The odds of holecards both being suited

[Cazz]

I believe N*p is the average number of opponents that have 2 cards of the same suit as you.

Let's pretend p = 0.20 (some other condition).
6*p = 1.20. This cannot be the percentage of times
that at least one guy has met the condition. It is
more the 100%.

The probability is
1 - (no one meets the condition)

The chance that one player *misses* (doesn't meet the condition) is 1-p.
The change that N players miss is (1-p)^N .
The probably this doesn't happen is
1- (1-p)^N