Monty hall problem tree diagram

When possibilities are mutually exclusive, their probabilities add up. Another key concept is possibilities that are exhaustive. Here are more examples of exhaustive possibilities:. In our tree diagrams, each branch-point always uses a set of possibilities that is both exclusive and exhaustive.

Monty hall problem tree diagram: MIT J Mathematics for

The first split on the three doors covers all the possibilities for where the prize might be, and only one of those possibilities can be the actual location of the prize. Likewise for the second stage of the diagram. On the bottom branch for example, the host must open either door B or door C given the rules, but he will only open one or the other.

There can be more than one way to partition the space of possibilities. When possibilities form a partition, their probabilities must add up to 1. In a way, the fundamental principle of probability is that probabilities over a partition must add up to 1. Tree diagrams follow a few simple rules based on these concepts. The parts of a tree are called nodesbranchesand leaves : see Figure 1.

Rule 1. Each node must branch into a partition.

Monty hall problem tree diagram: The probability tree diagram is

The subpossibilities that emerge from a node must be mutually exclusive, and they must include every way the possibility from which they emerge could obtain. Rule 2. Rule 3. The probability on a branch is conditional on the branches leading up to it. Consider the bottom path in the Monty Hall problem.

Monty hall problem tree diagram: This gives us the

Rule 4. The probability of a leaf is calculated by multiplying across the branches on the path leading to it. This number represents the probability that all possibilities on that path occur. Notice, Rule 4 is how we got the final probabilities we used to solve the Monty Hall problem the numbers in bold. In the version of the Monty Hall problem with a hundred doors, after the host opens every door except door 1 your door and door 59, the chance the prize is behind door 59 is:.

Imagine three prisoners, A, B, and C, are condemned to die in the morning. But the king decides to pardon one of them first. He makes his choice at random and communicates it to the guard, who is sworn to secrecy. Once again this is a tree diagram to explain the possible doors the host opens when a contestant starts by opening Door 1. We can see that if the car is behind Door 4 that the host has a choice of 2 doors to open Door 2 and Door 3.

This is shown in the bottom branch. We can then notice a pattern in our tree diagram to see that for 5 doors with 4 goats and 1 car we have a probability of swapping given by:. We can also use this general formula to prove why we should always swap.

Monty hall problem tree diagram: In our tree diagrams, each

There is also a general equation for when there are n doors and the host opens p other goat doors. This also follows from our tree diagram ideas:. You could try and make your own gameshow and work out the probability of someone winning. IB teacher? Hundreds of IB worksheets, unit tests, mock exams, treasure hunt activities, paper 3 activities, coursework support and more.

You decide whether you want to switch doors or stay with the one you chose. The probability that the car is behind the door unopened by your host depends on which choice you made originally. If the car is behind the door you chose first, the probability that it is behind the other door is 0. If the door you chose first has a goat behind it, then the car is most definitely behind the other door, and the probability that it is behind the door left unopened by your host is 1.