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Gambling, Probability, and Risk  (Basic Probability and Counting Methods)
First: Your class data Starting with politics…
 
 
Feelings about math and writing…
Optimism…
A gambling experiment ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
What do you want to bet? ,[object Object],[object Object],[object Object]
Rational strategy ,[object Object],[object Object],[object Object],[object Object]
Probability  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Assessing Probability ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Computing  theoretical  probabilities:counting methods ,[object Object],[object Object],Note:  these are called “counting methods” because we have to  count   the number of ways A can occur and the number of total possible outcomes.
Counting methods: Example 1 Example 1:  You draw one card from a deck of cards. What’s the probability that you draw an ace?
Counting methods: Example 2 Example 2.  What’s the probability that you draw 2 aces when you draw two cards from the deck? This is a “joint probability”—we’ll get back to this on Wednesday
Counting methods: Example 2 Numerator:   A  A  , A  A  , A  A  ,  A  A  ,  A  A  , A  A  ,  A  A  ,  A  A  ,  A  A  , A  A  , A  A  , or A  A   = 12 Two counting method ways to calculate this: 1. Consider order:   Denominator = 52x51 = 2652  -- why? . . .     52 cards 51 cards . . .  
Numerator:  A  A  , A  A  , A  A  ,  A  A  ,  A  A  ,  A  A   = 6 Denominator =  Counting methods: Example 2 2.  Ignore order: Divide out order!
Summary of Counting Methods Counting methods for computing probabilities With replacement Without replacement Permutations—order matters! Combinations— Order doesn’t matter Without replacement
Summary of Counting Methods Counting methods for computing probabilities Permutations—order matters! With replacement Without replacement
Permutations—Order matters! ,[object Object],[object Object],[object Object]
Summary of Counting Methods Counting methods for computing probabilities With replacement Permutations—order matters!
Permutations—with replacement With Replacement  – Think coin tosses, dice, and DNA.   “ memoryless” – After you get heads, you have an equally likely chance of getting a heads on the next toss (unlike in cards example, where you can’t draw the same card twice from a single deck).   What’s the probability of getting two heads in a row (“HH”) when tossing a coin?   H H T T H T Toss 1: 2 outcomes Toss 2: 2 outcomes 2 2  total possible outcomes: {HH, HT, TH, TT}
Permutations—with replacement What’s the probability of 3 heads in a row? H H T T H T Toss 1: 2 outcomes Toss 2: 2 outcomes Toss 3: 2 outcomes H T H T H T H T HHH HHT HTH HTT THH THT TTH TTT
Permutations—with replacement When you roll a pair of dice (or 1 die twice), what’s the probability of rolling 2 sixes? What’s the probability of rolling a 5 and a 6?
Summary: order matters, with replacement ,[object Object]
Summary of Counting Methods Counting methods for computing probabilities Without replacement Permutations—order matters!
Permutations—without replacement ,[object Object],[object Object]
Permutation—without replacement ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Quickly becomes a pain! Easier to figure out patterns using a the  probability tree!
Permutation—without replacement # of permutations = 5 x 4 x 3 x 2 x 1 = 5! There are 5! ways to order 5 people in 5 chairs (since a person cannot repeat) E B A C D E A B D A B C D …… . Seat One: 5 possible Seat Two: only 4 possible Etc….
Permutation—without replacement What if you had to arrange 5 people in only 3 chairs (meaning 2 are out)? E B A C D E A B D A B C D Seat One: 5 possible Seat Two: Only 4 possible E B D Seat Three: only 3 possible
Permutation—without replacement Note this also works for 5 people and 5 chairs:
Permutation—without replacement How many two-card hands can I draw from a deck when order matters (e.g., ace of spades followed by ten of clubs is different than ten of clubs followed by ace of spades) . . .     52 cards 51 cards . . .  
Summary: order matters, without replacement ,[object Object]
Practice problems: ,[object Object],[object Object]
Answer 1 ,[object Object],P(success) =  1 (there’s only way to get it right!) / total # of guesses she could make   Total # of guesses one could make randomly: glass one: glass two: glass three:   glass four: 4 choices 3 vintages left  2 left    no “degrees of freedom” left  P(success) = 1 / 4!  = 1/24 = .04167 = 4 x 3 x 2 x 1 = 4!
