Probability question from TV, cracking a phone password

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In summary: But that's because there are three times as many possibilities for passwords with three digits as there are with four. Each of the three digits can appear in any of the six positions, so there are six times as many possibilities.
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gswani
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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.
 
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gswani said:
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).
 

FAQ: Probability question from TV, cracking a phone password

What is the probability of guessing a 4-digit phone password correctly on the first try?

The probability of guessing a 4-digit phone password correctly on the first try is 1 in 10,000 or 0.01%. This is because there are 10,000 possible combinations from 0000 to 9999 and only one of those combinations will be the correct password.

If someone has a 6-digit phone password, what is the probability of guessing it correctly within 10 tries?

The probability of guessing a 6-digit phone password correctly within 10 tries is approximately 10%. This can be calculated by taking the total number of possible combinations (1 million) and dividing it by the number of tries (10), giving us a probability of 0.00001. We can then subtract this from 1 to get the probability of not guessing the password in 10 tries, which is 0.99999. Finally, we can subtract this from 1 again to get the probability of guessing the password in 10 tries, which is approximately 10%.

How does the probability change if the phone password is a combination of numbers and letters?

The probability of guessing a phone password that includes both numbers and letters is significantly lower than guessing a password that only includes numbers. This is because the number of possible combinations increases exponentially when letters are added, making it much harder to guess the correct password. For example, a 4-digit password with numbers and letters has 36 possible characters (26 letters and 10 numbers), resulting in 1 in 1.6 million possible combinations.

Is there a way to increase the probability of guessing a phone password?

While there is no guaranteed way to increase the probability of guessing a phone password, there are a few strategies that can improve your chances. These include using common number combinations (such as birthdates or anniversaries), trying commonly used passwords (1234, 0000, 1111, etc.), or using software or tools specifically designed to crack passwords. However, it is important to note that attempting to guess someone else's phone password without their permission is a violation of privacy and may be illegal.

What is the likelihood of someone guessing my phone password if it is randomly generated?

If your phone password is truly randomly generated, the likelihood of someone guessing it is extremely low. This is because the number of possible combinations is so large that it would be virtually impossible to guess the correct password. However, it is important to note that truly random passwords can be difficult to remember and may be easier to crack using advanced software or tools. It is recommended to use a combination of random characters and words to create a strong and secure password.

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