the last four digits. When requesting new cards from randomly generated card numbers, what is the probability that the next card issued will end in the same four digits of a card already in use by another employee?

If you can write the formula with the following, that will help:
x = probablity
y = number of existing distinct values
a = floor of values (currently zero)
b = ceiling of values (currently 9999)

e.g. a + b = number of possible distinct values

Top prize of "Thanks, that is really helpful!" will be awarded to the first correct response.

Go!
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Re: Math (probablility) Problem

#2 Post by Bob78164 » Thu Sep 08, 2022 2:59 pm

BackInTex wrote:
Thu Sep 08, 2022 2:57 pm
I need the formula to calulate x where x is the probablity that a randomly generated number between and including 0 and 9999 already exists in a population of y distinct numbers between 0 and 9999.

Real world explanation:

We issue company credit cards to employees. They are tracked/identified in the system by the last four digits. When requesting new cards from randomly generated card numbers, what is the probability that the next card issued will end in the same four digits of a card already in use by another employee?

If you can write the formula with the following, that will help:
x = probablity
y = number of existing distinct values
a = floor of values (currently zero)
b = ceiling of values (currently 9999)

e.g. a + b = number of possible distinct values

Top prize of "Thanks, that is really helpful!" will be awarded to the first correct response.

Go!
x=y/(b+1-a). --Bob
"Question with boldness even the existence of a God; because, if there be one, he must more approve of the homage of reason than that of blindfolded fear." Thomas Jefferson

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Re: Math (probablility) Problem

#3 Post by Beebs52 » Thu Sep 08, 2022 3:00 pm

I like pizza.
Well, then

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Re: Math (probablility) Problem

#4 Post by Vandal » Thu Sep 08, 2022 3:09 pm

Image


The answer is 42
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Re: Math (probablility) Problem

#5 Post by Beebs52 » Thu Sep 08, 2022 3:25 pm

Vandal wrote:
Thu Sep 08, 2022 3:09 pm
Image


The answer is 42
Don't panic.
Well, then

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Re: Math (probablility) Problem

#6 Post by BackInTex » Thu Sep 08, 2022 3:36 pm

Bob78164 wrote:
Thu Sep 08, 2022 2:59 pm
x=y/(b+1-a). --Bob
Thanks, that is really helpful!
..what country can preserve it’s liberties if their rulers are not warned from time to time that their people preserve the spirit of resistance? let them take arms.
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Re: Math (probablility) Problem

#7 Post by silverscreenselect » Thu Sep 08, 2022 3:44 pm

BackInTex wrote:
Thu Sep 08, 2022 2:57 pm
What is the probability that the next card issued will end in the same four digits of a card already in use by another employee?
If you're the person in charge of handing out these credit cards and you're going to get in trouble if there's a screwup, then the probability of there being one sometime while you're responsible is 100% (or 1 as we like to refer to it in probability language).

The probability of your continuing to be in charge after that is 0.

Isn't probabibility simple?
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Re: Math (probablility) Problem

#8 Post by BackInTex » Fri Sep 09, 2022 10:42 am

silverscreenselect wrote:
Thu Sep 08, 2022 3:44 pm
BackInTex wrote:
Thu Sep 08, 2022 2:57 pm
What is the probability that the next card issued will end in the same four digits of a card already in use by another employee?
If you're the person in charge of handing out these credit cards and you're going to get in trouble if there's a screwup, then the probability of there being one sometime while you're responsible is 100% (or 1 as we like to refer to it in probability language).

The probability of your continuing to be in charge after that is 0.

Isn't probabibility simple?
I'm not in charge of the cards, just the system that processes the charges. The software only uses the 4 digits to identtify the card holder. We are getting duplicates and the charges are going to whomever has those 4 digits assigned to them. I needed the probabilities to convince the software developer we need additonal identifers such as employee number included in the design.
..what country can preserve it’s liberties if their rulers are not warned from time to time that their people preserve the spirit of resistance? let them take arms.
~~ Thomas Jefferson

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Re: Math (probablility) Problem

#9 Post by Bob78164 » Fri Sep 09, 2022 11:33 am

BackInTex wrote:
Fri Sep 09, 2022 10:42 am
silverscreenselect wrote:
Thu Sep 08, 2022 3:44 pm
BackInTex wrote:
Thu Sep 08, 2022 2:57 pm
What is the probability that the next card issued will end in the same four digits of a card already in use by another employee?
If you're the person in charge of handing out these credit cards and you're going to get in trouble if there's a screwup, then the probability of there being one sometime while you're responsible is 100% (or 1 as we like to refer to it in probability language).

The probability of your continuing to be in charge after that is 0.

Isn't probabibility simple?
I'm not in charge of the cards, just the system that processes the charges. The software only uses the 4 digits to identtify the card holder. We are getting duplicates and the charges are going to whomever has those 4 digits assigned to them. I needed the probabilities to convince the software developer we need additonal identifers such as employee number included in the design.
Then you actually wanted the probability that if z numbers are randomly assigned, at least two of them are duplicates. That's much, much higher than most people realize. For example, the solution to the birthday problem tells you that if b=365 and a=1, then once you've assigned 23 numbers the odds are better than 50-50 that you've assigned at least one pair of duplicates. (The probability that you've avoided duplication is (365*364*363* . . . * 343)/365^23, which turns out to be less than 0.5.) For your actual problem, my intuition is that your odds of a duplicate reach 50-50 at around 300 assignments, but it's easy enough to set up a spreadsheet to figure out the actual number. --Bob
"Question with boldness even the existence of a God; because, if there be one, he must more approve of the homage of reason than that of blindfolded fear." Thomas Jefferson

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Re: Math (probablility) Problem

#10 Post by Estonut » Sat Sep 10, 2022 1:23 am

BackInTex wrote:
Fri Sep 09, 2022 10:42 am
I'm not in charge of the cards, just the system that processes the charges. The software only uses the 4 digits to identify the card holder. We are getting duplicates and the charges are going to whomever has those 4 digits assigned to them. I needed the probabilities to convince the software developer we need additional identifiers such as employee number included in the design.
Just the fact that you are getting duplicates should be enough to convince a good software developer that additional identifiers are required for uniqueness. It was a lame idea to use only the last 4 digits and expect to tie back to a single card holder.
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