While calculating: Why do I get $1-P(A^c)\neq P(A)$?

I have one single dice. And I throw this dice 4 times. What is the probability to get at least once a "6"? Let $A$ denote the desired event.

Sure this is easy, but I got my problems trying to understand why this

$$P(A)=1-P(A^c)=1-\left(\frac{5}{6}\right)^4$$

is correct. What I do understand is that $\frac{5}{6}$ is the probability that the dice does not show a "6" after being thrown. But why the power of 4? Because the events "to throw a six (or not)" are independent?

Assuming my guess is correct. I tryed to calculate the probability without considering the complement $A^c$:

$$P(A)=\left(\frac{1}{6}\right)^4$$

Clearly this is not the same result. I would be glad if someone could enlighten me. I clearly do miss something. :-)

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You are correct, the power of 4 is because each dice roll is independent.

However, the number $\left(\frac{1}{6}\right)^4$ represents the probability of getting four 6's (or, in fact, any specific outcome - in other words, $\left(\frac{1}{6}\right)^4$ is also the probability of the outcomes being 1, then 2, then 5, then 3). You want to calculate the probability of getting at least one 6, which is much higher. Naively calculating, we might think $$P(\text{at least one }6)=P(6\text{ on 1st roll})+P(6\text{ on 2nd roll})+P(6\text{ on 3rd roll})+P(6\text{ on 4th roll})$$ $$P(\text{at least one }6)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{4}{6}$$ But! We have double-counted some occurrences. For example, (6,6,2,1) is one of the possible combinations with at least one 6, but we've counted it in both $P(6\text{ on 1st roll})$ and in $P(6\text{ on 2nd roll})$. So we need to subtract off the amount we double-counted. $$P(\text{at least one }6)=\frac{4}{6}-\left(P(\text{6 on 1st and 2nd rolls})+\cdots+P(6\text{ on 3rd and 4th rolls})\vphantom{g^{G}}\right)$$ $$P(\text{at least one }6)=\frac{4}{6}-\left(\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2\right)=\frac{1}{2}$$ But! We now have double-subtracted some occurrences. For example, (6,6,6,2) is one of the possible combinations with at least one 6. We counted it 3 times initially (in $P(6\text{ on 1st roll})$, $P(6\text{ on 2nd roll})$, and $P(6\text{ on 3rd roll})$), but now have subtracted it three times (in $P(6\text{ on 1st and 2nd roll})$, $P(6\text{ on 1st and 3rd roll})$, and $P(6\text{ on 2nd and 3rd roll})$). We need to add the occurrences with three 6's back in now. $$P(\text{at least one }6)=\frac{1}{2}+\left(P(\text{6 on 1st, 2nd, 3rd rolls})+\cdots+P(6\text{ on 2nd, 3rd, 4th rolls})\vphantom{g^{G}}\right)$$ $$\frac{1}{2}+\left(\left(\frac{1}{6}\right)^3+\left(\frac{1}{6}\right)^3+\left(\frac{1}{6}\right)^3+\left(\frac{1}{6}\right)^3\right)=\frac{14}{27}$$ Finally, due to the same problem, we must subtract off the probability of the sole combination with four sixes, (6,6,6,6). $$P(\text{at least one }6)=\frac{14}{27}-P(6\text{ on 1st, 2nd, 3rd, 4th roll})=\frac{14}{27}-\frac{1}{1296}=\frac{671}{2196}$$ which is the correct answer. But consider what we've just calculated: $$4\left(\frac{1}{6}\right)-6\left(\frac{1}{6}\right)^2+4\left(\frac{1}{6}\right)^3-\left(\frac{1}{6}\right)^4=1-\left(1-\frac{1}{6}\right)^4=1-\left(\frac{5}{6}\right)^4$$ by the binomial theorem. So computing this probability via the complement is really a very clever, time-saving trick!

In general, the reasoning I gave above is referred to as the "inclusion-exclusion principle", and it is very useful in keeping track of combinatorics or probability arguments. For us, the sets $A$, $B$, $C$, and $D$ were the sets of possible dice rolls where a 6 was rolled in the first, second, third, or fourth roll, respectively. Then $|A\cup B\cup C\cup D|$ is the number of rolls with at least one 6, and we calculated it using inclusion-exclusion.

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$1-\left(\frac 5 6\right)^4$ is the probability that you throw at least one six in four throws (i.e. one minus the probability that you don't throw a six in any of them). On the other hand, $\left(\frac 1 6\right)^4$ is the probability that you throw a six every time in four throws. Clearly the latter probability must be considerably smaller than the former.

