I was reading through an elementary mathematics book lately. The author claims that every probability question, no matter how intimidating it may look, can be solved using a four-step method:
1) Finding the sample space
2) Defining Events of Interest
3) Determining Outcome Probabilities
4) Computing Event Probabilities
He then proceeds to apply this technique to well-known problems like Monty Hall etc. In all of those examples, he uses the tree diagram to demonstrate the method. He finds it a handy tool when the data is short and the problem is well-structured.
The examples he presents look fine and I understand what's going on. But then he poses this question:
The player chooses a number from $1$ to $6$, then rolls $3$ fair dice. He
wins the game if and only if the number he picked shows up at least in
one of the dice. Is this a fair game?
In solving this question, he intentionally avoids the tree diagram, saying that he prefers to wing it instead, and uses a formula from set theory and demonstrates the method using the venn diagram:
$$P(A\cup B\cup C)= P(A)+ P(B) + P(C) – P(A\cap B) – P(A\cap C)- P(B\cap C) + P(A\cap B\cap C)$$
and he finally concludes that, contrary to our intuition, it is not a fair game, and the chance of winning is around 40 percent. I understand what he is doing, and I see that the events overlap and we need to extract the common parts.
The thing is, he makes this silence leap from his structured four-step method plus the tree diagram to this formulaic technique without really explaining why the tree diagram here doesn't work or is not efficient. He abruptly dives into set theory. I don't find this pedagogically convincing.
I want to know if there is a relatively easy way to solve this using the neat four step method and tree diagram without having to use the above formula. Any tips on how and when the tree diagram works or stops working is also appreciated.
Best Answer
The tree analysis is simple as long as we do not insist on equally likely outcomes. It is a small binary tree, with at each node success (probability $1/6$) and failure (probability $5/6$). To find the probability of failure, we follow a single path, and find that the probability of three failures in a row is $(5/6)^3$.
Remarks: $1.$ A sample space with equally likely outcomes is, in many situations, not an appropriate model.
$2.$ Many techniques are needed and used in solving probability problems. Inclusion/Exclusion is a powerful idea, and it is a good thing that it has been introduced at least once.