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.

  • $\begingroup$ This is not [set-theory]. $\endgroup$
    – Asaf Karagila
    Apr 22, 2013 at 22:49
  • $\begingroup$ Thanks for your edit. But it does involve the application of set theory in solving a probability question. $\endgroup$
    – Omid
    Apr 22, 2013 at 22:55
  • $\begingroup$ Did the original question include any information on the cost of playing and the pay-out for winning? Both are essential factors in determining whether a game is "fair". $\endgroup$
    – DJohnM
    Apr 23, 2013 at 0:06
  • $\begingroup$ No, and I think by "fair" he only means whether our chance of winning the game is, at least, 50 percent. $\endgroup$
    – Omid
    Apr 23, 2013 at 0:35

2 Answers 2


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.

  • $\begingroup$ Thanks. Yeah, I actually wrote down the tree and it is pretty simple as it doesn't branch out after the number shows up, so it stays pretty small. I was wondering why the author had intentionally avoided it. Now I see, maybe he just wanted to expand the tool set, introducing other techniques too. $\endgroup$
    – Omid
    Apr 23, 2013 at 0:30

It’s not hard to see why the tree diagram is a bit impractical: it has $6^3=216$ nodes on the final level. You can think in terms of it, but actually drawing it, while not completely out of the question, is clearly not going to be much fun. If you think in terms of the tree analysis, you end up with something like the following calculation.

Suppose that the target number is $d$. The tree branches $6$ ways at the first level, corresponding to rolling $1,2,3,4,5$, or $6$ on the first die. One of the $6$ resulting nodes is labelled $d$; every final node descending from that one represents a successful outcome. Since the remaining two dice can come up in $6^2=36$ different ways, there are $36$ nodes on the final level that descend from $d$, all of which represent successful outcomes.

Each of the other $5$ nodes at the first level branches $6$ ways to produce $6$ nodes at the second level, correponding to the $6$ possible outcomes on the second die. One of these $5$ nodes is labelled $d$. Every final node descending from one of those $d$ nodes is a successful outcome. There are $5$ of these $d$ nodes, and each has $6$ final descendent, one for each possible outcome on the third die. Thus, these $d$ nodes have altogether $5\cdot 6=30$ descendents on the third and last level, all corresponding to successful outcomes.

Finally, each of the $5\cdot 5=25$ second-level nodes that are neither labelled $d$ nor children of the $d$ node in the first level has one descendent labelled $d$.

The total number of successful outcomes is therefore $36+30+25=91$. There are $6^3=216$ outcomes altogether, and they are equally likely, so each has probability $\frac1{216}$, and the probability of a successful outcome is therefore


In this case the inclusion-exclusion calculation based on the Venn diagram is simpler and shorter, and the difference becomes even more pronounced as the number of dice increases.

Added: The calculation with the Venn diagram is just another way of carrying out the author’s four steps. He’s still finding the sample space, defining events of interest, determining outcome probabilities, and computing event probabilities; he’s just using doing it in a slightly less cumbersome way. As Byron Schmuland notes in the comments below, this probability is much more easily calculated using an approach that does not follow this four-step organization at all, but rather makes use of the fact that the probabilities of complementary events sum to $1$. We can easily use the author’s approach to calculate the probability of a failure, and then subtract that from $1$. Using the tree model, for instance, we note that a failure requires not getting $d$ at the first level ($5$ nodes), not getting $d$ at the second level ($5$ children of each of those $5$, or $5^2=25$ nodes), and not getting $d$ at the third and final level ($5$ children of each of those $25$ nodes, or $5^3=125$ nodes). Thus, the probability of failure is $\frac{125}{216}$, and the probability of success is $1-\frac{125}{216}=\frac{91}{216}$.

  • $\begingroup$ For this problem, it is even easier to look at the complementary event. The probability that his number doesn't show up in three die rolls is $(5/6)^3=125/216$, so the solution is $(216-125)/216=91/216$. $\endgroup$
    – user940
    Apr 22, 2013 at 23:26
  • $\begingroup$ @Byron: Yep. I’m guessing that the author sees the inclusion-exclusion count as just a different way of organizing his four-step program, while looking at the complementary event is a fundamentally different approach. $\endgroup$ Apr 22, 2013 at 23:37
  • $\begingroup$ Yes, I agree. $\quad$ $\endgroup$
    – user940
    Apr 22, 2013 at 23:43
  • $\begingroup$ I don't agree that it is complicated or impractical. I actually wrote down the graph myself and it seems quite easy to work out. Remember that the game stops whenever the number shows up in the dice. So I don't have to actually draw the whole tree. There's a 1/6 chance that it may show up in the first try, in which case the game is over and the player wins. Otherwise another numbers shows up, the chance of which is 5/6. I only have to analyze this branch further and it turns out there are only two cases and it rounds up to this: (1/6) + (5/6 * 1/6) + (5/6*5/6*1/6) which is roughly 0.413 $\endgroup$
    – Omid
    Apr 23, 2013 at 0:16
  • $\begingroup$ I still find the solution relatively easy, and I don't understand why the author had to go out of his way to use a formula from set theory. $\endgroup$
    – Omid
    Apr 23, 2013 at 0:21

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