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There is a very similar question here, and in both that question and this question I'm asking right now, the accepted answer involves using a complement. The same thing occurs in this textbook question:

"In tossing four fair dice, what is the probability of at least one three?"

The solution is $1 - \big{(}\frac{5}{6}\big{)}^4$.

But I don't quite understand why, in certain cases, using the compliment will lead to a correct answer and when it's better to just calculate the straight probability, as in $\big{(}\frac{1}{6}\big{)}^4$ (which is wrong).

Is there a rule of thumb on which way to go? What should we be looking for when we confront these questions?

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Monica Heddneck
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    It's just a matter of computational simplicity. To do it "straight" as you put it, you'd need to add the probabilities of getting exactly one, exactly two, exactly three, and exactly four. Perfectly doable, but it is a lot faster to compute the "exactly zero" case and subtract. – lulu Jun 17 '16 at 22:48
  • What would the 'straight' answer look like? – Monica Heddneck Jun 17 '16 at 22:48
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    Let $X$ be the number of $3$'s. The "direct" approach has us finding $Pr(X=1)+\Pr(X=2)+\Pr(X=3)+\Pr(X=4)$. A lot more work! – André Nicolas Jun 17 '16 at 22:49
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    $\binom 41\times \left(\frac 16\right)^1\times\left(\frac 56\right)^3+\binom 42\times \left(\frac 16\right)^2\times\left(\frac 56\right)^2+\binom 43\times \left(\frac 16\right)^3\times\left(\frac 56\right)^1+\binom 44\times \left(\frac 16\right)^4\times\left(\frac 56\right)^0$ – lulu Jun 17 '16 at 22:51
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    Just to stress: if you were throwing $1000$ times then the straight sum would have $999$ terms, but the answer would still just be $1-\left(\frac 56 \right)^{1000}$. – lulu Jun 17 '16 at 22:56

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You are rolling four dice, and you want to know the chance of at least one three. Because it is at least one there are a few ways this could happen- namely there could be one three, two threes, three threes or four threes. Now, we can work all of these out using something called the binomial coefficients (technically speaking the number of heads is a binomial random variable) - the formula is the probability of a three $(1/6)$ raised to the power of how many, times by the probability of not a three (5/6) times by how many aren't threes, and then multiplied by the number of ways to select the dice (again, binomial coefficients). So the chance of one three is $4*(1/6)(5/6)^3=500/1296$. The chance of two threes is $6*(1/6)^2*(5/6)^2=150/1296$. The chance of three threes is $4*(1/6)^3*(5/6)=20/1296$ and the chance of four threes is $1*(1/6)^4*(5/6)^0=1/1296$.

If we add these all together, we get $(500+150+20+1)/1296=671/1296$ which is the the same as $1-(5/6)^4$ as expected. This is the formal way of working it out, but the short cut is the complement. The way I target these questions is to work out which has the least leg work. You are asked for the chance of at least one which includes the options 1,2,3,4, where as the complement is just 0. If you were asked for the chance of at least two, this is 2,3,4 and the complement is 0 or 1 (which is easier to work out). It is only when you're asked for 3 or more, that it's easier the formal way.

In general, if $X$ is the number of something with maximum value $Y$ (number of heads in Y coin tosses, threes on a dice roll, etc) and you want to calculate the probability that $X>n$ (written $P(X>n)$) it easier to use the complement if n

Dan Barry
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  • $1-\left(\dfrac{5}{6}\right)^4$ is the probability of not getting $0$ threes

which can also be calculated directly as $4 \times \dfrac{1}{6}\times \left(\dfrac{5}{6}\right)^3 +6 \times \left(\dfrac{1}{6}\right)^2\times \left(\dfrac{5}{6}\right)^2+4 \times \left(\dfrac{1}{6}\right)^3\times \dfrac{5}{6}+ \left(\dfrac{1}{6}\right)^4 $ as the probability of getting $1$, $2$, $3$ or $4$ threes

  • $\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}$ is the expected number of threes. This is more that the first value, as you might get more than $1$ three, which counts once towards the first question, but more than once towards the expectation

which can also be calculated as $1\times 4 \times \dfrac{1}{6}\times \left(\dfrac{5}{6}\right)^3 +2 \times 6 \times \left(\dfrac{1}{6}\right)^2\times \left(\dfrac{5}{6}\right)^2+3 \times 4 \times \left(\dfrac{1}{6}\right)^3\times \dfrac{5}{6}+ 4 \times\left(\dfrac{1}{6}\right)^4 $

Henry
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  • When am I supposed to know "probability of getting at least one" should be translated in my brain to "all probabilities except the probability of NOT getting NONE", the latter of which is frustratingly gymnastic and non-intuitive to the human brain? – Monica Heddneck Jun 17 '16 at 22:47
  • It comes with practice. One way is to find the probability of all the outcomes, which should add up to $1$, and then either add up the probabilities you are interested in, or take $1-$ the sum of all the probabilities your are not interested in – Henry Jun 17 '16 at 22:51
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"It is the case that at least one is a three" is equivalent to saying "It is not the case that none are a three"

Here, we use the axioms of probability that $Pr(A\cup B) = Pr(A)+Pr(B)$ when $A$ and $B$ are mutually exclusive events, that $A\cup A^c = \Omega$ the sample space, and that $Pr(\Omega)=1$ to conclude that $Pr(\text{At least 1 three}) = 1-Pr(\text{no threes})$

If you wanted to, you could calculate it directly as follows:

Break into cases based on the number of threes. "At least one three" is in this case equivalent to "Exactly one three or exactly two threes or exactly three threes or exactly four threes"

Each of these are disjoint events, so $Pr(\text{at least one three}) = Pr(\text{exactly one three})+Pr(\text{exactly two threes})+Pr(\text{exactly three threes})+Pr(\text{exactly four threes})$

We calculate each individually. In the case of exactly one three, pick which location the three occupies (4 choices) and pick the number for each other location (5 choices each). Divide by the sample space size. Alternatively, approach via binomial distribution. Either way, you will arrive at $Pr(\text{exactly one three})=4(\frac{1}{6})(\frac{5}{6})^3$

Similarly, we have the others as $\binom{4}{k}(\frac{1}{6})^k(\frac{5}{6})^{4-k}$ for the probability of getting exactly $k$ threes.

Adding, you have $Pr(\text{at least one three}) = \frac{4\cdot 5^3 + 6\cdot 5^2+4\cdot 5^1+4}{6^4}$, which after calculation you will find equals $1-(\frac{5}{6})^4$

The mistake of using $(\frac{1}{6})^4$ is that you calculated $Pr(\text{exactly four threes})$ which is only a small portion of the event at least one three.

JMoravitz
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