Answer 2 ,[object Object],[object Object],[object Object]
Summary of Counting Methods Counting methods for computing probabilities Combinations— Order doesn’t matter Without replacement
2. Combinations—Order doesn’t matter ,[object Object],Spoken: “ n  choose  r ” Written as:
Combinations How many two-card hands can I draw from a deck when order does  not  matter (e.g., ace of spades followed by ten of clubs is  the same  as ten of clubs followed by ace of spades) . . .     52 cards 51 cards . . .  
Combinations How many five-card hands can I draw from a deck when order does  not  matter? . . .     52 cards 51 cards 50 cards 49 cards 48 cards . . .   . . .   . . .   . . .  
Combinations   How many repeats total?? … . 1.  2.  3.
Combinations   i.e., how many different ways can you arrange 5 cards…? 1.  2.  3.  … .
Combinations   That’s a permutation without replacement. 5! = 120
Combinations ,[object Object],[object Object],[object Object],[object Object]
Combinations ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],6 permutations of the 3 men (=3!) x 2 permutations of the women (=2!) = 12 12 permutations    1 gender-based seating arrangement
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],   5! possible arrangements of A, B, C, D, and E are reduced to 5!/12 or 5!/(3!2!)   6 permutations of the 3 men (=3!) x 2 permutations of the women (=2!) = 12
Summary ,[object Object],[object Object]
Summary: combinations If   r  objects are taken from a set of  n  objects without replacement and disregarding order, how many different samples are possible?  Formally, “order doesn’t matter” and “without replacement”    use choosing     
Examples—Combinations  ,[object Object],Which of course means that your probability of winning is 1/13,983,816!
Examples How many ways can you get 3 heads in 5 coin tosses?
Summary of Counting Methods Counting methods for computing probabilities Combinations— Order doesn’t matter Without replacement: With replacement:  n r   Permutations—order matters! Without replacement: n(n-1)(n-2)…(n-r+1)=
Gambling, revisited ,[object Object],[object Object],[object Object],[object Object],[object Object]
Pair of the same color? ,[object Object],Numerator = red aces, black aces; red kings, black kings; etc.…= 2x13 = 26
Any old pair? ,[object Object]
Two cards of same suit?
Two cards of same color? Numerator:  26 C 2  x 2 colors = 26!/(24!2!) = 325 x 2 = 650 Denominator = 1326  So, P(pair of the same color) = 650/1326 = 49% chance A little non-intuitive? Here’s another way to look at it… 26x25 RR 26x26 RB 26x26 BR 26x25 BB 50/102 Not quite 50/100 . . .     52 cards 26 red branches 26 black branches From a Red branch: 26 black left, 25 red left . . .   From a Black branch: 26 red left, 25 black left
Rational strategy? ,[object Object],[object Object],[object Object]
Rational strategy? ,[object Object],[object Object],[object Object]
Rational strategy? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Rational strategy… ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Practice problem: ,[object Object],[object Object]
In-Class Exercises: Answer ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Use SAS as a calculator ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
For class of 30? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
In this class? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Gambling, Probability, and Risk

  • 1. Gambling, Probability, and Risk (Basic Probability and Counting Methods)
  • 2. First: Your class data Starting with politics…
  • 3.  
  • 4.  
  • 5. Feelings about math and writing…
  • 7.
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  • 13. Counting methods: Example 1 Example 1: You draw one card from a deck of cards. What’s the probability that you draw an ace?
  • 14. Counting methods: Example 2 Example 2. What’s the probability that you draw 2 aces when you draw two cards from the deck? This is a “joint probability”—we’ll get back to this on Wednesday
  • 15. Counting methods: Example 2 Numerator: A  A  , A  A  , A  A  , A  A  , A  A  , A  A  , A  A  , A  A  , A  A  , A  A  , A  A  , or A  A  = 12 Two counting method ways to calculate this: 1. Consider order: Denominator = 52x51 = 2652 -- why? . . .     52 cards 51 cards . . .  
  • 16. Numerator: A  A  , A  A  , A  A  , A  A  , A  A  , A  A  = 6 Denominator = Counting methods: Example 2 2. Ignore order: Divide out order!
