| 
X | 
X | 
 X | 
t | 
t | |||||
| 
X | 
X | 
X | 
m | 
m | |||||
| 
b | 
b | ||||||||
| 
l | 
c | 
r | |||||||
| 
l | 
c | 
r | |||||||
For each influence, I've color coded the result.
Black is no effect
Green is pair
Blue is deduction
Red is impossible/invalid
So in this first chart we can see that out of six occasions, three have a result and one can never happen in the game.  So 3/5ths or 60% of the time we get at least a pair.
But of course, the "crowd" in the square mighte be in a different configuration.
How about something that looks like this:
| 
X | 
X | 
 X | 
t | 
t | |||||
| 
X | 
X | 
m | 
m | ||||||
| 
X | 
b | 
b | |||||||
| 
l | 
c | 
r | |||||||
| 
l | 
c | 
r | |||||||
Here, our results are different.  There are no invalid interactions, and while two of the interactions have no result, two of them (b) will give us a hit, not just a pair.  If 2/3rds of our interactions give us a result here, that's 66%.  
We can create different charts with three spaces, but are they functionally equivalent to this one?
Nothing to do but test it.
| 
X | 
X | 
t | 
t | ||||||
| 
X | 
X | 
m | 
m | ||||||
| 
X | 
X | 
b | 
b | ||||||
| 
l | 
c | 
r | |||||||
| 
l | 
c | 
r | |||||||
And clearly the answer is no.  We're looking at one result with no influence, but we've raised our pairs to four and reduced our hits to one.  
5/6ths into percentages is more than my brain is up to just now.  But I've got one more chart for us to investigate.
This time, we'll break things up a bit.
| 
X | 
  X | 
t | 
t | ||||||
| 
X | 
X | 
m | 
m | ||||||
| 
X | 
X | 
b | 
b | ||||||
| 
l | 
c | 
r | |||||||
| 
l | 
c | 
r | |||||||
And we get a similar result.  One dud, one hit, and four pairs.
What makes this arrangement functionally equivalent to the one above it?
Something to think about.
 
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