Question text is in black, solutions in blue.
Q1: 10 points Q2: 10 points Q3: 10 points Q4: 10 points Q5: 20 points Q6: 40 points Total: 100 points
TRUE. This is the Coupon Collector's Problem with n = 6, and by our analysis the expected value is 6*H6 = 6[1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6] = 14.7, greater than 12. (The true/false question can be answered by just observing that 1 + 1/2 + 1/3 + 1/4 is greater than 2.) Some of you remembered the general result that E(CCn) is approximately n*ln(n), and asserted that the expected value here was 6*ln(6), which is less than 12 because e2 is between 7 and 8. But in fact hn is equal to ln(n) plus a number that is between 0 and 1 (approaching Euler's Constant), and the difference is important here -- you must use Hn rather than ln(n) to get the right answer.
TRUE. The probability is 66/66 = 6*5*4*3*2*1/6*6*6*6*6*6 = 20/6*6*6*6 = 20/1296, less than 2% and thus less than 5%.
TRUE. There are five possible ways to make three of a kind, where the player's hand is a pair of the same rank as one of the five table cards. For each such rank there are (3 choose 2) possible pairs, so there are 5*(3 choose 2) hands making three of a kind out of the (47 choose 2) possible hands.
TRUE. The given probability is the chance that both the player's cards have
rank different from any of the five cards on the table. But the specified
probability is less, because there is a positive chance that the player's two
cards form a pair with each other. The correct probability is
(8 choose 2)*4*4/(47 choose 2), thinking of choosing two unused ranks and then
a card of each rank, or equivalently (32/47)*(28/46), thinking of choosing
any card of an unused rank first and then any card whose rank is not used on
the table or by the first card. The number (8 choose 2)*4*4 evaluates to 448,
while (32 choose 2) is 496.
By the way, I chose boolean answers for the four true/false questions
randomly and uniformly, so it just happened that all four were true.
E(W) = 600(1/1000) + 0(999/1000 = 0.6 dollars expected winnings.
There are 1000 total sequences and 103 = 10*9*8 = 720 no-repeat sequences, probability 720/1000 = 0.72 that the random sequence has no repeats.
The expected winnings from the all-different digit bet are 600*(1/720) = 5/6 dollars. The expected winnings from the not-all-different-digit bet are 600*(1/280) = 15/7 dollars. If I bet the dollar regardless, choosing my number using my secret knowledge, my expected winnings are (0.72)(5/6) + (0.28)(15/7) = 1.2 dollars. In fact, if I am told whether the winning number is in some set S, such that both S and its complement are nonempty, and I choose a uniform random element of S or of S-bar for my bet, my expected winnings are 1.2 dollars no matter what size S may be. (This is because there are now two winning numbers for me out of the 1000 total, so my overall winning probability is now 2/1000 rather than 1/1000.)
There are 213 = 8192 possible subsets of a 13-element set. Many of
you correctly noted that the total is the sum for i from 0 through 13 of
(13 choose i), which can be evaluated by adding the elements of the 13th row
of Pascal's Triangle, which in turn are easily computed from the 12th row which
I gave you. But you forgot one of the basic facts about the Triangle, which is
that the i'th row adds up to 2i.
To choose a set uniformly from all possible sets, Customer A can flip a
fair coins 13 times to choose whether to include each of the 13 vegetables.
Some of you chose a size for her set uniformly before choosing the set, but
this is wrong because there are many more sets of size 6 or 7 (1716 of each,
as you should have noted for part (c)) than there are of size 0 or 13 (1 of
each) and thus her set is far more likely to have size 6 or 7.
There are (13 choose 5) = 1287 subsets of size 5. (You could compute the 1287 directly as 13*12*11*10*/1*2*3*4*5 or add the entries 495 and 792 on the given Pascal's Triangle.) Customer B could choose a set uniformly by taking all the cards of one suit from the deck, shuffling, and dealing out five of them. (He first assigns one vegetable type to each rank.) Equivalently, he could take the entire deck, shuffle it, deal out cards until five different ranks have occurred, and order the vegetables corresponding to those ranks.
The probability is [(13 choose 6) + (13 choose 7)]/8192 = (1716 + 1716)/8192 (using the Triangle and adding the entries 792 and 924) = 3432/8192 = 429/1024 = about 41.9%. I was disappointed that a few people answered 2/14 on the claim that the 14 set sizes were equally likely.
From our analysis of binomial random variables, the mean is np and the variance npq where here n = 13, p = 1/2, and q = 1/2, giving a mean of 13/2 and a variance of 13/4. It is also true, but tedious to use, that the expected value is also the sum for i from 0 to 13 of i times (13 choose 1)/213. The variance could be calculated as E(NV2) - E(NV)2, with E(NV2) calculated from a similar sum, but this is much more work than using the fact that NV is the sum of 13 independent variables, each with mean 1/2 and variance 1/4.
Let Xk be the random variable that is the sum of k independent
values of NV, and hence the total
number of vegetables ordered by Customer
A on k visits to Zane's. The average number of vegetables is thus
Xk/k. Because the variances of independent random variables add,
the variance of Xk is 13k/4, using the result of part (d).
By the Normal Approximation to the Binomial and the Law of Large Numbers,
X_k is well approximated by a normal random variable with mean 13k/2 and
variance 13k/4. (In fact Xk can be thought of as just the sum of
13k independent flips of a fair coin.) To have the average between 6 and 7,
Xk must between 6k and 7k, within k/2 of its mean of 13k/2. So we
need k/2 to be two standard deviations of Xk to have this chance
be about 95%. Equating k/2 with twice
the square root of 13k/4 and squaring both
sides, we get k2/4 = 13k or k = 52. Thus any number greater than
52 will work.
Last modified 14 March 2009