CS 201: Make-Up Test #2

This test is a 24-hour take-home test. You are welcome to use your notes, your text, enve the library if you are so inclined. However, you are not to communicate with anyone about this test, except with me. I will be in my office all day Thursday, reachable on the phone (277-5699) or by email (moret@cs.unm.edu); after 5pm, us email, which I will check until 10 or 11pm on Thursday night. The test is due Friday morning in class (8am); you can also email it to me at anytime on Thursady or Friday morning until 8am. Your score on this test will be combined with your score on Test #2 to derive a composite score and letter grade.


There are 5 problems, all of equal worth, although not all of equal difficulty. I would expect most everyone to solve 3 of the 5 problems, a fair number to solve 4 of the 5, and a few to solve all 5; two problems (3 and 5) have a subquestion identified as "harder" -- these are strictly bonus work. Show your work! Just writing down a few numbers leaves no margin for error, whereas writing down your reasoning will get you some type of partial credit in case your answer is wrong. Some of the problems (particularly Problem 3) have answers that involve very large numbers -- just give the formula.


Problem 1. If a 5-card hand is dealt from a well shuffled deck of 52 cards, what is the probability that the hand forms a full house? (A full house is made of a triple and a pair -- 3 cards of equal value and 2 cards of equal value; for instance K-K-K-4-4 and 9-9-3-3-3 are full houses.)


Problem 2. You roll some fair six-sided dice. What is the probability that the dice all show different values when you roll

Problem 3. You are arranging red, white, and blue beads in a row. You have a total of 40 beads: 20 red beads, 10 blue beads, and 10 white beans. Problem 4. In the (very British) game of cribbage, scores are tallied by counting every possible combination in your hand that can contribute. Thus, for instance, if pairs, triples, and quadruples all earn you something, a triple would contribute not just the earnings of a triple, but also the earnings of the three pairs you can make from the three cards in the triple.

Suppose we do the same type of counting for the following simple game. You are dealt a hand of 5 cards from a well shuffled deck of 52 cards; you pay $5 for the hand. Every pair of cards of equal value is worth $2; every triple (viewed just as the triple) is worth $3; and every fourtuple (again viewed just as such) is worth $4. Nothing else is worth anything. Thus, for instance, a full house (see Prob. 1 for a definition) gives you 4 possible different pairs and a triple, for a total return of $11; a hand with a triple and two other different cards gives you 3 possible pairs and a triple, for a total return of $9; etc.

What is your expected loss or gain per hand in this game?
(Hint: start by thinking about the possible kinds of hands that have some value, compute the worth of each type, then compute the number of different hands of each type, get the resulting probability, and finish by computing your expected return. Be sure to show intermediate work, since it's easy in this problem to get one of the several types of hands counted wrong.)


Problem 5. In a very simplified version of Bingo, everyone buys "cards" on which are written 10 distinct integers chosen at random between 1 and 100. An announcer has a bag of 100 beads numbered from 1 to 100. She repeatedly pulls a new bead from the bag (at random) and announces its value, upon which announcement each gambler whose card includes that value crosses it from the card. The first gambler to "fill a card" (i.e., cross all 10 values on a card) wins; if several gamblers cross their last value at the same time, the first to shout "Bingo!" wins.