GOING LARGE
Finding out the values for 10,000 really only requires us to plug 10,000 into our formulas rather than 50.
  This sounds easy and in theory it is easy except for that even in Derive it took so long to compute the sums that I had to leave and come back.

Zac and I were able to get some information, though:
Total Draws: C(10000,3) = 166,61,66,70,000
Total Triangles: 83,29,58,37,500
Total Acute:  35,740,274,335
Total Obtuse: 47555548565
Total Rights: 14,600

Click here to see the formulas used to solve these.

for 10,000 straws:
P(drawing a triangle) = .4999
P(drawing an acute) = .2145
P(drawing an obtuse) = .2854
P(drawing a right) = .0000000876
P(acute given triangle) = .429
P(obtuse given triangle) = .5709
P(right given triangle) = .000000175

In comparison to 50:
P(drawing a triangle) = .48
P(acute) = .198        or         P(acute given triangle) = .407
P(obtuse) = .286        or        P(obtuse given triangle) = .589
P(right) = .001        or        P(right given triangle) = .002

I am guessing that the probability of drawing a triangle will go to .5 as the number of straws goes to infinity,
but I don't know quite what to think about the rest.

Let's determine probabilities of certain outcomes for the game:

The probability of you winning if both got a triangle is equal to:
P(right given t.)*(P(acute given t.) + P(obtuse given t.)) + P(acute given t.)*(P(obtuse given t.)
= .000000175*(.429+.5709)+.429*.5709 = .2449

The probability of you tying is:
P(right given t.)^2 + P(acute given t.) ^2 + P(obtuse given t.) = .000000175^2 + .429^2 + .5709^2 = .5099

That's a high probability.  You would have to break the tie when that happens.  
One thing we always will know is that someone has to have the smallest length because no two pieces are of the same length.  
This is a sure fire way to break a tie.

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