The Monty Hall problem based on Harvard Fat chance

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The Monty Hall problem

key question

How to understand Monty Hall problem intuitively and with conditional probability and total probability of sum?

video links

original courses p51-53

my note videos p50-52

Problem

3 doors, 1 car, 2 goats; you pick one, host open a door with a goat; should you switch to the remaining door?

Intuitive analysis

total = 3 doors or options

if not switch, winning outcome = pick1(car) = only 1 car to pick from = 1

• prob(win without switch) = 1/3

if switch, winning outcome = pick1(goat first, switch to car) = 2 goat to pick from = 2

• prob(win with switch) = 2/3

Formula analysis

total = n doors or options, and 1 cars inside

if not switch,

• winning outcome = pick1(car) from only 1 car exists
• losing outcome = pick1(goat) from n-1 goats
• openning a door with a goat affect nothing
• prob(win without switch) = pick1(car) = 1/n
• prob(lose without switch) = pick1(goat) = n-1/n

if switch and win,

• how many doors left = n - 1 - 1
• then the original pick must a goat
  • prob(second pick car | original pick goat and open a goat door and switch) = 1/(n-1-1)
• prob(second pick car after switch and original pick goat) = prob(original goat) x prob(second pick car | original goat, open a goat door, switch) = $\frac{n-1}{n} \times \frac{1}{n-2} = \frac{n-1}{n-2} \times \frac{1}{n}$ ⚛️ = A
• A > 1/n so switch is always better

Extended Analysis

total = n doors, k doors have cars

prob(car) = k/n, prob(goat) = (n-k)/n

if original pick = car, and then switch after open a door with a goat

• prob(second pick car | original pick car and switch) = (k-1)/(n-2)
• prob(second pick car | original pick goat and switch) = k/(n-2)
• prob(second pick car and original pick car) = prob(original pick car) x prob(second pick car | original pick car, open goat door, switch) = k/n x (k-1)/(n-2) = A ⚛️
• prob(second pick car and original pick goat) = prob(original pick goat) x prob(seocnd pick car | original pick goat, open goat door, switch) = (n-k)/n x k/(n-2) = B
• A + B = prob(pick car) = prob(second pick car with or without original pick car) = $\frac{k(k-1)+k(n-k)}{n(n-2)} = \frac{k(n-1)}{n(n-2)} = \frac{k}{n} \frac{n-1}{n-2} \gt \frac{k}{n}$

Practice

problem

n doors, k doors have cars, monty opens a goat door, do you switch? (always switch)

• prob(original pick car) = k/n = A
• prob(original pick goat) = (n-k) / n = B
• prob(second pick goat | original pick car, open goat door, switch) = (how many goat left = (n-k)-1 ) / (how many doors left = n - 1 - 1) = (n-k-1) / (n-2) = C|A
• prob(second pick goat | original pick goat, open goat door, switch) = (how many goat left = n-k-1-1) / (how many doors left = n - 1 - 1) = (n-k-2) / (n-2) = C|B
• prob(second pick goat & first pick car) = A x C|A
• prob(seocnd pick goat & first pick goat) = B x C|B
• prob(second pick goat) = prob(lose) = (A and C) + (B and C) = C = prob(lose when always switch)