This question is from DEGROOT's PROBABILITY and STATISTICS.
Rolling Dice Repeatedly:
Suppose that two dice are to be rolled repeatedly and the
sum $T$ of the two numbers is to be observed for each roll. We shall determine the
probability $p$ that the value $T =7$will be observed before the value $T =8$ is observed.
Solution:
The desired probability $p$ could be calculated directly as follows: We could assume that the sample space $S$ contains all sequences of outcomes that terminate as
soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the
sum of the probabilities of all the sequences that terminate when the value $T = 7$ is
obtained.
However,there is a simpler approach in this example. We can consider the simple
experiment in which two dice are rolled. If we repeat the experiment until either the
sum $T = 7$ or the sum $T = 8$ is obtained, the effect is to restrict the outcome of the
experiment to one of these two values. Hence,the problem can be restated as follows:
Given that the outcome of the experiment is either $T = 7$ or $T = 8$, determine the
probability $p$ that the outcome is actually $T = 7$.
If we let $A$ be the event that $T = 7$ and let $B$ be the event that the value of $T$ is
either $7$ or $8$, then $A ∩ B = A$ and
$p = Pr(A|B) =Pr(A ∩ B)/
Pr(B)
=Pr(A)/
Pr(B)$.
But $Pr(A) = 6/36$ and
$Pr(B) = (6/36) + (5/36) = 11/36$. Hence, $p = 6/11$.
Now,my doubts are
1.) Why the author says "We could assume that the sample space $S$ contains all sequences of outcomes that terminate as
soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the
sum of the probabilities of all the sequences that terminate when the value $T = 7$ is
obtained." ?
2.)How can we go from lengthy sequences of outcomes that terminate as
soon as either the sum $T = 7$ or the sum $T = 8$ is obtained to just the outcome of the experiment for which either $T = 7$ or $T = 8$ ?
Please help.Thank you.
Best Answer
For 1, once you get either a $7$ or an $8$, you know the result of the run. Any rolls we make subsequently will not affect the outcome, so we might as well ignore them. As we roll $7$ or $8$ with probability $1$, to find the chance that a $7$ is last we just add them up.
2 is the more subtle point. If you like thinking of Markov processes, we have one with just three states. You have a start state and two ending states. If you throw a $7$ you move to one end state. If you throw $8$ you move to the other end state. If you throw anything else, your return to start. We are interested in the chance we end at $7$. This is the chance of $7$ given that we threw either $7$ or $8$. As the cycles back to start leave us in the same place, we can ignore them.