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CMU 15-251
Probability I
Teachers:
Anil Ada
Ariel Procaccia (this time)
Course overview
2
Gambling 101
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3
7
71
7
7
1
5
55
5
5
5
3
33
9
9
3
8
28
2
2
8
Gambling 101
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• 1 −5
6
4= 0.518
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Gambling 101
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• 1 −35
36
24= 0.491
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Gambling 101
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Pennies and gold
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1. 1/6
2. 1/3
3. 2/3
4. 1
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Language of probability
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Language of probability
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𝑆
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𝑝
𝑥∈𝑆
𝑝 𝑥 = 1
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0.10.5
𝑆
0.2 00.2
Language of probability
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𝐸 ⊆ 𝑆
• Pr 𝐸 = 𝑥∈𝐸 𝑝(𝑥)
• 𝑥 ∈ 𝑆
Pr 𝐸 =
𝑥∈𝐸
𝑝 𝑥 =|𝐸|
|𝑆|
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0.1
0.2
0.5
00.2
𝑆
Pr 𝐸 = 0.6
Language of probability
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• 𝑆 = { 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)}
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1. 1/9
2. 2/9
3. 3/9
4. 4/9
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Conditional probability
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𝐴 𝐵
Pr 𝐴 𝐵 = Pr[𝐴∩𝐵]Pr[𝐵]
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𝐴 ∩ 𝐵𝐵
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0.1
0.2
0.5
0
0.2
𝑆
𝐴
𝐵
𝐴 ∩ 𝐵
Pennies and gold, revisited
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• 𝐺𝑖 𝑖 ∈ {1,2}
• Pr 𝐺1 =1
2, Pr 𝐺1 ∩ 𝐺2 =
1
3
• Pr 𝐺2 𝐺1 =1/3
1/2=2
3
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Conditional probability
• Pr 𝐴 ∩ 𝐵 = Pr 𝐵 × Pr[𝐴|𝐵]
• 𝐴 𝐵 𝐵𝐴 𝐵
• Applying iteratively:Pr 𝐴1 ∩⋯∩ 𝐴𝑛 = Pr 𝐴1 × Pr 𝐴2 𝐴1 ×⋯Pr[𝐴𝑛|𝐴1, ⋯ , 𝐴𝑛−1]
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bayes’ rule
• Pr B × Pr A B = Pr 𝐴 ∩ 𝐵 = Pr 𝐴 × Pr[𝐵|𝐴]
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Pr 𝐴 𝐵 =Pr 𝐴 Pr[𝐵|𝐴]
Pr[𝐵]
Monty Hall problem
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Monty Hall problem
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Monty Hall problem
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• Pr 𝑃3 𝑂2 =Pr[𝑃3] Pr[𝑂2|𝑃3]
Pr[𝑂2]
• Pr[𝑃3] =1
3, Pr 𝑂2 𝑃3 = 1,
Pr[𝑂2] = 1/2
• Pr 𝑃3 𝑂2 = 2/3
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1. 3/15
2. 4/15
3. 5/15
4. 6/15
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Pr 𝐴 𝐵 =Pr 𝐴 Pr[𝐵|𝐴]
Pr[𝐵]
Independence
• 𝐴 𝐵Pr 𝐴 𝐵 = Pr[𝐴]
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The birthday paradox
• 𝑚
• 𝑆 = 1,… , 365 𝑚, s 𝑥 = (𝑥1, … 𝑥𝑚)
• 𝐸 = 𝑥 ∈ 𝑆 ∃𝑖, 𝑗, s.t. 𝑥𝑖 = 𝑥𝑗}
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The birthday paradox
• 𝐸
• 𝐸
• 𝐴𝑖 𝑖1,… , 𝑖 − 1
• 𝐸 = 𝐴1 ∩⋯∩ 𝐴𝑛•
Pr 𝐸 = Pr 𝐴1 × Pr 𝐴2 𝐴1 ×⋯Pr[𝐴𝑛|𝐴1, ⋯ , 𝐴𝑛−1]
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Pr 𝐴𝑖 𝐴1, … , 𝐴𝑖−1 ?
The birthday paradox
• 𝐴1 ∩⋯∩ 𝐴𝑖−1 𝑖 − 1
• 𝑖 − 1 𝑖
• Pr 𝐴𝑖 𝐴1, … , 𝐴𝑖−1 =365−(𝑖−1)
365= 1 −
𝑖−1
365
• Pr 𝐸 = 1 × 1 −1
365×⋯× 1 −
𝑚−1
365
• Pr 𝐸 = 1 − Pr[ 𝐸]
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The birthday paradox
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Pr 𝐸 :
Pr 𝐸 :
The birthday paradox
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1. 1/2
2. 0.75
3. 0.99999999999997
4. 1
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Birthday attack *
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𝑆 𝑘 𝑓(𝑆)
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𝑆1, 𝑆2 𝑓 𝑆1 = 𝑓 𝑆2•
o 𝑚
o 𝑓(𝑚)
o
𝑚𝑚′ 𝑓 𝑚 = 𝑓(𝑚′)
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Birthday attack *
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1,… , 2160
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• 2160 = 280
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Birthday attack *
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263
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Summary
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o
o
o
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o
o
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