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« on: March 28, 2017, 05:42:17 pm »
First question:
The cap of 2 Events or Landmarks per kingdom actually makes this easier. Consider drawing 10 cards then looking to see if you have 0, 1, 2, etc. events/landmarks. If you've drawn 0, you are done. If you have exactly one event/landmark, draw one more card. If it's an event/landmark, you max out at 2; if it's not, you've filled the kingdom and have 1. If you have drawn 2 or more, you max out at 2 and discard all events/landmarks until you fill out the 10 cards for the kingdom.
Let t = number of kingdom cards (not including events and landmarks) and let w = number of events and landmarks. (I am thinking "tall"/portrait orientation cards and "wide"/landscape orientation cards)
The expected number of events or landmarks in a 10 card kingdom is equal to ....
0 TIMES Probability of drawing 10 tall cards
PLUS 1 TIMES Probability of drawing 9 tall cards and 1 wide card, then 1 more tall card
PLUS 2 TIMES Probability of drawing 9 tall cards and 1 wide card, then 1 more wide card
PLUS 2 TIMES Probability of drawing at most 8 tall cards
In math, that becomes:
0 * C(t,10)*C(w,0)/C(t+w,10)
+ 1 * C(t,9)*C(w,1)/C(t+w,10) * (t-9)/(t+w-10)
+ 2 * C(t,9)*C(w,1)/C(t+w,10) * (w-1)/(t+w-10)
+ 2 * (1 - C(t,10)*C(w,0)/C(t+w,10) - C(t,9)*C(w,1)/C(t+w,10))
Simplifying, the expected number of event or landmarks is
2 - 2 * C(t,10)/C(t+w,10) - C(t,9)*w/C(t+w,10) * (t-9)/(t+w-10)
For t = 330 275 and w = 55 we get approximately 1.273 1.416.
C(n,k) is "n choose k".
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Note: When sampling n cards without replacement from a deck with two types (of size t and w), the probability of drawing exactly k of the first type is C(t,k)*C(w,n-k)/C(t+w,n).
Edit: Correct number of kingdom cards not including landmarks and events is evidently 275. Also, the probability on the last line, probability of at most 8 tall cards, is just 1 minus probability of exactly 10 tall cards minus the probability of exactly 9 tall cards, as Mic mentions below. That shortcut is what the formula above reflects.
Edit 2: Simplified formula.