CMU 15-251 Probability I Teachers: Anil Ada Ariel Procaccia (this - PowerPoint PPT Presentation

CMU 15-251 Probability I Teachers: Anil Ada Ariel Procaccia (this time)

Course overview 2

Gambling 101 • • • 5 7 8 3 5 5 5 1 7 1 8 2 8 3 3 3 5 7 2 9 5 7 2 9 3

Gambling 101 • • 4 5 • 1 − = 0.518 6 4

Gambling 101 • • 24 35 • 1 − = 0.491 36 • 5

Gambling 101 6

Pennies and gold • • • 1/6 1. 1/3 2. 2/3 3. 1 4. 7

Language of probability 8

Language of probability • 𝑇 0.5 0.1 • 𝑞 0.2 0.2 0 𝑞 𝑦 = 1 𝑇 𝑦∈𝑇 9

Language of probability • 𝐹 ⊆ 𝑇 0.5 0.1 • Pr 𝐹 = 𝑦∈𝐹 𝑞(𝑦) 𝑦 ∈ 𝑇 • Pr 𝐹 = 0.6 0.2 0.2 0 𝑞 𝑦 = |𝐹| Pr 𝐹 = 𝑇 |𝑇| 𝑦∈𝐹 10

Language of probability • • 𝑇 = { 1,1 , 1,2 , 1,3 , 1,4 , 1,5 , 1,6 , 2,1 , 2,2 , 2,3 , 2,4 , 2,5 , 2,6 , 3,1 , 3,2 , 3,3 , 3,4 , 3,5 , 3,6 , 4,1 , 4,2 , 4,3 , 4,4 , 4,5 , 4,6 , 5,1 , 5,2 , 5,3 , 5,4 , 5,5 , 5,6 , 6,1 , 6,2 , 6,3 , 6,4 , 6,5 , (6,6)} • 1/9 1. 2/9 2. 3/9 3. 4/9 4. 11

Conditional probability • 0.5 𝐵 𝐶 0.1 𝐵 Pr 𝐵 𝐶 = Pr[𝐵∩𝐶] 𝐵 ∩ 𝐶 Pr[𝐶] 0.2 • 𝐵 ∩ 𝐶 𝐶 0.2 0 𝐶 𝑇 12

Pennies and gold, revisited • • • 𝐻 𝑗 𝑗 ∈ {1,2} 1 1 • Pr 𝐻 1 = 2 , Pr 𝐻 1 ∩ 𝐻 2 = 3 1/3 2 • Pr 𝐻 2 𝐻 1 = 1/2 = 3 13

Conditional probability • Pr 𝐵 ∩ 𝐶 = Pr 𝐶 × Pr[𝐵|𝐶] 𝐵 𝐶 𝐶 • 𝐵 𝐶 • Applying iteratively: Pr 𝐵 1 ∩ ⋯ ∩ 𝐵 𝑜 = Pr 𝐵 1 × Pr 𝐵 2 𝐵 1 × ⋯ Pr[𝐵 𝑜 |𝐵 1 , ⋯ , 𝐵 𝑜−1 ] 14

bayes’ rule • Pr B × Pr A B = Pr 𝐵 ∩ 𝐶 = Pr 𝐵 × Pr[𝐶|𝐵] Pr 𝐵 𝐶 = Pr 𝐵 Pr[𝐶|𝐵] Pr[𝐶] 15

Monty Hall problem • • • • 16

Monty Hall problem 17

Monty Hall problem • Pr[𝑄 3 ] Pr[𝑃 2 |𝑄 3 ] • Pr 𝑄 3 𝑃 2 = Pr 𝐵 𝐶 = Pr 𝐵 Pr[𝐶|𝐵] Pr[𝑃 2 ] Pr[𝐶] 1 • Pr[𝑄 3 ] = 3 , Pr 𝑃 2 𝑄 3 = 1, Pr[𝑃 2 ] = 1/2 Pr 𝑄 3 𝑃 2 = 2/3 • • 3/15 1. 4/15 2. 5/15 3. 6/15 4. 18

Independence 𝐵 𝐶 • Pr 𝐵 𝐶 = Pr[𝐵] • 19

The birthday paradox • 𝑛 • 𝑇 = 1, … , 365 𝑛 , s 𝑦 = (𝑦 1 , … 𝑦 𝑛 ) • 𝐹 = 𝑦 ∈ 𝑇 ∃𝑗, 𝑘, s.t. 𝑦 𝑗 = 𝑦 𝑘 } 20

The birthday paradox • 𝐹 𝐹 • 𝐵 𝑗 𝑗 • 1, … , 𝑗 − 1 • 𝐹 = 𝐵 1 ∩ ⋯ ∩ 𝐵 𝑜 • Pr 𝐹 = Pr 𝐵 1 × Pr 𝐵 2 𝐵 1 × ⋯ Pr[𝐵 𝑜 |𝐵 1 , ⋯ , 𝐵 𝑜−1 ] Pr 𝐵 𝑗 𝐵 1 , … , 𝐵 𝑗−1 ? 21

The birthday paradox • 𝐵 1 ∩ ⋯ ∩ 𝐵 𝑗−1 𝑗 − 1 • 𝑗 − 1 𝑗 = 1 − 𝑗−1 • Pr 𝐵 𝑗 𝐵 1 , … , 𝐵 𝑗−1 = 365−(𝑗−1) 365 365 1 𝑛−1 • Pr 𝐹 = 1 × 1 − 365 × ⋯ × 1 − 365 • Pr 𝐹 = 1 − Pr[ 𝐹] 22

The birthday paradox Pr 𝐹 : Pr 𝐹 : 23

The birthday paradox • 1/2 1. 0.75 2. 0.99999999999997 3. 1 4. 24

Birthday attack * • 𝑇 𝑙 𝑔(𝑇) • 𝑇 1 , 𝑇 2 𝑔 𝑇 1 = 𝑔 𝑇 2 • 𝑛 o 𝑔(𝑛) o o 𝑛 𝑔 𝑛 = 𝑔(𝑛 ′ ) 𝑛′ 25

Birthday attack * • • • 1, … , 2 160 • 2 160 = 2 80 • 26

Birthday attack * • • 2 63 • 27

Summary • o o o • o o 28