  • 17. Summary of Counting Methods Counting methods for computing probabilities With replacement Without replacement Permutations—order matters! Combinations— Order doesn’t matter Without replacement
  • 18. Summary of Counting Methods Counting methods for computing probabilities Permutations—order matters! With replacement Without replacement
  • 19.
  • 20. Summary of Counting Methods Counting methods for computing probabilities With replacement Permutations—order matters!
  • 21. Permutations—with replacement With Replacement – Think coin tosses, dice, and DNA.   “ memoryless” – After you get heads, you have an equally likely chance of getting a heads on the next toss (unlike in cards example, where you can’t draw the same card twice from a single deck).   What’s the probability of getting two heads in a row (“HH”) when tossing a coin? H H T T H T Toss 1: 2 outcomes Toss 2: 2 outcomes 2 2 total possible outcomes: {HH, HT, TH, TT}
  • 22. Permutations—with replacement What’s the probability of 3 heads in a row? H H T T H T Toss 1: 2 outcomes Toss 2: 2 outcomes Toss 3: 2 outcomes H T H T H T H T HHH HHT HTH HTT THH THT TTH TTT
  • 23. Permutations—with replacement When you roll a pair of dice (or 1 die twice), what’s the probability of rolling 2 sixes? What’s the probability of rolling a 5 and a 6?
  • 24.
  • 25. Summary of Counting Methods Counting methods for computing probabilities Without replacement Permutations—order matters!
  • 26.
  • 27.
  • 28. Permutation—without replacement # of permutations = 5 x 4 x 3 x 2 x 1 = 5! There are 5! ways to order 5 people in 5 chairs (since a person cannot repeat) E B A C D E A B D A B C D …… . Seat One: 5 possible Seat Two: only 4 possible Etc….
  • 29. Permutation—without replacement What if you had to arrange 5 people in only 3 chairs (meaning 2 are out)? E B A C D E A B D A B C D Seat One: 5 possible Seat Two: Only 4 possible E B D Seat Three: only 3 possible
  • 30. Permutation—without replacement Note this also works for 5 people and 5 chairs:
  • 31. Permutation—without replacement How many two-card hands can I draw from a deck when order matters (e.g., ace of spades followed by ten of clubs is different than ten of clubs followed by ace of spades) . . .     52 cards 51 cards . . .  
  • 32.
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  • 34.
  • 35.
  • 36. Summary of Counting Methods Counting methods for computing probabilities Combinations— Order doesn’t matter Without replacement
  • 37.
  • 38. Combinations How many two-card hands can I draw from a deck when order does not matter (e.g., ace of spades followed by ten of clubs is the same as ten of clubs followed by ace of spades) . . .     52 cards 51 cards . . .  
  • 39. Combinations How many five-card hands can I draw from a deck when order does not matter? . . .     52 cards 51 cards 50 cards 49 cards 48 cards . . .   . . .   . . .   . . .  
  • 40. Combinations   How many repeats total?? … . 1. 2. 3.
  • 41. Combinations   i.e., how many different ways can you arrange 5 cards…? 1. 2. 3. … .
  • 42. Combinations   That’s a permutation without replacement. 5! = 120
  • 43.
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  • 48. Summary: combinations If r objects are taken from a set of n objects without replacement and disregarding order, how many different samples are possible? Formally, “order doesn’t matter” and “without replacement”  use choosing   
  • 49.
  • 50. Examples How many ways can you get 3 heads in 5 coin tosses?
  • 51. Summary of Counting Methods Counting methods for computing probabilities Combinations— Order doesn’t matter Without replacement: With replacement: n r Permutations—order matters! Without replacement: n(n-1)(n-2)…(n-r+1)=
  • 52.
  • 53.
  • 54.
  • 55. Two cards of same suit?
  • 56. Two cards of same color? Numerator: 26 C 2 x 2 colors = 26!/(24!2!) = 325 x 2 = 650 Denominator = 1326 So, P(pair of the same color) = 650/1326 = 49% chance A little non-intuitive? Here’s another way to look at it… 26x25 RR 26x26 RB 26x26 BR 26x25 BB 50/102 Not quite 50/100 . . .     52 cards 26 red branches 26 black branches From a Red branch: 26 black left, 25 red left . . .   From a Black branch: 26 red left, 25 black left
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