Best strategy to pick a lock which opens if at least two of its three decimal digit wheels are dialed correctly? Suppose you want to open a lock with three digits code, the lock is a little special because it can be opened if you have two digits guessed right. To clarify, if the correct three digit is 123, then guess 124 or 153 can open the box. The lock looks like this:

Question is: what is best strategy to open the box? The best strategy is a strategy which requires least attempts at average. You should also find the average.
The strategies I came up with are:
First strategy: hold first digit, and constantly change second and third while keep them equal when changing them. For example, I will try: 111, 122...199,211,222,...,299,....
Second strategy: hold second and third equal, and constantly change first one. For example: 111, 211,...,911,122,222,...
I don't know if these two are best, nor do I know if they are equivalent efficient.
Edit
Here is a program to calculate the average number of trails for a strategy from an ardent comment. To use it, replace the line '// Input your list here' with your test list and press run.
 A: This is a problem with entropy, in each try you want to maximally lowerize the entropy.
Since I forgot the formulas for entropy I'll go this way... Maximize the probability to unlock in each try.(it's kinda same)
Let n be order of the try and p(n) probability to unlock, then:
$p(1)=\frac{3*9+1}{1000}=\frac{28}{1000}$
After first try you discriminate 28 combinations, so if you failed with 111, you also know that 112,113,...,121,... are not the combination
Lets try your second strategy for the next try, if you pick 211, you will discriminate another 18 combinations (212,213,...210,221,231,...230 but combinations 311,411... you already discriminated, so p(2) with the 2nd strategy is $p(2)=\frac{18}{972}$ 
Lets try 1st strategy, so second pick to be 122, this way you discriminate another 26 combinations cos 121 and 112 were already disc.
So I would decide for strategy to change all three digits, to discriminate another 28 combinations, so $p(2)=\frac{28}{972}$
That's what would you do for the first 10 tries, and for the rest... try on your own :)
A: So, what if you reduced the question to just 3 possible numbers allowed for each digit?  
From $000$ to $222$
In that case, you can cover any of the 27 permutations with just 6 guesses
$$000$$
$$111$$
$$222$$
$$010$$
$$020$$
$$101$$
So that's 6 guesses to get $3^3$.  I'm not sure how this would generalize though. 
Is it $n^2\frac{n-1}{n}=n(n-1)$ which would be 90 in your problem? (This generalization also works for the near-trivial example of only 2 numbers, 0 and 1, in which case 000 and 111 covers any permutation)  
Also, this method is trying to minimize worst case, exhaustive method, not the average number of tries.
Edit: Confirmed $n(n-1)$ also covers every options in base 4 (from 000 to 333). 
000, 111, 222, 333, 
012, 013, 021, 
102, 103, 120, 
210, 
301
And from what I gather from brute force methods, in your original question, cycling through the 000, 111, 222, 333....999 will take out 28 permutations each, for 280. After that, you will be able to make 80 guesses that will take out 9 permutations each, for a total of 90 guesses that cover all 1000 permutations. 
Also, after 29 guesses, you've taken out over 500 of the permutations. 
A: 1) 000: Rules out more than 1 zero
2-4) 011;101;110: Rules out one and zero and more than one 1
25-28) 022;202;220;....099;909;990:  Rules out 0 and more than 1 of any number.
29-34) 123,213,231,132,312,321: Rules any two of 1,2,3
35-52)145...167...189: Rules out any 1s; and any pair of (4,5)(6,7),(89)
53-58)246....642: Rules out 2s with 4or 6 and any pair of 4 or 6.
59-64) 257....752: Rules out 2s with 5 or 7; so only 2s with 8 or 9 are left.  But 8 and 9 can't pair so this is impossible so, rules out 2s.
65- 70)347...743: Rules out 3s with anything less than 5s.  
71-76) 356..653: Rules out 3s with anything other than 8 and 9, which is impossible so rules out 3.
4s are only possible with 8 and 9 so that's impossible.
6s are only possible with 8 and 9.  7 is only possible with 8 or 9.
So any combination we would have guessed in 76 guesses.
I'm not sure how to generalize.
