# Monte Hall Problem - Question on Decision Tree Construction (Conditional Probability)

(The Monte Hall Problem, also known as the 3 door problem): On the game show Let's Make a Deal, there are 3 doors hiding 3 prizes. Two of the prizes are goats, 1 prize is an expensive new automobile. Naturally you would like the car, but only Monte knows which door it is behind. You select one of the doors. Rather than opening the door you chose, Monte opens one of the two other doors to reveal a goat. You are now given the option to stick with your original selection or to switch to the other opened door.

(a) What is the conditional probability that you win the car if you stick with the original choice of doors?

(b) What is the (conditional) probability that you win the car if you change yoour mind and select the other opened door?

I read about the Monte Hall problem and understand the principle behind it: The door you choose is random, but the door Monte chooses is NOT. This is why switching doors gives you a higher probability. What I really had a question on is the construction of the decision tree for part (b).

(a): I understand this solution is 1/3.

(b) This question is asking you $Pr(win_{auto}|\Delta doors)=\cfrac{Pr(win_{auto}) \cap Pr(\Delta doors)}{Pr(\Delta doors)}$. I was just confused on the decision tree provided as a solution. Originally, I thought a decision tree gives you ALL possible outcomes in the sample space. But in this problem, a decision tree is drawn ONLY IF you change doors. The solutions tree is below:

Above you can correctly calculate the Probability by summing the branches: (1/3)*1*1 *2 = 2/3.

Now, lets say I said draw the tree assuming that you always keep your initial door choice. What is the probability that you win an auto given that you always stay with your initial door (I know its 1-2/3 just go with me for a second)? I thought I would draw something like this:

Above you see the probability of winning is = 2*(1/3* 0) + 1/3 * 1 = 1/3 (which is the correct answer).

Is it correct to state that decision trees do not necessarily show the entire sample space of a problem? Instead, sometimes they only show a specific scenario or subset of the entire sample space when calculating probabilities?

Thank you!

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There are only two moves in the game so the tree has depth $2$. Your first move is to guess where the prize is. The two alternatives are that you guessed correctly or you did not. Thus, one branch corresponds to a correct guess with a probability of $\frac{1}{3}$ and the other corresponds to an incorrect guess with a probability of $\frac{2}{3}$.
The second move is to decide whether to stay or switch. On the branch corresponding to an initially correct guess, the "stay" branch is labeled with a $1$ meaning if you were originally correct and you stay you win all the time. The "switch" branch is labeled with a $0$. The opposite values are given for "stay" and "switch" on the branch corresponding to an initially incorrect guess.