Suppose I have to select 10 random numbers from 1-10 to match some previously selected random number. I know the probability of any single set of numbers being selected correctly is 1 in 10 billion. However, what if I knew the set of numbers fulfills one of the following two requirements:
There will be 1 number that repeats itself at least 3 times (not necessarily consecutively) and 1 number that repeats itself at least twice. For example: 7775243389 (7 repeats 3 times, 3 repeats twice) and 6678709436 (6 repeats 3 times, 7 repeats twice) are valid sets/selections. 6677832154 is not valid (missing single number that repeats 3 times).
There will be 3 numbers that repeat twice (again not necessarily consecutively). Example of valid set: 6648492102 (6,4 and 2 repeat twice). Example of non valid set: 6648492105 (only 6 and 4 repeat twice).
Note: In both cases a number may repeat more than two or three times, these are both minimum values.
I know that in both of these cases the probability of me choosing a correct set is far better than one in 10 billion (since there are much less numbers to choose from) however I am not quite sure what the exact formula is to calculate the probability in each case.
Any help is much appreciated.
If you have $n$ total digits the number of sequences with one digit three times, one digit twice, and no other repeats is ${n \choose 3}{n-3 \choose 2}\frac {10!}{(13-n)!}$ where the first factor chooses the locations of the three matching digits, the second chooses the locations of the two matching digits, and the last chooses the digits. This has a maximum at $n=13$ of $46\ 702\ 656\ 000$. You can get rather close to $1$ in $10$ billion with $n=12$ where it is about $14.37$ billion.
For three pairs and length $n$ you have $\frac 1{3!}{n \choose 2}{n-2 \choose 2}{n-4 \choose 4}\frac {10!}{(13-n)!}$ where the logic is similar but you get the leading $\frac 1{3!}$ because you can choose the three numbers to pair in $3!$ equivalent ways. This gives about $9.34$ billion at $n=13$, which is very close.
The choosing of positions for the duplicated digits brings you a lot of extra possibilities compared to $10!=3\ 628\ 800$.