extrapolate full social security based on last 4 digits

[y3tter]

I have an employee worksheet that I am using to get open enrollment processed. Our insurance provider provided us with a a sheet with employee names and their redacted social security number(ex xxx-xx-1234). I am needing to have the full SS number displayed. I have a another sheet that lists employee names and their full SS number. How can I do a lookup to match the last four digits of the redacted number to and return the full SS number?

 

[Scott T]

If the names are the same in both files you can do a lookup on the name otherwise you would need to lookup the last 4. Note that the last for of a SSN is not unique so it is possible that you could have an employee with 123-45-6789 and another employee with 231-13-6789. What is in the file that you get from the insurance provider is it only 4 digits or xxx-xx-1234?

 

[y3tter]

Right, but we do not have enough employees for the 1/10,000 chance of occurrence. The issue with employee name lookup is that there are several relatives that work together and their beneficiaries.

The sheet I got from insurance is xxx-xx-1234, but I did text to columns to remove the xxx-xx- and am just left with the last four digits.

 

[Scott T]

Excel 2010

	A	B	C
1	File form insurance
2	Beth Green	1234	123-13-1234
3	Rick Grimes	3214	312-74-3214
4
5
6	your file
7			helper column
8	Beth Green	123-13-1234	1234
9	Rick Grimes	312-74-3214	3214

<tbody>
</tbody>

Sheet4​

Worksheet Formulas Cell Formula C8 =RIGHT(B8,4) C9 =RIGHT(B9,4) <tbody> </tbody>

<tbody>
</tbody>

Array Formulas Cell Formula C2 {=INDEX($B$8:$B$9,MATCH(B2,RIGHT($C$8:$C$9,4)+0,0))} C3 {=INDEX($B$8:$B$9,MATCH(B3,RIGHT($C$8:$C$9,4)+0,0))} <tbody> </tbody> Entered with Ctrl+Shift+Enter. If entered correctly, Excel will surround with curly braces {}. Note: Do not try and enter the {} manually yourself

<tbody>
</tbody>

 

[y3tter]

Excel 2010

	A	B	C
1	File form insurance
2	Beth Green	1234	123-13-1234
3	Rick Grimes	3214	312-74-3214
4
5
6	your file
7			helper column
8	Beth Green	123-13-1234	1234
9	Rick Grimes	312-74-3214	3214

<tbody>
</tbody>

Sheet4​

Worksheet Formulas Cell Formula C8 =RIGHT(B8,4) C9 =RIGHT(B9,4) <tbody> </tbody>

<tbody>
</tbody>

Array Formulas Cell Formula C2 {=INDEX($B$8:$B$9,MATCH(B2,RIGHT($C$8:$C$9,4)+0,0))} C3 {=INDEX($B$8:$B$9,MATCH(B3,RIGHT($C$8:$C$9,4)+0,0))} <tbody> </tbody> Entered with Ctrl+Shift+Enter. If entered correctly, Excel will surround with curly braces {}. Note: Do not try and enter the {} manually yourself

<tbody>
</tbody>

Perfect, thank you SO MUCH!!

 

[Joe4]

Right, but we do not have enough employees for the 1/10,000 chance of occurrence.

You have fallen prey to some bad math! That is not how statistics work.
Actually, if you have 118 people, you have a 50% chance that two will have the same last four digits.
So, if you have over 118 people, you have more than a 50% chance that you will duplicates!
At 214 people, the percentage goes up to 90%!

So, I would recommend by matching on more than just the last four digits. Maybe add last name to your match too.

 

[Scott T]

If you want to avoid the helper column you can used this instead.

=INDEX($B$8:$B$9,MATCH("*-"&B3,$B$8:$B$9,0))

 

[y3tter]

You have fallen prey to some bad math! That is not how statistics work.
Actually, if you have 118 people, you have a 50% chance that two will have the same last four digits.
So, if you have over 118 people, you have more than a 50% chance that you will duplicates!
At 214 people, the percentage goes up to 90%!

So, I would recommend by matching on more than just the last four digits. Maybe add last name to your match too.

Cool, thanks for the info! Out of 655, there were no duplicates. Could I have just won the proverbial statistical lottery? Statistical analysis vs probability...

 

[StephenCrump]

Out of 655, there were no duplicates. Could I have just won the proverbial statistical lottery? Statistical analysis vs probability...

Are you sure?

Scott T's formula relies on there being no duplicates of the last 4 digits. If there are duplicates, the formula will return the first match, which may not be the correct one.

If the last 4 digits are totally random (which is not exactly true), then extrapolating out Joe4's analysis you have of the order of 1 in a billion chance of having 655 different 4 digit sequences.

Based on those odds, my bet is that you do have duplicates.

 

[y3tter]

Are you sure?

Scott T's formula relies on there being no duplicates of the last 4 digits. If there are duplicates, the formula will return the first match, which may not be the correct one.

If the last 4 digits are totally random (which is not exactly true), then extrapolating out Joe4's analysis you have of the order of 1 in a billion chance of having 655 different 4 digit sequences.

Based on those odds, my bet is that you do have duplicates.

Yes, data was sorted by last 4 digits and then reviewed for duplicates. Scott T's formula was the best option due to not every family falling into the traditional model(stepchildren, unmarried, children with different address). Joe4 scared me with his reminder of the birthday paradox. Apparently social security numbers aren't completely random and rely on dob, place of birth, etc. Still, I was probably lucky to not have witnessed any duplications.