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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