You decide to take up a life of crime and plan to break into a bank. You and your buddies have to decide which bank to choose from based on their locked door. One bank has a lock that requires four letters (repetition is not allowed) and the other bank has a lock that requires 6 digits (repetition is allowed). You want to break into the bank with the door that gives you the higher probability of breaking the code. Of those two options, which one has the greater probability of you guessing the code?

To break the code for the first bank which uses four letters, since there are 26 letters in the alphabet, and require a four letter code, the first letter can be any of the 26 letters. Since one letter has been taken, we are left with 25 letters for the second position, 24 for the third position and 24 for the fourth position.

So, the total number of ways of selecting the 4 letter code without repetition is ²⁶P₄ = 26 × 25 × 24 × 23 = 358800

Since we have only 1 correct code, this is our required outcome. So, the probability of breaking the code is P = required outcome/total outcome = 1/358800 = 2.787 × 10⁻⁶

To break the code for the second bank which uses 6 digits with repetition, since there are 10 digits, and we require a 6 digit code, each digit can be filled with 10 digits each. Since there is repetition, the total number of ways we can select the 6 digits is (¹⁰P₁)⁶ = 10 × 10 × 10 × 10 × 10 × 10 = 10⁶.

Since there is going to be only 1 correct code, the number of ways of selecting the correct code is 1.

The probability P' of cracking the second bank code is P' = required outcome/total outcome = 1/10⁶ = 1 × 10⁻⁶

Since P = 2.787 × 10⁻⁶ > P' = 1 × 10⁻⁶, the first bank with the four letter code has the grater probability of me guessing the code.