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For the Monty hall problem, with 3 doors, two of which have sheep and 1 has a car. I calculated the probability of getting the car if you swap being 2/6 instead of 2/3. I have drawn this tree diagram of how I calculated it:

And from it I get that the probability of getting a car if you swap is 2/6 and if you stay it's 1/6, which is not the same as the actual answer 2/3 and 1/3. I have included the probabilities of getting sheep, as well as my overall probabilities, add up so I want to know what is wrong with my tree diagram, which resulted in this wrong answer. In my tree diagram I haven't included the host revealing a door, so could this be a factor as to why my answer is wrong?

It's simply because your diagram assumes you stay with probability $1/2$. In that case the probability of switching and getting the car is indeed $2/6$, and the probability of switching and not getting the car is $1/6$.

But if you always switch, you're twice as likely to get each of these outcomes - instead of a $1/2$ probability of switching, you have probability $1$ of switching - so they become $2/3$ and $1/3$ respectively.

That indicates that instead of using a strategy, you are randomly choosing whether to swap or stay.   The probability of winning a car by doing that is and unsurprising $1/2$.

$$tfrac 13tfrac 12+tfrac 23tfrac 12=tfrac 12$$

The point of the scenario is that if you use strategy of always swapping, the probability of winning a car becomes: $2/3$.

$$tfrac 13tfrac 01+tfrac 23tfrac 11=tfrac 23$$

You're calculating the probability wrong. If you decide at random with equal probability to swap or stay, in $tfrac{2}{6}$ of all cases you will have swapped and gotten the car.

But that is not the probability of if you decide to swap, would you get the car?

Note that in exactly $tfrac{3}{6}=tfrac{1}{2}$ of all cases, you will have swapped, and in $tfrac{1}{2}$ of all cases you will have stayed. But look at those cases.

Of the $tfrac{3}{6}$ cases where you will have swapped, there is a $tfrac{2}{6}$ chance of having swapped and gotten a car while there is a $tfrac{1}{6}$ of having swapped and gotten a goat sheep.

On the other hand there is a $tfrac{1}{6}$ chance of having stayed and gotten a car while there is a $tfrac{2}{6}$ of having swapped and gotten a sheep.

So if you swap, the probability of getting a car is twice that of getting a sheep, while if you stay, the chance of getting a sheep is twice that of getting a car.

Which is the exact outcome of the canonical Monty Hall problem.

The key to the solution of the Monty Hall problem is that Monty Hall reveals one of the incorrect solutions after stage one. The problem ends up being exclusively about the door that you choose versus the one that is left.

This means that the probabilities are as follows:

We will assume Door 1 is the car and Door 2 and 3 are sheep.

Choose Door 1:
Monty can reveal door 2 or 3, so you have a choice between 1 and whichever of those is left, both being incorrect. Swapping will give you sheep in either case (each possibility with a 1/6th chance, for a 1/3rd chance total)

Choose door 2:
Monty reveals door 3 as incorrect. You WILL get the car if you swap for Door 1. (1/3rd chance).

Choose Door 3:
Monty reveals door 2 as incorrect. You WILL get the car if you swap for Door 1. (1/3rd chance)

If you choose to always stay, you will only have a 1/3rd chance of a car.

How is this related to your question?

If you look at all six probabilities for staying and swapping at once, you will get the answer you saw, and you will see a 1/2 chance of getting the car (2/6 if you do swap plus 1/6 chance if you don't swap). That doesn't show the impact of making a choice.

Using your chart, just compare the red "swap" probabilities to each other to see the impact of always swapping. If you ALWAYS choose to swap, you have a 2/3rd chance of being correct. If you ALWAYS choose to stay, you have a 1/3rd chance to be correct. The possible choices have to be examined independently to assess individual merit of a given choice.

Your question states as if you always choose to swap, but your examples show that you are accounting for both swapping and staying. Once you remove the chance of staying, you are only left with three possible results, not 6.