In-Class Exercises
In each class, we present a problem, question, or exercise that we will work on during class. After class, you will find those problems stated on this page -- most recent first -- perhaps along with some of the results of our work in class.
Exercise from Class #8
Answers to questions about Sol Lewitt's art:
1) Construct all possible squares containing two different arcs chosen from the 8 available. For example, the top left corner of the work contains arcs 1 and 2. There are 28 different ones of these "a/b" squares.2) Place a/b squares in the 4 x 7 array as shown.
Exercise from Class #7
Exercise from Class #6
Exercise from Class #5
Traveling Salesman Problem and Minimum Spanning Tree of a graph
Exercise from Class #4
Approach presented by Steve:
First, figure out how many different hands are possible. Than ask how many of those hands will be Royal Flushes. (Royal Flushes) divided by (possible hands) equals the odds against drawing a Royal Flush.
Q: How many 5-card hands in a deck of 52 cards?
A: There are 52x51x50x49x48 = 311,875,200 ways to pick a string of 5 cards, one at a time, from the deck. But each string of 5 cards can be rearranged (as you can do in a hand of cards) in 5x4x3x2x1 = 120 ways. So there are actually only 311,875,200/120 = 2,598,960 distinct 5-card hands.
Q; What are the odds of drawing 5 cards that give a Royal Flush?
A: There are 4 possible hands that are Royal Flushes, one in each suit. So the odds are 4/2,598,960 = 1/649,740 = .0000015390…
Approach presented by Gale:
“What is the probability of drawing a Royal Flush (5 card poker hand)”?
(A probability is a number between 0 (no chance) and 1 (lead-pipe cinch). So the probability of heads in a coin flip is 1/2, or "1 out of 2", or more obscurely, 0.5.)
A Royal Flush is 10-J-Q-K-A, all in the same suit.
Think about drawing the cards one at a time, and the probability that each draw leads one step toward the Royal Flush.
• 1st draw must be one of 10-J-Q-K-A, of any suit. In the deck of 52 cards, there are 20 of these "high cards", and any one of twenty cards (any suit) will get you off on the right foot, and also will determine the suit for the whole hand. So for the first draw, the probability, P1, that you draw a promising card is
P1 = 20/52 (20 out of 52, or 20 possible successes out of 52 tries).
• 2nd draw must be one of the (only) four remaining high cards in the same suit. From the remaining 51 cards, the probability of drawing one of these cards is
P2 = 4/51
• 3rd draw must be one of the three remaining high cards in the right suit from 50 cards, so
P3 = 3/50
• Continuing this line of thinking for the 4th draw,
P4 = 2/49
• 5th draw,
P = 1/48
These five events must occur in succession, so the overall probability of the jackpot is the product of P1 through P5:
P(Flush) = P1 x P2 x … x P5 = 20/52 x 4/51 x 3/50 x 2/49 x 1/48 = 480/311,875,200 =
1.539 x 10–6 or 0.000001539 or 1 Royal Flush out of 649,740 hands.
A long shot, but much much much much more probable than winning at Powerball.
It's not really so amazing that both methods give precisely the same results. Both methods reason with exactly the same numbers, but the different ways of thinking about the problem lead to grouping those numbers differently.
If you figured this out by another line of thinking, or by either one of these approaches, you should be flushed with pride.
Exercise from Class #3
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Exercise from Class #2
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Exercise from Class #1
Geometric intuition that the sum of odd numbers is a square number:
Example (blue) and proof (red) that the sum of the first n odd numbers always equals the square of n.
