Decision Analysis: The Mudslide Problem

ML Essentials

Vadim Sokolov

Case Study: Mudslide Threat

I live in a house at risk of a mudslide damage.

Option A: Build a protective wall ($10,000).
Damage Cost: $100,000 if the house is hit (and wall fails/absent).
Wall Effectiveness: 95% protection.
Probability of Mudslide: $P(\text{Slide}) = 0.01$.

What is the best course of action?

Decision Tree: Initial Options

graph LR
    Start((Decision)) --> Build[Build Wall]
    Start --> NoBuild[Don't Build]
    
    Build -- "$10,000" --> WallNode{Slide?}
    WallNode -- "0.01" --> FailNode{Wall Fails?}
    FailNode -- "0.05" --> Loss["$100,000 Cost"]
    FailNode -- "0.95" --> NoLoss["$0 Cost"]
    WallNode -- "0.99" --> NoSlide["$0 Cost"]
    
    NoBuild -- "$0" --> SlideNode{Slide?}
    SlideNode -- "0.01" --> Loss2["$100,000 Cost"]
    SlideNode -- "0.99" --> NoLoss2["$0 Cost"]

Comparison: No Test

Don’t Build

$EV = 0.01 \times \$100,000 = \$1,000$

Build (No Test)

$EV = \$10,000 + (0.01 \times 0.05 \times \$100,000) = \$10,050$

[!IMPORTANT] Based purely on expected cost, Don’t Build is the rational choice despite the high impact of a slide.

The Geological Test

A test is available to better estimate the risk.

Cost: $3,000
Accuracy:
- $P( T \mid \text{Slide} ) = 0.90$
- $P( \text{not } T \mid \text{No Slide} ) = 0.85$

Should we take the test?

Updating Probabilities: Bayes’ Rule

Probability of Positive Test $P(T)$:

$P(T) = (0.90 \times 0.01) + (0.15 \times 0.99) = 0.1575$

Posterior $P(\text{Slide} \mid T)$:

$P(\text{Slide} \mid T) = \frac{0.90 \times 0.01}{0.1575} \approx 0.0571$

Posterior $P(\text{Slide} \mid \text{not } T)$:

$P(\text{Slide} \mid \text{not } T) = \frac{0.1 \times 0.01}{0.8425} \approx 0.0012$

The Testing Strategy

If we test:

If $T$: Build the wall.
If not $T$: Don’t build.

Expected Cost with Test:

\[\begin{aligned} &\text{Test Cost} + P(T) \times \text{EV(Build} \mid T) \\ &\quad + P(\text{not } T) \times \text{EV(No Build} \mid \text{not } T) \end{aligned}\]

$= 3,000 + (0.1575 \times 10,285) + (0.8425 \times 120) \approx \$4,693$

Risk vs. Reward

Choice	Expected Cost	Risk of Loss	P
Don’t Build	$1,000	0.01	1 in 100
Build w/o test	$10,050	0.0005	1 in 2000
Test & Decide	$4,693	0.00146	1 in 700

Conclusion

Lowest Expected Cost: Don’t Build ($1,000).
Lowest Risk of Catastrophe: Build without testing (0.0005).
Middle Ground: Testing ($4,693) significantly reduces risk compared to “Don’t Build” without the full $10k upfront cost.

Decision? It depends on your utility function (risk tolerance).

View Python Implementation (Notebook)

Saint Petersburg Paradox

Imagine a gambling game where a fair coin is flipped repeatedly until it lands on heads. The payoff for the game is $2^N$, where $N$ is the number of tosses needed for the coin to land on heads.

The expected value of this game is infinite:

\[ E(X) = \frac{1}{2} \cdot 2 + \frac{1}{4} \cdot 4 + \frac{1}{8} \cdot 8 + \ldots = \infty \]

This means that, in theory, a rational person should be willing to pay any finite amount to play this game. However, in reality, most people would be unwilling to pay a large amount.

Expected Utility Resolution

Bernoulli argued that people do not maximize expected monetary value but rather expected utility $U(x)$.

\[ E[U(X)] = \sum^\infty_{k=1} 2^{-k} U(2^k) \]

For the log utility case, $U(x) = \log(x)$, the expected utility is $2 \log(2)$. To find the certain dollar amount $x^*$ (certainty equivalent) that provides the same utility:

\[ \log(x^*) = 2\log(2) = \log(2^2) = \log(4) \implies x^* = 4 \]

Under log utility, a rational player would pay at most $4 to play, despite the infinite expected monetary value.