There is a flaw in Ed Sheldon's calculation of P(N), that is the
probability that a group of N cards would have zero sets, for values
of N>=5. To start the explanation of what Sheldon gets wrong, start
with the case where N=5. Sheldon is right that when N=4 there are 6
base pairs in that instance. However, this does not necessarily mean
that there would be 6 cards that could complete a set. If two of
those pairs share a complement, that is a card that completes a set,
then there would only be 5 cards that would complete a set, one of
them completing two sets. (When N=4, it is impossible for there to be
more than one pair of base pairs sharing one card to complete a set,
thus no fewer than 5 cards would complete a set when N=4, the proof
of which would distract from the calculation of the subsequent
probabilities.) Thus, P(5) would not equal P(4)*71/77 as Sheldon
indicates, rather, it the correct equation would be P(5)=P(4)*
(71a+72b)/77 where a is the probability that no two base pairs share
a common complement and b is the probability that there is one
instance of two pairs sharing one complement and a+b=1 by identity.


Now to calculate a and b. There are 81 cards in the deck, each of
which complement 40 pairs of cards. Thus, there are 81*[40]C[2],
where [m]C[n]=m!/(m-n)!/n!, ways that two base pairs can share one
complement out of [81]C[4] ways to select 4 cards. Of the total ways
to select 4 cards, 78*[81]C[2]/3 would include a set. Thus, b=81*[40]
C[2]/([81]C[4]-78*[81]C[2]/3)=13/325. Putting everything together, P
(5)=75/79*((71*312+72*13)/325)/77 which equals roughly 0.87588.


For higher values of N, Sheldon's method requires considering the
possibility of more than two pairs with one complement or more than
one group of pairs each sharing one complement. Thus, P(6) would
equal P(5)*(66a+67b+68c)/76 where a and b are as defined above and c
is probability of two groups of pairs each sharing one complement and
the identity a+b+c=1 holds. Likewise, P(7)=P(6)*(60a+61b+62c+63d+64e)
/75 though my guess is that e=0, but I haven't gone through the
possibilities enough to prove it. I have not extended this process
all the way to the point where P(N)=0, so I would not speculate if
the proposer is correct that P(20)>0, or even that Sheldon is wrong
that P(14)=0, only that Sheldon's calculations of P(N) for 5<=N<=13
are smaller than their correct values.


Another method of calculating P(N) is to calculate separate values
for S(N), the number of possible groups of N cards containing at
least one set, and C(N), the number of possible groups of N cards. S
(1) and S(2) are obviously 0 and C(N)=[81]C[N] for all values of N. S
(3) is the number of sets in the deck. Each set has three pairs, thus
the number of sets in the deck is the number of pairs, or [81]C[2],
divided by 3. Substituting, S(3)/C(3)=1/79, or P(3)=78/79. S(4) is
the number of ways to select a set multiplied by the number of ways
to select an additional card, or S(3)*78. Thus, S(4)/C(4)=4/79 or P
(4)=75/79.


S(5) is more complicated because of the possibility of their being
two sets. This can be simplified by breaking S(5) into S[1](5)+S[2]
(5) where S[x](N) is the number of ways to select x sets with N
cards. S[1](5)=S(3)*78*74/2 for 78*74/2 ways to select a pair of
cards that do not form a set with a group of 3 cards and S[2](5)=81*
[40]C[2]. Adding together and dividing by C(5) yields 39260/316316 or
P(5) roughly equals 0.87588.


The method can be extended to larger values of N, but requiring more
terms as N gets larger both values of x in S[x ]and for each
component S[x].