And yes, the $\,^4$ is basically there because the throws are independent and identically distributed — if the probability of getting a six was different for different throws, you'd have to multiply all the different probabilities together; and if they weren't even independent, you'd end up with a complicated expression involving conditional probabilities.

Specifically, $1-\left(\frac 5 6\right)^4$ is just shorthand for $1 - \left( \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \right)$ $= 1 - P(A_1^c) \cdot P(A_2^c) \cdot P(A_3^c) \cdot P(A_4^c)$, where the event $A_n$ denotes getting a six on the $n$-th throw.

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So if I don't want to consider $A^c$: $(1/6)^4$ denotes only the probability of throwing the sequence $(6,6,6,6)$. That means I have to consider all the probabilitys of the other sequences like $(6,6,x,x)$ or $(6,x,x,x)$ or $(x,6,x,6)$ (and all the other combinations) and sum them up, right? – Aufwind Jul 19 '11 at 12:14
Yes, that will also work; you just have to be careful to count all the possible combinations once and none of them twice. There are two basic approaches you can try: either a) enumerate all the possible basic outcomes like (6,6,6,6), (6,6,6,1), (6,6,1,6), etc. and sum their probabilities (you can save some work by only considering the throws "six" and "not six"), or b) use the inclusion–exclusion principle like Zev Chonoles does in his answer. Both, however, are a lot more work than just using $P(A)=1-P(A^c)$. – Ilmari Karonen Jul 19 '11 at 12:23

The following is another approach to the "at least one $6$" problem, grossly less efficient than the approach through first finding the probability of the complement. But it introduces ideas that will be very important later, so it is worth thinking about.

We will be happy if we get at least one $6$. What is the probability that we will be happy? Well, we will be happy if we get exactly $1$ $6$. We will also be happy if we get exactly $2$ $6$'s. Also if we get exactly $3$ $6$'s. And also with exactly $4$ $6$'s.

These events are (pairwise) disjoint. For example, it is impossible to get exactly $2$ $6$'s and (simultaneously) also exactly $4$ $6$'s.

Now I will introduce a bit of notation that looks initially mysterious, and that is not really needed here, but will become important to you later. Let $X$ be the number of $6$'s that we get. Then, for example, $P(X=2)$ means the probability that we get exactly $2$ $6$'s.

We will be happy if $X=1$, also if $X=2$, and so on up to if $X=4$. Since these events are pairwise disjoint, the probability we will be happy is $$P(X=1)+P(X=2)+P(X=3)+P(X=4).$$

So "all" we need to do is to calculate the probabilities of these four events, and add up!

Let us, for instance, find $P(X=2)$. Think about about our $4$ tosses. Write Y (for yes) if on a toss we get a $6$, and N is we don't. So for example NYYN means we got a non-$6$, then a $6$, then a $6$, then a non-$6$. Note that on this sequence of tosses, the total number of $6$ is $2$.

The probability of NYYN is is $(5/6)(1/6)(1/6)(5/6)$. There are several other patterns in which we end up with exactly $2$ $6$'s. For example, there is YYNN, YNYN, some others. The probability of YYNN is $(1/6)(1/6)(5/6)(5/6)$, which is exactly the same as the probability of NYYN. In fact the probability of any single pattern with exactly $2$ $6$'s is $(1/6)^2(5/6)^2$.

So to find $P(X=2)$, we find the number of patterns with exactly $2$ $6$'s, and multiply that by $(1/6)^2(5/6)^2$. The number of patterns is the number of ways of choosing where the two Y will go.

There are $\dbinom{4}{2}$ such ways (I assume you have already met this). So $$P(X=2)=\binom{4}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2.$$

We can do a similar analysis for exactly $1$ $6$, exactly $3$, exactly $4$. The probability we will be happy is therefore $$\binom{4}{1}\left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^3+ \binom{4}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2+ \binom{4}{3}\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1+ \binom{4}{4}\left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^0.$$

Now calculate! Really, I almost mean it. Getting the right answer, which you already know from the "quick" method, is a useful test of accuracy and attention to detail.

Of course the above was an inefficient way of calculating the probability. But the same set of ideas shows, for example, that if we toss the die $20$ times, the probability of exactly $4$ $6$'s is $\dbinom{20}{4}(1/6)^4(5/6)^{16}$. The general idea has many applications, and not only to gambling!

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