This question appears on page 244 of "Statistics, 4th ed" by David Freedman.

The text of the question is:

A die is rolled 6 times. The chance that the first roll is an ace or the last roll is an ace equals ______. Answer choices are: $(1/6 + 1/6)$, $(1/6 * 1/6)$, and "neither of these".

The correct answer, according to the book is "neither of these".

I tried calculating the probability assuming that any number of aces can appear in the six rolls but that the first and the sixth rolls must result in aces.

I got ~0.278 as the answer.

$(1/6)^2*(5/6)^4 + (1/6)^3*(5/6)^3*4!/3! + (1/6)^4*(5/6)^2 * 4! / (2!*2!) + (1/6)^5*(5/6)*4!/3! + (1/6)^6$

Is this correct?

My question is: how can one infer from this question that we are to consider all cases of an "ace" appearing on any of the six rolls. Initially, I thought I could consider the six rolls as independent.

  • 3
    $\begingroup$ The correct answer is "that depends on what you mean by the word 'or'". $\endgroup$ – Mark Dec 26 '15 at 3:25

You are correct that the six rolls are independent. So the probability that the first roll is an ace is $1/6$, and similarly, the probability that the last roll is an ace is $1/6$. However, the probability of either event happening is not just the sum: you need to subtract the probability of their intersection. The probability of both the first and last roll being aces is $1/6^2$, so the final answer is $1/6 + 1/6 - 1/6^2 = 11/36$.

  • $\begingroup$ brilliant. that makes a lot of sense. btw, that should be -1/6^2 $\endgroup$ – user3837690 Dec 26 '15 at 2:05

The probability that neither the first or last die is an ace is $(5/6)(5/6) = 25/36$, so the complementary event has probability $1 - 25/36 = 11/36$.

  • $\begingroup$ thanks. i usually fail to consider the approach of subtracting the probability of unwanted events from 1. $\endgroup$ – user3837690 Dec 26 '15 at 2:07

I just wanted to present some notation for practice, in case you are not familiar. Let $A = \{\text{First roll is one}\}, B = \{\text{Last roll is one}\}$. Then, recalling the inclusion-exclusion rule, \begin{align*} P(A\text{ OR }B) &= P(A)+P(B)-P(AB)\\ &=P(A)+P(B)-P(A)P(B) \\ &= \frac{1}{6}+\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}\\ &=\frac{11}{36}, \end{align*} where the $P(AB)$ is the probability of $A$ and $B$ occurring and it can be expressed as the product $P(A)P(B)$ since the events are independent. Alternatively, $$P(A\text{ OR }B) = 1 -P(A^cB^c) = 1-P(A^c)P(B^c) = 1-\frac{5}{6}\cdot\frac{5}{6} = \frac{11}{36},$$ by the complement rule.

Final note, if $A$ and $B$ were disjoint (mutually exclusive), then $P(AB) = 0$ and $$P(A\text{ OR }B) = P(A)+P(B)-P(AB) = P(A)+P(B)-0 = P(A)+P(B).$$ But this is not the case.

  • $\begingroup$ thank you for your helpful post. $\endgroup$ – user3837690 Dec 29 '15 at 16:48

The probability that the first roll is a 1 is 1/6. The probability that the last roll is a 1 is 1/6. Since you're looking at the first OR the last role is a 1, you add the two probabilities. So the answer is 1/6 + 1/6 = 2/6 = 1/3.

  • 1
    $\begingroup$ Consider what that would mean for tossing a coin twice and asking for the probability that either result was heads. $\endgroup$ – Neil Slater Dec 26 '15 at 7:24
  • $\begingroup$ Dear god @NeilSlater, if I flip a coin twice 1/2 + 1/2 = 1! I'll always land on heads! Genius! (Or tails) (Wait, crap!) $\endgroup$ – Insane Dec 26 '15 at 7:38

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