In a game where there are
d cards from Dark Ages and
p cards from Prosperity, the probabilities are:
Method A (Draw 2 cards, 1st to check for Shelters and 2nd to check for Colonies):
P(Shelters + Colonies) = d*p/90
P(Shelters, no Colonies) = d*(9-p)/90
P(Colonies, no Shelters) = (9-d)*p/90
P(No Shelters or Colonies) = ((9-d)*(9-p)+9)/90
Method B (Draw 1 card for Shelters, reshuffle then draw 1 card for Colonies):
P(Shelters + Colonies) = d*p/100
P(Shelters, no Colonies) = d*(10-p)/100
P(Colonies, no Shelters) = (10-d)*p/100
P(No Shelters or Colonies) = (10-d)*(10-p)/100
Method C (Draw 2 cards, check both for Shelters and Colonies):
P(Shelters + Colonies) = 2d*p/90 = d*p/45
P(Shelters, no Colonies) = (19d-dē-2dp)/90
P(Colonies, no Shelters) = (19p-pē-2dp)/90
P(No Shelters or Colonies) = (10-d-p)*(9-d-p)/90
So if we have 5 each of Dark Ages and Prosperity:
Method: | A | B | C |
S+C | 27.8% | 25.0% | 55.6% |
S, no C | 22.2% | 25.0% | 22.2% |
C, no S | 22.2% | 25.0% | 22.2% |
No S or C | 27.8% | 25.0% | 0.0% |
And if we have 1 of each, and 8 of either:
Method: | A | B | C |
S+C | 1.1% | 1.0% | 2.2% |
S, no C | 8.9% | 9.0% | 17.8% |
C, no S | 8.9% | 9.0% | 17.8% |
No S or C | 81.1% | 81% | 62.2% |
So Method C will give a much higher chance of at least one, and of both being in the game compared to the other two, whereas method B tends to smooth out the chances of those while giving a reasonable chance of neither being in. I'd say Method A is a decent compromise between fairness/accuracy to the rules and being easy to implement (only have to shuffle the randomisers once).