What is more likely?

Produce a Markov chain of the possible states.

First, we categorize the possible states:

• Event A has occurred before Event B. Call this state $A$ .
• Event B has occurred before Event A. Call this state $B$ .
• Neither event has occurred; previous roll wasn't a sum of 7. Call this state $N0$ .
• Neither event has occurred; previous roll was a sum of 7 (and hence the second previous roll wasn't). Call this state $N1$ .

Now we draw the Markov chain, just to get a visual aid.

Suppose we want to compute the probability that we end up in state $B$ . Thus we assign $P(B|B)=1$ ; that is, if we start on state $B$ , we will obviously end up there. Also, $P(B|A)=0$ ; that is, starting from $A$ , we won't reach $B$ . Let $P(B|N0)=x$ and $P(B|N1)=y$ ; we're looking for $x$ .

A property of Markov chains is that the probability of an event occurring on a state can be computed as the sum of weighted probabilities of its possible next states after one step. In other words, $P(B|N0) = \frac{29}{36} P(B|N0) + \frac{1}{36} P(B|A) + \frac{1}{6} P(B|N1)$ , since there's $\frac{29}{36}$ probability of remaining on $N0$ , $\frac{1}{36}$ probability to go to $A$ , and $\frac{1}{6}$ probability to go to $N1$ . Similarly, $P(B|N1) = \frac{29}{36} P(B|N0) + \frac{1}{36} P(B|A) + \frac{1}{6} P(B|B)$ . Simplifying, we have:

$\begin{cases} x = \frac{29}{36} \cdot x + \frac{1}{36} \cdot 0 + \frac{1}{6} \cdot y \\ y = \frac{29}{36} \cdot x + \frac{1}{36} \cdot 0 + \frac{1}{6} \cdot 1 \end{cases}$

Substituting the second equation to the first equation and simplifying gives $\frac{13}{6} x = 1$ , or $x = \frac{6}{13}$ . Thus the probability of ending up in state $B$ is $\frac{6}{13}$ .

Similarly, we can compute $P(A|N0)$ , by letting $P(A|A) = 1, P(A|B) = 0$ , and using the same method to solve it. We get $P(A|N0) = \frac{7}{13}$ , so the probability of ending up in state $A$ is $\frac{7}{13}$ .

This implies Event A is more likely than Event B .

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Or by using intuition (but not rigorous), observe that although Event B has probability $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$ of occurring for any two consecutive rolls, and Event A has probability $\frac{1}{36}$ of occurring on any roll, the difference is that Event B takes two rolls to execute. Event A has two chances to appear during the two rolls, so it should be more likely.

For event A, since the probability of both dice showing $6$ 's is $\frac{1}{36}$ the expected number of rolls until this event occurs is $36.$

For event B, let $P$ be the expected number of rolls until the second of two consecutive rolls that each give a sum of $7$ . If the first roll does not give a sum of $7$ , (an event that has probability $\frac{5}{6}$ ), then this "costs" us one roll. If the first roll does give a sum of $7$ , then there is a $\frac{1}{6}$ chance that the next roll will lead to event B occurring, and a $\frac{5}{6}$ chance that the next roll will send us back to start the count over again after "costing" us $2$ rolls. In equation form, we then have

$P = \dfrac{5}{6}(1 + P) + \dfrac{1}{6}(\dfrac{1}{6}*2 + \dfrac{5}{6}(P + 2))$

$\Longrightarrow P = \dfrac{5}{6} + \dfrac{5}{6}P + \dfrac{1}{18} + \dfrac{5}{36}P + \dfrac{5}{18}$

$\Longrightarrow P = \dfrac{35}{36}P + \dfrac{21}{18} \Longrightarrow \dfrac{1}{36}P = \dfrac{21}{18} \Longrightarrow P = 42.$

As it is then expected that event B will take more rolls to occur than event A, we can conclude that event A is more likely to occur first.

I simply figured the probabilities using AND or OR operators. In the first case it is simple (only AND, thus multiplying the probabiliies of each occurrence) In the second case, the probability of a single roll of 7 is higher because of the OR operator between the possible combinations. However, because the condition says it has to occur in two consecutive rolls, I used the AND to relate first and second roll., thereby reducing the probability. Was that approach too simplistic?