Probability of $3$ of $4$ numbers matching numbers in same position on license plate License plates in our area end in a 4-digit number $0000$ through $9999$. What is the probability of my second car's plate matching $3$ of the $4$ digits in the identical position as my first car's plate? This actually happened and it seems the probability would be fairly low. I think the odds are $1:10,000$ that all four numbers would match, so it must be somewhat higher probability that any three would match in the same position, e.g., $0123$ and $0523$.
What is the probability, and what is the equation to obtain it?
 A: We need to calculate in how many ways can three digits match (but not the last one). Let me give you an example - suppose your plate was previously $0000$. The "matching at three positions" license plates are:
$$1000,2000,\ldots,9000$$
$$0100,0200,\ldots,0900$$
$$0010,0020,\ldots,0090$$
$$0001,0002,\ldots,0009$$
In other words, pick the "mismatching" position (in $4$ ways) and then pick the mismatching digit (in $9$ ways). The total number of ways to get a mismatch of that sort is $4\times 9=36$. Thus, the probability for this to happen (if all the combinations of digits are equally probable) is $\frac{36}{10,000}=0.0036=0.36\%$.
A: Suppose one license plate is fixed. Choose one of the four digits that will not match. Once that is chosen, there are nine digits you can change it to. So, there are a total of 36 license plates that will match exactly three digits out of all possible license plates. That is:
$$\dfrac{4\cdot 9}{10^4} = \dfrac{9}{2500} = 0.0036$$
A: When you fix one (of four) position and want to count how many combinations are there that match the given number at three other positions but the one you fixed you get $\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{9}{10}$ out of the total number of combinations. The last factor is that what it is because you allow all digits to happen on that place but the one that is written on the original four-digit number. That leaves $9$ possibilities. Now circle through all four positions, fixing them one at the time, and add up the combinations that match at all but fixed place. There are four identical summands. So the total sum is $4\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{9}{10}$, which is $\frac{36}{10000}=\frac{9}{2500}$.
