Solution to Problem #4



Philippe Fondanaiche of Paris (France) solved the first four questions. Darryl K. Nester of Bluffton College, Robin Stokes of the University of New England (Australia), and Ryan Moats of Omaha NE solved all of the problems. Here is Darryl Nester's solution:

The answers are:
  (a) 53 / 5                         = 10.6
  (b) 689 / 45                       = 13.531...
  (c) 3551 / 195                     = 18.210...
  (d) 67151 / 3315                   = 20.256...
  (e) 76,949,045,039 / 2,748,462,675 = 27.997...

In the following, let C(n,k) = (n choose k) = n! / [k! (n-k)!].

Lemma: If a<b, then sum(C(a,k) / C(b,k), k, 0, a) = (b+1) / (b-a+1).

Proof: This is a fairly standard exercise; the key observation is that
C(a,k) / C(b,k) = C(b-k,a-k) / C(b,a), so that the given sum equals
  sum(C(b-k,a-k), k, 0, a) / C(b,a) = C(b+1,a) / C(b,a)
which simplifies to (b+1) / (b-a+1).

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Now, for (a), let N1 be the number of cards flipped over up to and including
the first ace.  Then using standard hypergeometric probabilities:
  P[N1>k] = P[no ace in the first k cards] = C(4,0) C(48,k) / C(52,k)
Now use the fact (valid for any random variable with nonnegative integer
values) that E(N1) = sum(P[N1>k], k, 0, infinity).  In this case the sum
extends only up to k=48 (all probabilities beyond that equal 0), so that
  E(N1) = sum(C(48,k) / C(52,k), k, 0, 48) = 53/5
using the formula from the lemma.

For (b), let N2 be the number of cards flipped over before a king and an ace
have been revealed.  Let A=A(k)={no ace in the first k cards}, and K={no
king in the first k cards}; then
  P[N2>k] = P[A or K] = P[A] + P[K] - P[A and K]
But P[A] = P[K] = C(4,0) C(48,k) / C(52,k), while P[A and K] = C(8,0)
C(44,k) / C(52,k), so
  E(N2) = sum(P[N>k],k) = 2(53/5) - (53/9) = 689/45.

For (c), let Q={no queen in the first k cards}; then we have
  P[N3>k] = P[A or K or Q]
          = P[A] + P[K] + P[Q]
             - P[A and K] - P[A and Q] - P[K and Q]
             + P[A and K and Q]
Since P[A]=P[K]=P[Q]=C(4,0) C(48,k) / C(52,k), while all the two-card
combinations equal C(8,0) C(44,k) / C(52,k), and the three-card combination
equals C(12,0) C(40,k) / C(52,k), we find
  E(N3) = sum(P[N>k],k) = 3(53/5) - 3(53/9) + (53/13) = 3551/195.

In general, probabilities for the waiting time Nj until at least one of each
of j cards have been flipped over can be found like so:
  P[Nj>k] = sum( C(j,n) (-1)^(n-1) C(52-4n,k) / C(52,k), n, 1, j)
so that
  E(Nj) = sum(P[N>k],k) = sum( C(j,n) (-1)^(n-1) * 53/(4n+1), n, 1, j)

Then for (d), E(N4) = 4(53/5)-6(53/9)+4(53/13)-1(53/17) = 67151 / 3315,
while for (e),
  E(N13) = sum( C(13,n) (-1)^(n-1) * 53/(4n+1), n, 1, 13)
         = 76,949,045,039 / 2,748,462,675.



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