Puzzler's Edge

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I'm going to make a simplifying assumption that sampling with and without replacement are similar, as there is a ∏_n=999001^n=1000000 / 1000000^1000≈.6067 chance of no repeats, and any repeats will likely not impact the size of the sample, so the binomial distribution is a fair assumption.

For triangles, there is a 300/1000=.3 chance of a success, so the odds of getting 500 successes in 1000 trials is 3.479×10^-40.
For the squares, given 500 triangles, there is a 200/700=.28 chance of a success, so the odds of getting 50 successes in 500 trials is 2.532×10^-24.
For the pentagons, given the others matched, there is a 350/500=.7 chance of a success, so the odds of getting 50 successes in 450 trials is 1.129×10^-150.
For the hexagons, given the others matched, there is a 100/150=.67 chance of a success, so the odds of getting 100 successes in 400 trials is 4.028×10^-65.
For the kites, given the others matched, there is a 48/50=.96 chance of a success, so the odds of getting 290 successes in 300 trials is .105892.
These are all independent, so the odds are all those terms multiplied together, or approximately 4.24×10^-279

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EDIT:
Well, guess who looked up the multivariate hypergeometric distribution. It's surprisingly easier than what I thought it would be. It's simply P(X_i=x_i)=∏(n_iCx_i)/(nCx). Which I can easily plug into Wolfram Alpha and get 1.80886819768606391029406770736646900211436086948272×10^-269. So I was off by a factor of 10 billion. Whoops. Granted, still ridiculously unlikely, but not as much. So yes, that's much better than what you had before.