Discrete math urn question without replacement. An urn contains 30 purple gems, 25 green gems, 20 blue gems, 10 orange gems, and 5 red gems. Randomly picking 3 matching gems, without replacement, wins the game. What is the probability of picking the 3 winning purple, green, blue, orange and red gems respectively?
My method of calculation is not matching my simulation. They way I calculated the problem is as follows. I first listed out all possible winning outcomes by number of gems chosen for each color. There are 405 outcomes total as at least one color must have 3 gems and the pick of the remaining gems can be either 0,1 or 2. 3 to the power of 4 results in 81, multiplied by the 5 different colors (or positions if you will) gives us 405 different unique winning outcomes.
I then calculated the number of "ways" you can pick the order of the winning outcomes. For example there is only 1 way to pick 0 red, 0 blue, 0 green, 0 orange and 3 purple gems. While there is 3 ways to pick 0 purple, 0 green, 0 blue, 1 orange and 3 red gems. (ORRR, RORR, RROR)
Then I calculated the total hits. For example the total hits for picking 1 orange and 3 red is 600,  (5 x 4 x 3=60) for the red hits multiplied by 10 for the orange hit. I then calculated the sample space by taking the permutation of the total number of gems by the number of gems picked. Dividing the total hits by the sample space and then multiplying by the number of "ways" results in my probability for each unique winning outcome. Summing the wins for each color gave me the following probabilities.

*

*Purple: 0.4530

*Green: 0.3132

*Blue: 0.1941

*Orange: 0.0355

*Red: 0.0042

I wrote a quick simulation in C# which gives me the following probabilities after 10 million sims (rounded to 4 decimal places).

*

*Purple: 0.4434

*Green: 0.3109

*Blue: 0.1971

*Orange: 0.0414

*Red: 0.0072

I do think the error is in my theoretical calculation as it's not a very complicated problem to sim. Any help is greatly appreciated!
 A: After the update OP provided I don't understand the source of all the confusion. At the end of a "game" you have 3 of one color gem, and up to 3 (exclusive) of every other color. This gives you $5*3^4=405$ possible states to look at. Listing these out in Excel makes the problem a lot easier. First let $<0,3,2,1,2>$ represent 0 purples, 3 green, 2 blue, 1 orange, and 2 red. Removing one green (this has to be picked last as it ends the game) and dividing off repetitions there are $6!/(2!2!2!)=630$ arrangements of these 6 gems. Picking one of these arrangements (to later multiply by the number of arrangements) we can count the number of ways this can be achieved. Using permutations you have
$P(30,0)$ for purple
$P(25,2)$ for green
$P(20,2)$ for blue
$P(10,1)$ for orange
$P(5,1)$ for red
There are also N-2 ways to pick the last color (where N is the number of that color in the urn), so 23 for green. Multiplying this out:
$23*P(30,0)*P(25,2)*P(20,2)*P(10,1)*P(5,1)=1,048,800,000$
Multiplying by the number of arrangements:
$1,048,800,000*630= 660,744,000,000$.
There are $P(90,8)=3,125,425,824,259,200$ ways to select 8 gems, giving the probability of $660744000000/3125425824259200$. Doing this for all possible states I matched your original probabilities:
Purple: 0.4530
Green: 0.3132
Blue: 0.1941
Orange: 0.0355
Red: 0.0042
Guessing your simulation is off.
