A slight variant of the Monty Hall Problem.

Question

Suppose that there are 4 doors having 2 cars and 2 goats. You arbitrarily choose a door, then Monty shows a door with a goat. What is the probability of getting a car if you switch. Apparently the answer is 3 / 4. I got it right but the line of thinking I had seems to be wrong. Can anyone give me a very sound explanation how it arrived to 3 / 4? I really want to wrap my head around this concept. Thanks

Answer 1

Use the same logic as computing 2/3 in original Monty Hall. take cases.

Case 1 Goat 1

P(switch gets car in this case) = 1

Case 2 Goat 2

P(switch gets car in this case) = 1

Case 3 Car 1

P(switch gets car in this case) = 1/2

Case 3 Car 2

P(switch gets car in this case) = 1/2

Each case has 1/4 probability of happening.

Hence P(switch gets car) = (1/4)(1+1+1/2+1/2) = 3/4

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Btw I'm guessing there maybe a pattern.

In original MH problem, people think the probability is 1/2 when it is really 2/3.

Here, people might think the probability is 2/3 (leftover doors after revealing one goat are 2 cars and 1 goat) when it is really 3/4.

Answer 2

Our original choice was either a car or a goat. If our original choice was a goat, then the probability that we choose a car once we switch is $$ \Bbb P(\text{car}\mid\text{goat})=1 $$ If our original choice was a car, then the probability that we choose a car once we switch is $$ \Bbb P(\text{car}\mid\text{car})=\frac{1}{2} $$ Now, given that we decide to switch we have \begin{align*} \Bbb P(\text{switch to car}) &=\Bbb P(\text{car}\mid\text{goat})\cdot\Bbb P(\text{goat first})+\Bbb P(\text{car}\mid\text{car})\cdot\Bbb P(\text{car first}) \\ &= 1\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2} \\ &= \frac{3}{4} \end{align*} Since the original probability of choosing a car is $\frac{1}{2}$ we see that switching is the best strategy.