Multiplayer Rock-Paper-Scissors
Python, 41 bytes



lambda r,s,p:[s&gt;p&lt;=r&gt;0,p&gt;r&lt;=s&gt;0,r&gt;s&lt;=p&gt;0]


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Takes counts as (r,s,p). Outputs a three-element list with True at the position of the winner, or all False's if there's a tie or only a single figure appearing.

Rock:     [True, False, False]
Scissors: [False, True, False]
Paper:    [False, False, True]
Tie:      [False, False, False]


We use an alternate characterization without multiplication (except when only one figure is present). Below is the condition for Rock winning, with Scissors and Paper having similar conditions.


  Rock wins if: Paper has the fewest, strictly fewer than Scissors but possibly the same as Rock.


We can write this in Python as s&gt;p&lt;=r using inequality chaining.

Unfortunately, this doesn't correctly handle the special case when r=p=0, saying that Rock wins even though only Scissors is present. To fix this, we strengthen the condition for Rock to win to include r&gt;0 via s&gt;p&lt;=r&gt;0, which makes the only-Scissors case give all False for every condition, matching a tie. 



44 bytes



lambda r,s,p:r*s==s*p==p*r or[s&gt;p&lt;=r,p&gt;r&lt;=s]


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Takes counts as (r,s,p). Outputs as:

Rock:     [True, False]
Scissors: [False, True]
Paper:    [False, False]
Tie:      True    


The "Tie" case also includes where there's only a single figure present.