05AB1E, 19 bytes

0 -> Ore
1 -> Brick
2 -> Log
3 -> Wheat
4 -> Sheep

Returns 0 when false, and 1 otherwise.

{γvyDĀi¼¨}g4÷}¾)O3›

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Explanation:

{γvyDĀi¼¨}g4÷}¾)O3› Implicit input, e.g. 0030201
{                   Sort -> 0000123
 γ                  Split into chunks of consecutive elements: [0000, 1, 2, 3]
  vy                For each chunk...
    DĀ                 ...is different than 0?
      i¼¨}                ...if true: increment the counter by 1, and 
                              remove 1 element from the chunk
          g4÷         ...divide the number of elements by 4
             }      End For
              ¾     Push the counter
               )    Wrap the entire stack in a list
                O   Sum of that list
                 3> True if > 3
                    Implicit output

Non-competive solution: 17 bytes

There was a bug in 05AB1E when I first submitted that solution, where some operators badly handled empty inputs. This resulted in this solution answering 1 on an empty input. This has now been fixed, so this solution works just fine.

The difference here is that we add an ore prior to removing one of each resource, indiscriminately, counting the number of resources removed that way. We then decrement the counter by 1 to get the correct number of B, L, W and S.

0«{γε¨g4÷¼}O¾<+3›

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