fairness vs rules

I'm gonna respond to this because I disagree with "= P[X1...X7] + P[not X1...X7]*P[Xw]" shouldn't it be = P[X1...X7] + P[not Y1...Y7]*P[Xw]" ? because AAAABBAAC (3^9-77) carry the 1 = imconfusedsomuch. lol impressive if that actually makes any sense, i feel so dumb.

No, it is correct. What I meant was, two things can happen: 1) either you have X1,X2,..X7, in this case it doesn't matter whether or not one of the cards is replaced by a wildcard (this is P[X1...X7]), or 2) you don't have a combination (P[not X1...X7]) so you need a wildcard, which makes the probability P[not X1...X7]*P[Xw] The total probability is the sum of the two. Here's a small python code to test the numbers numerically. I am a physicist, and I used to teach probability theory. I might carry this a bit too far in game and my intuition for probabilities is clearly not very good given the moves I tend to make :D [spoiler=python script] [i]#!/usr/bin/python3 import collections,random x1 = [1,1,1,2,2,2] x2 = [1,1,1,3,3,3] x3 = [2,2,2,3,3,3] x4 = [1,1,2,2,3,3] x5 = [1,1,1,1,1,1] x6 = [2,2,2,2,2,2] x7 = [3,3,3,3,3,3] compare = lambda x, y: collections.Counter(x) == collections.Counter(y) n = [0]*8 maxiter = int(1e5) for i in range(maxiter): l=[] for j in range(6): l.append(random.randint(1,3)) n[0] += 1 if compare(l,x1) : n[1] += 1 elif compare(l,x2) : n[2] += 1 elif compare(l,x3) : n[3] += 1 elif compare(l,x4) : n[4] += 1 elif compare(l,x5) : n[5] += 1 elif compare(l,x6) : n[6] += 1 elif compare(l,x7) : n[7] += 1 print(n) print("P[X1] = %5.2f %% (AAA BBB)"%(float(n[1])/n[0]*100)) print("P[X2] = %5.2f %% (AAA CCC)"%(float(n[2])/n[0]*100)) print("P[X3] = %5.2f %% (BBB CCC)"%(float(n[3])/n[0]*100)) print("P[X4] = %5.2f %% (ABC ABC)"%(float(n[4])/n[0]*100)) print("P[X5] = %5.2f %% (AAA AAA)"%(float(n[5])/n[0]*100)) print("P[X6] = %5.2f %% (BBB BBB)"%(float(n[6])/n[0]*100)) print("P[X7] = %5.2f %% (CCC CCC)"%(float(n[7])/n[0]*100)) print("probability of any 2 pairs from 6 cards = %5.2f %%"%(float(sum(n)-n[0])/n[0]*100))[/i] [/spoiler]

