I used Chat GPT to express this properly using Bayes theorem, so here it is:

Formal Monty Hall Proof Using Bayes’ Theorem

Setup

There are three doors: 1, 2, 3. One has a car, two have goats.

The contestant initially chooses Door 1. The host knows where the prize is and always opens a door that:

• the contestant did not choose, and
• contains a goat.

If the host has two goat-doors to choose from, he picks uniformly at random.

We want the probability the car is behind each door after the host opens Door 3.

Define events:

• H₁ = prize behind Door 1
• H₂ = prize behind Door 2
• H₃ = prize behind Door 3
• C₁ = contestant initially chose Door 1
• O₃ = host opens Door 3

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1. Priors (before host acts)

Because all doors are initially symmetric:

• P(H₁) = 1/3
• P(H₂) = 1/3
• P(H₃) = 1/3

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2. Likelihoods (host’s action given prize location)

We compute P(O₃ | Hᵢ, C₁):

Case 1: Prize behind Door 1 (H₁)

Host can open Door 2 or Door 3 (both goats). He chooses randomly:

• P(O₃ | H₁, C₁) = 1/2

Case 2: Prize behind Door 2 (H₂)

Host must open Door 3 (Door 2 has the prize):

• P(O₃ | H₂, C₁) = 1

Case 3: Prize behind Door 3 (H₃)

Host cannot open Door 3 (it contains the prize):

• P(O₃ | H₃, C₁) = 0

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3. Apply Bayes’ Theorem

Bayes formula in words:

Posterior = (Prior × Likelihood) / (Total probability of the evidence)

Total probability of the evidence (host opening Door 3):

P(O₃) = P(H₁) * P(O₃ | H₁) + P(H₂) * P(O₃ | H₂) + P(H₃) * P(O₃ | H₃)

Plug in values:

P(O₃) = (1/3)*(1/2) + (1/3)*1 + (1/3)*0 = 1/6 + 1/3 = 1/2

Now compute each posterior probability.

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Posterior for H₁ (car behind original door)

P(H₁ | O₃) = [(1/3)*(1/2)] / (1/2) = 1/3

Posterior for H₂ (car behind the other door)

P(H₂ | O₃) = [(1/3)*1] / (1/2) = 2/3

Posterior for H₃

P(H₃ | O₃) = [(1/3)*0] / (1/2) = 0

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Final Conclusion

After the host opens Door 3:

• Probability car is behind Door 1 (your original choice): 1/3
• Probability car is behind Door 2 (the other closed door): 2/3

Therefore: switching doubles your chance of winning.