MHB Probability question from TV, cracking a phone password

[gswani]

I'm watching a TV show and it is explained that a teenager found a phone and figured out the password "using finger smudges". This means he determined which digits are in the passcode by looking at the smudges on the face of the phone. Assuming that the passcode is 4 numbers, this should mean 24 possibilities if the "smudges" indicate that four distinct numbers are used.

4 x 3 x 2 x 1 = 24

What I can't figure out is what if the phone only had a smudge on three numbers. So that would mean a 4-digit code made up of three numbers, one of them repeated. I feel like the answer is 18 but I don't know why. Let's say the numbers are 2, 4 and 6. If you happened to select the repeated number on the first try, the math would be:

3 x 3 x 2 x 1 = 18
1st#: 2, 4, 6, 6 (two of them are 6 for this example)
2nd#: 2, 4, 6, 6
3rd#: 2, 4, 6, 6
4th#: 2, 4, 6, 6

That would be 3 possibilities on the first number since there are only three options. Then for the second number, you would still have three options, and then two and one. But what if you selected one of the digits that doesn't repeat? Then it would be:

3 x 2 x 2 x 1 = 12
1st#: 2, 4, 6, 6 (two of them are 6 for this example too)
2nd#: 2, 4, 6, 6
3rd#: 2, 4, 6, 6
4th#: 2, 4, 6, 6

Obviously I could add another scenario where you choose 2, and 4 first and then you only have one number left for the last two options.

3 x 3 x 1 x 1 = 9

But there has to be a way to determine which of these is right. I'd like to know the math that backs this up. I hope this is interesting enough (and clear enough) to interest someone. Thanks in advance for taking the time.

 

[Opalg]

I'm watching a TV show and it is explained that a teenager found a phone and figured out the password "using finger smudges". This means he determined which digits are in the passcode by looking at the smudges on the face of the phone. Assuming that the passcode is 4 numbers, this should mean 24 possibilities if the "smudges" indicate that four distinct numbers are used.

4 x 3 x 2 x 1 = 24

What I can't figure out is what if the phone only had a smudge on three numbers. So that would mean a 4-digit code made up of three numbers, one of them repeated. I feel like the answer is 18 but I don't know why. Let's say the numbers are 2, 4 and 6. If you happened to select the repeated number on the first try, the math would be:

3 x 3 x 2 x 1 = 18
1st#: 2, 4, 6, 6 (two of them are 6 for this example)
2nd#: 2, 4, 6, 6
3rd#: 2, 4, 6, 6
4th#: 2, 4, 6, 6

That would be 3 possibilities on the first number since there are only three options. Then for the second number, you would still have three options, and then two and one. But what if you selected one of the digits that doesn't repeat? Then it would be:

3 x 2 x 2 x 1 = 12
1st#: 2, 4, 6, 6 (two of them are 6 for this example too)
2nd#: 2, 4, 6, 6
3rd#: 2, 4, 6, 6
4th#: 2, 4, 6, 6

Obviously I could add another scenario where you choose 2, and 4 first and then you only have one number left for the last two options.

3 x 3 x 1 x 1 = 9

But there has to be a way to determine which of these is right. I'd like to know the math that backs this up. I hope this is interesting enough (and clear enough) to interest someone. Thanks in advance for taking the time.

Hi gswani, and welcome to MHB!

There are two possible answers to your problem. To decide which of them is correct, your teenage sleuth is going to have to look more carefully at the three smudges on the phone. The key question is: is one of those smudges heavier than the other two? If so, you can guess that it must correspond to the repeated digit. Otherwise, any of the three digits could be the repeated one, and the answer will be three times as large.

Taking your example where the three digits are 2, 4 and 6, suppose that the 2 has the heavier smudge, so that it occurs twice. There are six possible places where the two 2s can occur in the password $****$, as follows: $$2\,2**,$$ $$2*2\,*,$$ $$2**\,2,$$ $$*\,2\,2\,*,$$ $$*\,2*2,$$ $$**2\,2.$$ For each of these there are two possible passwords, depending whether the $4$ comes before or after the $6$ in the remaining $*$ positions. That gives a total of 12 possible passwords.

But if we don't know which digit is repeated, that gives an additional 12 possible passwords with a repeated $4$, and 12 with a repeated $6$, for a total of 36 altogether.

That last result seems surprising, because there are more possible passwords with three digits (36) than there were with four digits (24).