Nice one. Haven't reviewed the math properly (too lazy) but the python code seems fine (or maybe not - see edit 2). I also edited your post to put the python script in a spoiler for you, saves some space. I also thought the 7 card double turn in was more like 93% (again, too lazy to do the math) - not that it matters much, in both cases enough to rely on, but nice to know anyways. [b]Edit:[/b] I modified your script a bit to work for 7 cards, and found that it is more like 94% chance of having no double turn in (ignoring wild cards). [spoiler=Python script] #!/usr/bin/python3 import collections,random def six_cards(): x1 = [1,1,1,2,2,2] x2 = [1,1,1,3,3,3] x3 = [2,2,2,3,3,3] x4 = [1,1,2,2,3,3] x5 = [1,1,1,1,1,1] x6 = [2,2,2,2,2,2] x7 = [3,3,3,3,3,3] compare = lambda x, y: collections.Counter(x) == collections.Counter(y) n = [0]*8 maxiter = int(1e5) for i in range(maxiter): l=[] for j in range(6): l.append(random.randint(1,3)) n[0] += 1 if compare(l,x1) : n[1] += 1 elif compare(l,x2) : n[2] += 1 elif compare(l,x3) : n[3] += 1 elif compare(l,x4) : n[4] += 1 elif compare(l,x5) : n[5] += 1 elif compare(l,x6) : n[6] += 1 elif compare(l,x7) : n[7] += 1 print(n) print("P[X1] = %5.2f %% (AAA BBB)"%(float(n[1])/n[0]*100)) print("P[X2] = %5.2f %% (AAA CCC)"%(float(n[2])/n[0]*100)) print("P[X3] = %5.2f %% (BBB CCC)"%(float(n[3])/n[0]*100)) print("P[X4] = %5.2f %% (ABC ABC)"%(float(n[4])/n[0]*100)) print("P[X5] = %5.2f %% (AAA AAA)"%(float(n[5])/n[0]*100)) print("P[X6] = %5.2f %% (BBB BBB)"%(float(n[6])/n[0]*100)) print("P[X7] = %5.2f %% (CCC CCC)"%(float(n[7])/n[0]*100)) print("probability of any 2 pairs from 6 cards = %5.2f %%"%(float(sum(n)-n[0])/n[0]*100)) def seven_cards(): x1 = [1,1,1,1,1,2,2] x2 = [1,1,1,1,1,3,3] x3 = [2,2,2,2,2,1,1] x4 = [2,2,2,2,2,3,3] x5 = [3,3,3,3,3,1,1] x6 = [3,3,3,3,3,2,2] compare = lambda x, y: collections.Counter(x) == collections.Counter(y) n = [0]*7 maxiter = int(1e5) for i in range(maxiter): l=[] for j in range(7): l.append(random.randint(1,3)) n[0] += 1 if compare(l,x1) : n[1] += 1 elif compare(l,x2) : n[2] += 1 elif compare(l,x3) : n[3] += 1 elif compare(l,x4) : n[4] += 1 elif compare(l,x5) : n[5] += 1 elif compare(l,x6) : n[6] += 1 print(n) print("P[X1] = %5.2f %% (AAA BBB)"%(float(n[1])/n[0]*100)) print("P[X2] = %5.2f %% (AAA CCC)"%(float(n[2])/n[0]*100)) print("P[X3] = %5.2f %% (BBB CCC)"%(float(n[3])/n[0]*100)) print("P[X4] = %5.2f %% (ABC ABC)"%(float(n[4])/n[0]*100)) print("P[X5] = %5.2f %% (AAA AAA)"%(float(n[5])/n[0]*100)) print("P[X6] = %5.2f %% (BBB BBB)"%(float(n[6])/n[0]*100)) print("probability of no 2 pairs from 7 cards = %5.2f %%"%(float(sum(n)-n[0])/n[0]*100)) seven_cards() [/spoiler] [b]Edit 2:[/b] For six cards: What about the case AAA ABC (and likewise cases)?

Ah youre right, oops! Probability theory was too long ago. Always some forgotten cases! For 7 I only considered one case (AAAAA BB), of course there are also the other identical ones (AAAAA CC) etc, so it 6 times more likely, so indeed about 94%. And for 6 there is also AAA ABC as you said. There are 6x5x3=90 ways for this : 6 places for B, 5 for C, and the the major can be A, B or C. P[AAA ABC] = 30/729 ~ 4% So this adds another ~12 %, giving a total probability of exactly 1/3 to hit 2 pairs with 6 cards without wildcards.

All that mathematics looks very impressive, but there is another factor involved and that is, when attempting to acquire 2 sets from 7 cards, you have cards that you [b]can see[/b] and if you have 3 cards in your hand, the odds are very simple to work out. If those 3 cards are [b]all different colour[/b] or one is Black, your odds are 100% guaranteed to have 2 sets from 7 cards total. If those 3 cards are any other combination, your odds are 95%. The chances you refer to are just calculating possible combinations of sets of 3 from 7 cards using 3 colours & wild cards, not the actual possibilities with the advantage of being able to already see some of those cards. Also, for all your calculations, are you aware that on some maps there is an additionally assigned colour card where it is not possible to make an even distribution.